text
stringlengths 100
957k
| meta
stringclasses 1
value |
|---|---|
# OpenStax: General Chemistry
Source: OpenStax
Student Price: FREE
Get your students excited about solving General Chemistry problems by engaging them every step of the way with this interactive text by Openstax.Download EPUB
This content has been used by 4,822 students
# 17 Electrochemistry
## Chapter Outline
17.1 Balancing Oxidation-Reduction Reactions
17.2 Galvanic Cells
17.3 Standard Reduction Potentials
17.4 The Nernst Equation
17.5 Batteries and Fuel Cells
17.6 Corrosion
17.7 Electrolysis
## Introduction
Electrochemistry deals with chemical reactions that produce electricity and the changes associated with the passage of electrical current through matter. The reactions involve electron transfer, and so they are oxidation-reduction (or redox) reactions. Many metals may be purified or electroplated using electrochemical methods. Devices such as automobiles, smartphones, electronic tablets, watches, pacemakers, and many others use batteries for power. Batteries use chemical reactions that produce electricity spontaneously and that can be converted into useful work. All electrochemical systems involve the transfer of electrons in a reacting system. In many systems, the reactions occur in a region known as the cell, where the transfer of electrons occurs at electrodes.
## 17.1 Balancing Oxidation-Reduction Reactions
By the end of this section, you will be able to:
• Define electrochemistry and a number of important associated terms
• Split oxidation-reduction reactions into their oxidation half-reactions and reduction half-reactions
• Produce balanced oxidation-reduction equations for reactions in acidic or basic solution
• Identify oxidizing agents and reducing agents
Electricity refers to a number of phenomena associated with the presence and flow of electric charge. Electricity includes such diverse things as lightning, static electricity, the current generated by a battery as it discharges, and many other influences on our daily lives. The flow or movement of charge is an electric current (Figure 17.2). Electrons or ions may carry the charge. The elementary unit of charge is the charge of a proton, which is equal in magnitude to the charge of an electron. The SI unit of charge is the coulomb (C) and the charge of a proton is 1.602 × 10−19 C. The presence of an electric charge generates an electric field. Electric current is the rate of flow of charge. The SI unit for electrical current is the SI base unit called the ampere (A), which is a flow rate of 1 coulomb of charge per second (1 A = 1 C/s). An electric current flows in a path, called an electric circuit. In most chemical systems, it is necessary to maintain a closed path for current to flow. The flow of charge is generated by an electrical potential difference, or potential, between two points in the circuit. Electrical potential is the ability of the electric field to do work on the charge. The SI unit of electrical potential is the volt (V). When 1 coulomb of charge moves through a potential difference of 1 volt, it gains or loses 1 joule (J) of energy. Table 17.1 summarizes some of this information about electricity.
Electrochemistry studies oxidation-reduction reactions, which were first discussed in an earlier chapter, where we learned that oxidation was the loss of electrons and reduction was the gain of electrons. The reactions discussed tended to be rather simple, and conservation of mass (atom counting by type) and deriving a correctly balanced chemical equation were relatively simple. In this section, we will concentrate on the half-reaction method for balancing oxidation-reduction reactions. The use of half-reactions is important partly for balancing more complicated reactions and partly because many aspects of electrochemistry are easier to discuss in terms of half-reactions. There are alternate methods of balancing these reactions; however, there are no good alternatives to half-reactions for discussing what is occurring in many systems. The half-reaction method splits oxidation-reduction reactions into their oxidation “half” and reduction “half” to make finding the overall equation easier.
Electrochemical reactions frequently occur in solutions, which could be acidic, basic, or neutral. When balancing oxidation-reduction reactions, the nature of the solution may be important. It helps to see this in an actual problem. Consider the following unbalanced oxidation-reduction reaction in acidic solution:
We can start by collecting the species we have so far into an unbalanced oxidation half-reaction and an unbalanced reduction half-reaction. Each of these half-reactions contain the same element in two different oxidation states. The Fe2+ has lost an electron to become Fe3+; therefore, the iron underwent oxidation. The reduction is not as obvious; however, the manganese gained five electrons to change from Mn7+ to Mn2+
In acidic solution, there are hydrogen ions present, which are often useful in balancing half-reactions. It may be necessary to use the hydrogen ions directly or as a reactant that may react with oxygen to generate water. Hydrogen ions are very important in acidic solutions where the reactants or products contain hydrogen and/or oxygen. In this example, the oxidation half-reaction involves neither hydrogen nor oxygen, so hydrogen ions are not necessary to the balancing. However, the reduction half-reaction does involve oxygen. It is necessary to use hydrogen ions to convert this oxygen to water.
The situation is different in basic solution because the hydrogen ion concentration is lower and the hydroxide ion concentration is higher. After finishing this example, we will examine how basic solutions differ from acidic solutions. A neutral solution may be treated as acidic or basic, though treating it as acidic is usually easier.
The iron atoms in the oxidation half-reaction are balanced (mass balance); however, the charge is unbalanced, since the charges on the ions are not equal. It is necessary to use electrons to balance the charge. The way to balance the charge is by adding electrons to one side of the equation. Adding a single electron on the right side gives a balanced oxidation half-reaction:
You should check the half-reaction for the number of each atom type and the total charge on each side of the equation. The charges include the actual charges of the ions times the number of ions and the charge on an electron times the number of electrons.
If the atoms and charges balance, the half-reaction is balanced. In oxidation half-reactions, electrons appear as products (on the right). As discussed in the earlier chapter, since iron underwent oxidation, iron is the reducing agent. Now return to the reduction half-reaction equation:
The atoms are balanced (mass balance), so it is now necessary to check for charge balance. The total charge on the left of the reaction arrow is [(−1) × (1) + (8) × (+1)], or +7, while the total charge on the right side is [(1) × (+2) + (4) × (0)], or +2. The difference between +7 and +2 is five; therefore, it is necessary to add five electrons to the left side to achieve charge balance.
You should check this half-reaction for each atom type and for the charge, as well:
Mn: Does (1 × 1) = (1 × 1)? Yes.
H: Does (8 × 1) = (4 × 2)? Yes.
O: Does (1 × 4) = (4 × 1)? Yes.
Charge: Does [1 × (−1) + 8 × (+1) + 5 × (−1)] = [1 × (+2)]? Yes.
Now that this half-reaction is balanced, it is easy to see it involves reduction because electrons were gained when MnO4 − was reduced to Mn2+. In all reduction half-reactions, electrons appear as reactants (on the left side). As discussed in the earlier chapter, the species that was reduced, MnO4 in this case, is also called the oxidizing agent. We now have two balanced half-reactions.
It is now necessary to combine the two halves to produce a whole reaction. The key to combining the half-reactions is the electrons. The electrons lost during oxidation must go somewhere. These electrons go to cause reduction. The number of electrons transferred from the oxidation half-reaction to the reduction half-reaction must be equal. There can be no missing or excess electrons. In this example, the oxidation half-reaction generates one electron, while the reduction half-reaction requires five. The lowest common multiple of one and five is five; therefore, it is necessary to multiply every term in the oxidation half-reaction by five and every term in the reduction half-reaction by one. (In this case, the multiplication of the reduction half-reaction generates no change; however, this will not always be the case.) The multiplication of the two half-reactions by the appropriate factor followed by addition of the two halves gives
The electrons do not appear in the final answer because the oxidation electrons are the same electrons as the reduction electrons and they “cancel.” Carefully check each side of the overall equation to verify everything was combined correctly:
Fe: Does (5 × 1) = (5 × 1)? Yes.
Mn: Does (1 × 1) = (1 × 1)? Yes.
H: Does (8 × 1) = (4 × 2)? Yes.
O: Does (1 × 4) = (4 × 1)? Yes.
Charge: Does [5 × (+2) + 1 × (−1) + 8 × (+1)] = [5 × (+3) + 1 × (+2)]? Yes.
Everything checks, so this is the overall equation in acidic solution. If something does not check, the most common error occurs during the multiplication of the individual half-reactions.
Now suppose we wanted the solution to be basic. Recall that basic solutions have excess hydroxide ions. Some of these hydroxide ions will react with hydrogen ions to produce water. The simplest way to generate the balanced overall equation in basic solution is to start with the balanced equation in acidic solution, then “convert” it to the equation for basic solution. However, it is necessary to exercise caution when doing this, as many reactants behave differently under basic conditions and many metal ions will precipitate as the metal hydroxide. We just produced the following reaction, which we want to change to a basic reaction:
However, under basic conditions, MnO4 normally reduces to MnO2 and iron will be present as either Fe(OH)2 or Fe(OH)3. For these reasons, under basic conditions, this reaction will be
(Under very basic conditions MnO4 will reduce to MnO4 2, instead of MnO2.)
It is still possible to balance any oxidation-reduction reaction as an acidic reaction and then, when necessary, convert the equation to a basic reaction. This will work if the acidic and basic reactants and products are the same or if the basic reactants and products are used before the conversion from acidic or basic. There are very few examples in which the acidic and basic reactions will involve the same reactants and products. However, balancing a basic reaction as acidic and then converting to basic will work. To convert to a basic reaction, it is necessary to add the same number of hydroxide ions to each side of the equation so that all the hydrogen ions (H+ ) are removed and mass balance is maintained. Hydrogen ion combines with hydroxide ion (OH ) to produce water.
Let us now try a basic equation. We will start with the following basic reaction:
Balancing this as acid gives
In this case, it is necessary to add two hydroxide ions to each side of the equation to convert the two hydrogen ions on the left into water:
Note that both sides of the equation show water. Simplifying should be done when necessary, and gives the desired equation. In this case, it is necessary to remove one H2O from each side of the reaction arrows.
Again, check each side of the overall equation to make sure there are no errors:
Cl: Does (1 × 1) = (1 × 1)? Yes.
Mn: Does (2 × 1) = (2 × 1)? Yes.
H: Does (1 × 2) = (2 × 1)? Yes.
O: Does (2 × 4 + 1 × 1) = (3 × 1 + 2 × 2 + 2 × 1)? Yes.
Charge: Does [1 × (−1) + 2 × (−1)] = [1 × (−1) + 2 × (−1)]? Yes
Everything checks, so this is the overall equation in basic solution.
### Example 17.1Balancing Acidic Oxidation-Reduction Reactions
Balance the following reaction equation in acidic solution:
Solution
This is an oxidation-reduction reaction, so start by collecting the species given into an unbalanced oxidation half-reaction and an unbalanced reduction half-reaction.
Starting with the oxidation half-reaction, we can balance the chromium
In acidic solution, we can use or generate hydrogen ions (H+ ). Adding seven water molecules to the left side provides the necessary oxygen; the “left over” hydrogen appears as 14 H+ on the right:
The left side of the equation has a total charge of [2 × (+3) = +6], and the right side a total charge of [−2 + 14 × (+1) = +12]. The difference is six; adding six electrons to the right side produces a mass- and chargebalanced oxidation half-reaction (in acidic solution):
Checking the half-reaction:
Cr: Does (2 × 1) = (1 × 2)? Yes.
H: Does (7 × 2) = (14 × 1)? Yes.
O: Does (7 × 1) = (1 × 7)? Yes.
Charge: Does [2 × (+3)] = [1 × (−2) + 14 × (+1) + 6 × (−1)]? Yes.
Now work on the reduction. It is necessary to convert the four oxygen atoms in the permanganate into four water molecules. To do this, add eight H+ to convert the oxygen into four water molecules: reduction (unbalanced):
Then add five electrons to the left side to balance the charge:
Make sure to check the half-reaction:
Mn: Does (1 × 1) = (1 × 1)? Yes.
H: Does (8 × 1) = (4 × 2)? Yes.
O: Does (1 × 4) = (4 × 1)? Yes.
Charge: Does [1 × (−1) + 8 × (+1) + 5 × (−1)] = [1 × (+2)]? Yes.
Collecting what we have so far:
The least common multiple for the electrons is 30, so multiply the oxidation half-reaction by five, the reduction half-reaction by six, combine, and simplify:
Checking each side of the equation:
Mn: Does (6 × 1) = (6 × 1)? Yes.
Cr: Does (10 × 1) = (5 × 2)? Yes.
H: Does (11 × 2) = (22 × 1)? Yes. O: Does (11 × 1 + 6 × 4) = (5 × 7)? Yes.
Charge: Does [10 × (+3) + 6 × (−1)] = [5 × (−2) + 22 × (+1) + 6 × (+2)]? Yes.
This is the balanced equation in acidic solution.
Balance the following equation in acidic solution:
### Example 17.2Balancing Basic Oxidation-Reduction Reactions
Balance the following reaction equation in basic solution:
Solution
This is an oxidation-reduction reaction, so start by collecting the species given into an unbalanced oxidation half-reaction and an unbalanced reduction half-reaction
Starting with the oxidation half-reaction, we can balance the chromium
In acidic solution, we can use or generate hydrogen ions (H+ ). Adding one water molecule to the left side provides the necessary oxygen; the “left over” hydrogen appears as five H+ on the right side:
The left side of the equation has a total charge of [0], and the right side a total charge of [−2 + 5 × (+1) = +3]. The difference is three, adding three electrons to the right side produces a mass- and charge-balanced oxidation half-reaction (in acidic solution):
Checking the half-reaction:
Cr: Does (1 × 1) = (1 × 1)? Yes.
H: Does (1 × 3 + 1 × 2) = (5 × 1)? Yes.
O: Does (1 × 3 + 1 × 1) = (4 × 1)? Yes.
Charge: Does [0 = [1 × (−2) + 5 × (+1) + 3 × (−1)]? Yes.
Now work on the reduction. It is necessary to convert the four O atoms in the MnO4 minus the two O atoms in MnO2 into two water molecules. To do this, add four H+ to convert the oxygen into two water molecules:
Then add three electrons to the left side to balance the charge:
Make sure to check the half-reaction:
Mn: Does (1 × 1) = (1 × 1)? Yes.
H: Does (4 × 1) = (2 × 2)? Yes.
O: Does (1 × 4) = (1 × 2 + 2 × 1)? Yes.
Charge: Does [1 × (−1) + 4 × (+1) + 3 × (−1)] = [0]? Yes.
Collecting what we have so far:
In this case, both half reactions involve the same number of electrons; therefore, simply add the two half reactions together.
Checking each side of the equation:
Mn: Does (1 × 1) = (1 × 1)? Yes.
Cr: Does (1 × 1) = (1 × 1)? Yes.
H: Does (1 × 3) = (2 × 1 + 1 × 1)? Yes.
O: Does (1 × 4 + 1 × 3) = (1 × 4 + 1 × 2 + 1 × 1)? Yes.
Charge: Does [1 × (−1)] = [1 × (−2) + 1 × (+1)]? Yes.
This is the balanced equation in acidic solution. For a basic solution, add one hydroxide ion to each side and simplify:
Checking each side of the equation:
Mn: Does(1 × 1) = (1 × 1)? Yes.
Cr: Does (1 × 1) = (1 × 1)? Yes.
H: Does (1 × 1 + 1 × 3) = (2 × 2)? Yes.
O: Does (1 × 1 + 1 × 4 + 1 × 3) = (1 × 4 + 1 × 2 + 2 × 1)? Yes.
Charge: Does [1 × (−1) + 1 × (−1)] = [1 × (−2)]? Yes.
This is the balanced equation in basic solution.
Balance the following in the type of solution indicated.
(d) Identify the oxidizing agents in reactions (a), (b), and (c).
(e) Identify the reducing agents in reactions (a), (b), and (c).
## Exercises
### Question 17.1
17.1
1. If a 2.5 A current is run through a circuit for 35 minutes, how many coulombs of charge moved through the circuit?
### Question 17.2
17.2
No correct answers: No correct answer has been set for this question
2. For the scenario in the previous question, how many electrons moved through the circuit?
### Question 17.3
17.3
3. For each of the following balanced half-reactions, determine whether an oxidation or reduction is occurring. $(a) Fe^{3+} + 3e^− ⟶ Fe;$ $(b) Cr ⟶ Cr^{3+} + 3e^−;$ $(c) MnO_4^{2−} ⟶ MnO_4^− + e^−;$ $(d) Li^+ + e^− ⟶ Li;$
### Question 17.4
17.4
4. For each of the following unbalanced half-reactions, determine whether an oxidation or reduction is occurring. $(a) C^l− ⟶ Cl_2$ math⟶ MnO_2 $(c) H_2 ⟶ H^+$ $(d) NO_3− ⟶ NO$
### Question 17.5
17.5
5. Given the following pairs of balanced half-reactions, determine the balanced reaction for each pair of halfreactions in an acidic solution. $(a) Ca ⟶ Ca^{2+} + 2e^{−}, F_2 + 2e^{−} ⟶ 2F^−$ $(b) Li ⟶ Li^+ + e^−, Cl_2 + 2e^− ⟶ 2Cl^−$ $(c) Fe ⟶ Fe^{3+} + 3e^−, Br_2 + 2e^− ⟶ 2Br^−$ $(d) Ag ⟶ Ag^+ + e^−, MnO_4^− + 4H^+ + 3e^− ⟶ MnO_2 + 2H^2O$
### Question 17.6
17.6
6. Balance the following in acidic solution: (a)$H_2 O_2 + Sn^{2+} ⟶ H_2 O + Sn^{4+}$ (b)$PbO_2 + Hg ⟶Hg_2^{2+} + Pb^{2+}$ "(c) $(c) Al + Cr2O7^{2−} ⟶ Al^{3+} +Cr^{3+}$"
### Question 17.7
17.7
7. Identify the species that undergoes oxidation, the species that undergoes reduction, the oxidizing agent, and the reducing agent in each of the reactions of the previous problem.
### Question 17.8
17.8
8. Balance the following in basic solution: (a)$SO_3^{2−}(aq) +Cu(OH)_2(s) ⟶ SO_4^{2−}(aq) +Cu(OH)(s)$; (b)$O_2(g) + Mn(OH)_2(s) ⟶ MnO_2(s)$ (c)$NO_3^−(aq) + H_2(g) ⟶ NO(g)$ (d)$Al(s) + CrO_4^{2−}(aq) ⟶Al(OH)_3(s) + Cr(OH)_4^−(aq)$
### Question 17.9
17.9
9. Identify the species that was oxidized, the species that was reduced, the oxidizing agent, and the reducing agent in each of the reactions of the previous problem.
### Question 17.10
17.10
10. Why is it not possible for hydroxide ion$(OH^−)$ to appear in either of the half-reactions or the overall equation when balancing oxidation-reduction reactions in acidic solution?
### Question 17.11
17.11
11. Why is it not possible for hydrogen ion $(H^+)$ to appear in either of the half-reactions or the overall equation when balancing oxidation-reduction reactions in basic solution?
## 17.2 Galvanic Cells
By the end of this section, you will be able to:
• Use cell notation to describe galvanic cells
• Describe the basic components of galvanic cells
Galvanic cells, also known as voltaic cells, are electrochemical cells in which spontaneous oxidation-reduction reactions produce electrical energy. In writing the equations, it is often convenient to separate the oxidation-reduction reactions into half-reactions to facilitate balancing the overall equation and to emphasize the actual chemical transformations.
Consider what happens when a clean piece of copper metal is placed in a solution of silver nitrate (Figure 17.3). As soon as the copper metal is added, silver metal begins to form and copper ions pass into the solution. The blue color of the solution on the far right indicates the presence of copper ions. The reaction may be split into its two half-reactions. Half-reactions separate the oxidation from the reduction, so each can be considered individually.
The equation for the reduction half-reaction had to be doubled so the number electrons “gained” in the reduction halfreaction equaled the number of electrons “lost” in the oxidation half-reaction.
Galvanic or voltaic cells involve spontaneous electrochemical reactions in which the half-reactions are separated (Figure 17.4) so that current can flow through an external wire. The beaker on the left side of the figure is called a half-cell, and contains a 1 M solution of copper(II) nitrate [Cu(NO3)2] with a piece of copper metal partially submerged in the solution. The copper metal is an electrode. The copper is undergoing oxidation; therefore, the copper electrode is the anode. The anode is connected to a voltmeter with a wire and the other terminal of the voltmeter is connected to a silver electrode by a wire. The silver is undergoing reduction; therefore, the silver electrode is the cathode. The half-cell on the right side of the figure consists of the silver electrode in a 1 M solution of silver nitrate (AgNO3). At this point, no current flows—that is, no significant movement of electrons through the wire occurs because the circuit is open. The circuit is closed using a salt bridge, which transmits the current with moving ions. The salt bridge consists of a concentrated, nonreactive, electrolyte solution such as the sodium nitrate (NaNO3) solution used in this example. As electrons flow from left to right through the electrode and wire, nitrate ions (anions) pass through the porous plug on the left into the copper(II) nitrate solution. This keeps the beaker on the left electrically neutral by neutralizing the charge on the copper(II) ions that are produced in the solution as the copper metal is oxidized. At the same time, the nitrate ions are moving to the left, sodium ions (cations) move to the right, through the porous plug, and into the silver nitrate solution on the right. These added cations “replace” the silver ions that are removed from the solution as they were reduced to silver metal, keeping the beaker on the right electrically neutral. Without the salt bridge, the compartments would not remain electrically neutral and no significant current would flow. However, if the two compartments are in direct contact, a salt bridge is not necessary. The instant the circuit is completed, the voltmeter reads +0.46 V, this is called the cell potential. The cell potential is created when the two dissimilar metals are connected, and is a measure of the energy per unit charge available from the oxidation-reduction reaction. The volt is the derived SI unit for electrical potential
In this equation, A is the current in amperes and C the charge in coulombs. Note that volts must be multiplied by the charge in coulombs (C) to obtain the energy in joules (J).
When the electrochemical cell is constructed in this fashion, a positive cell potential indicates a spontaneous reaction and that the electrons are flowing from the left to the right. There is a lot going on in Figure 17.4, so it is useful to summarize things for this system:
• Electrons flow from the anode to the cathode: left to right in the standard galvanic cell in the figure.
• The electrode in the left half-cell is the anode because oxidation occurs here. The name refers to the flow of anions in the salt bridge toward it.
• The electrode in the right half-cell is the cathode because reduction occurs here. The name refers to the flow of cations in the salt bridge toward it.
• Oxidation occurs at the anode (the left half-cell in the figure).
• Reduction occurs at the cathode (the right half-cell in the figure).
• The cell potential, +0.46 V, in this case, results from the inherent differences in the nature of the materials used to make the two half-cells.
• The salt bridge must be present to close (complete) the circuit and both an oxidation and reduction must occur for current to flow.
There are many possible galvanic cells, so a shorthand notation is usually used to describe them. The cell notation (sometimes called a cell diagram) provides information about the various species involved in the reaction. This notation also works for other types of cells. A vertical line, │, denotes a phase boundary and a double line, ‖, the salt bridge. Information about the anode is written to the left, followed by the anode solution, then the salt bridge (when present), then the cathode solution, and, finally, information about the cathode to the right. The cell notation for the galvanic cell in Figure 17.4 is then
Note that spectator ions are not included and that the simplest form of each half-reaction was used. When known, the initial concentrations of the various ions are usually included.
One of the simplest cells is the Daniell cell. It is possible to construct this battery by placing a copper electrode at the bottom of a jar and covering the metal with a copper sulfate solution. A zinc sulfate solution is floated on top of the copper sulfate solution; then a zinc electrode is placed in the zinc sulfate solution. Connecting the copper electrode to the zinc electrode allows an electric current to flow. This is an example of a cell without a salt bridge, and ions may flow across the interface between the two solutions.
Some oxidation-reduction reactions involve species that are poor conductors of electricity, and so an electrode is used that does not participate in the reactions. Frequently, the electrode is platinum, gold, or graphite, all of which are inert to many chemical reactions. One such system is shown in Figure 17.5. Magnesium undergoes oxidation at the anode on the left in the figure and hydrogen ions undergo reduction at the cathode on the right. The reaction may be summarized as
The cell used an inert platinum wire for the cathode, so the cell notation is
The magnesium electrode is an active electrode because it participates in the oxidation-reduction reaction. Inert electrodes, like the platinum electrode in Figure 17.5, do not participate in the oxidation-reduction reaction and are present so that current can flow through the cell. Platinum or gold generally make good inert electrodes because they are chemically unreactive.
### Example 17.3Using Cell Notation
Consider a galvanic cell consisting of
Write the oxidation and reduction half-reactions and write the reaction using cell notation. Which reaction occurs at the anode? The cathode?
Solution
By inspection, Cr is oxidized when three electrons are lost to form Cr3+, and Cu2+ is reduced as it gains two electrons to form Cu. Balancing the charge gives
Cell notation uses the simplest form of each of the equations, and starts with the reaction at the anode. No concentrations were specified so:
Oxidation occurs at the anode and reduction at the cathode.
Using Cell Notation
Consider a galvanic cell consisting of
Write the oxidation and reduction half-reactions and write the reaction using cell notation. Which reaction occurs at the anode? The cathode?
Solution
By inspection, Fe2+ undergoes oxidation when one electron is lost to form Fe3+, and MnO4 is reduced as it gains five electrons to form Mn2+. Balancing the charge gives
Cell notation uses the simplest form of each of the equations, and starts with the reaction at the anode. It is necessary to use an inert electrode, such as platinum, because there is no metal present to conduct the electrons from the anode to the cathode. No concentrations were specified so:
Oxidation occurs at the anode and reduction at the cathode.
Use cell notation to describe the galvanic cell where copper(II) ions are reduced to copper metal and zinc metal is oxidized to zinc ions.
Answer: From the information given in the problem:
The overall reaction is:
which is represented in cell notation as:
## Exercises
### Question 17.13
17.13
13. Write the following balanced reactions using cell notation. Use platinum as an inert electrode, if needed. (a)$Mg(s) + Ni^{2+}(aq) ⟶ Mg^{2+}(aq) + Ni(s)$ (b)$2Ag^+(aq) + Cu(s) ⟶ Cu^{2+}(aq) + 2Ag(s)$ (c)$Mn(s) + Sn(NO_3)_2(aq) ⟶ Mn(NO_3)_2(aq)+Au(s)$ (d) $3CuNO_3(aq) + Au(NO_3)_3(aq) ⟶ 3Cu(NO_3)_2(aq) + Au(s)$
### Question 17.14
17.14
14. Given the following cell notations, determine the species oxidized, species reduced, and the oxidizing agent and reducing agent, without writing the balanced reactions. (a) $Mg(s)\ │\ Mg^{2+}(aq)\ ║\ Cu^{2}+(aq)\ │\ Cu(s)$ (b) $Ni(s)\ │\ Ni^{2+}(aq)\ ║\ Ag^{+}(aq)\ │\ Ag(s)$
### Question 17.15
17.15
15. For the cell notations in the previous problem, write the corresponding balanced reactions.
### Question 17.16
17.16
16. Balance the following reactions and write the reactions using cell notation. Ignore any inert electrodes, as theyare never part of the half-reactions. (a) $)Al(s) + Zr^{4+}(aq) ⟶ Al^{3+}(aq) + Zr(s)$ (b)$Ag^+(aq) + NO(g) ⟶ Ag(s) + NO_{3}^{−}(aq)$ (c) $SiO_{3}^{2−}(aq) + Mg(s) ⟶ Si(s) + Mg(OH)2(s)$ (d) $ClO_{3}^{−}(aq) + MnO_{2}(s) ⟶ Cl^−(aq) + MnO_{4}^{−}(aq)$
### Question 17.17
17.17
17. Identify the species oxidized, species reduced, and the oxidizing agent and reducing agent for all the reactions in the previous problem.
### Question 17.18
17.18
18. From the information provided, use cell notation to describe the following systems: (a) In one half-cell, a solution of $Pt(NO_3)_2$ forms Pt metal, while in the other half-cell, Cu metal goes into a $Cu(NO_3)_2$ solution with all solute concentrations 1 M. (b) The cathode consists of a gold electrode in a 0.55 M $Au(NO_3)_3$ solution and the anode is a magnesium electrode in 0.75 M $Mg(NO_3)_2$ solution. (c) One half-cell consists of a silver electrode in a 1 M $AgNO_3$ solution, and in the other half-cell, a copper electrode in 1 M $Cu(NO_3)_2$ is oxidized.
### Question 17.19
17.19
19. Why is a salt bridge necessary in galvanic cells like the one in Figure?
### Question 17.20
17.20
20. An active (metal) electrode was found to gain mass as the oxidation-reduction reaction was allowed to proceed. Was the electrode part of the anode or cathode? Explain.
### Question 17.21
17.21
21. An active (metal) electrode was found to lose mass as the oxidation-reduction reaction was allowed to proceed. Was the electrode part of the anode or cathode? Explain.
### Question 17.22
17.22
22. The mass of three different metal electrodes, each from a different galvanic cell, were determined before and after the current generated by the oxidation-reduction reaction in each cell was allowed to flow for a few minutes. The first metal electrode, given the label A, was found to have increased in mass; the second metal electrode, given the label B, did not change in mass; and the third metal electrode, given the label C, was found to have lost mass. Make an educated guess as to which electrodes were active and which were inert electrodes, and which were anode(s) and which were the cathode(s).
## 17.3 Standard Reduction Potentials
By the end of this section, you will be able to:
• Determine standard cell potentials for oxidation-reduction reactions
• Use standard reduction potentials to determine the better oxidizing or reducing agent from among several possible choices
The cell potential in Figure 17.4 (+0.46 V) results from the difference in the electrical potentials for each electrode. While it is impossible to determine the electrical potential of a single electrode, we can assign an electrode the value of zero and then use it as a reference. The electrode chosen as the zero is shown in Figure 17.6 and is called the standard hydrogen electrode (SHE). The SHE consists of 1 atm of hydrogen gas bubbled through a 1 M HCl solution, usually at room temperature. Platinum, which is chemically inert, is used as the electrode. The reduction half-reaction chosen as the reference is
E° is the standard reduction potential. The superscript “°” on the E denotes standard conditions (1 bar or 1 atm for gases, 1 M for solutes). The voltage is defined as zero for all temperatures.
A galvanic cell consisting of a SHE and Cu2+/Cu half-cell can be used to determine the standard reduction potential for Cu2+ (Figure 17.7). In cell notation, the reaction is
Electrons flow from the anode to the cathode. The reactions, which are reversible, are
The standard reduction potential can be determined by subtracting the standard reduction potential for the reaction occurring at the anode from the standard reduction potential for the reaction occurring at the cathode. The minus sign is necessary because oxidation is the reverse of reduction.
Using the SHE as a reference, other standard reduction potentials can be determined. Consider the cell shown in Figure 17.8, where
Electrons flow from left to right, and the reactions are
The standard reduction potential can be determined by subtracting the standard reduction potential for the reaction occurring at the anode from the standard reduction potential for the reaction occurring at the cathode. The minus sign is needed because oxidation is the reverse of reduction.
It is important to note that the potential is not doubled for the cathode reaction.
The SHE is rather dangerous and rarely used in the laboratory. Its main significance is that it established the zero for standard reduction potentials. Once determined, standard reduction potentials can be used to determine the standard cell potential, cell , for any cell. For example, for the cell shown in Figure 17.4,
Again, note that when calculating E°cell , standard reduction potentials always remain the same even when a halfreaction is multiplied by a factor. Standard reduction potentials for selected reduction reactions are shown in Table 17.2. A more complete list is provided in Appendix L.
### Example 17.4Cell Potentials from Standard Reduction Potentials
What is the standard cell potential for a galvanic cell that consists of Au3+/Au and Ni2+/Ni half-cells? Identify the oxidizing and reducing agents.
Solution
Using Table 17.2, the reactions involved in the galvanic cell, both written as reductions, are
Galvanic cells have positive cell potentials, and all the reduction reactions are reversible. The reaction at the anode will be the half-reaction with the smaller or more negative standard reduction potential. Reversing the reaction at the anode (to show the oxidation) but not its standard reduction potential gives:
The least common factor is six, so the overall reaction is
The reduction potentials are not scaled by the stoichiometric coefficients when calculating the cell potential, and the unmodified standard reduction potentials must be used.
From the half-reactions, Ni is oxidized, so it is the reducing agent, and Au3+ is reduced, so it is the oxidizing agent.
A galvanic cell consists of a Mg electrode in 1 M Mg(NO3)2 solution and a Ag electrode in 1 M AgNO3 solution. Calculate the standard cell potential at 25 °C.
## Exrecises
### Question 17.23
17.23
23. For each reaction listed, determine its standard cell potential at 25 °C and whether the reaction is spontaneous at standard conditions. (a) $Mg(s) + Ni^{2+}(aq) ⟶ Mg^{2+}(aq) + Ni(s)$ (b)$2Ag^+(aq) + Cu(s) ⟶ Cu^{2+}(aq) + 2Ag(s)$ (c) $Mn(s) + Sn(NO_3)_2(aq) ⟶ Mn(NO_3)_2(aq) + Sn(s)$ (d) $3Fe(NO_3)_2(aq) + Au(NO_3)_3(aq) ⟶ 3Fe(NO_3)_3(aq) + Au(s)$
### Question 17.24
17.24
24. For each reaction listed, determine its standard cell potential at 25 °C and whether the reaction is spontaneous at standard conditions. (a) $Mn(s) + Ni^{2+}(aq) ⟶ Mn^{2+}(aq) + Ni(s)$ (b)$3Cu^{2+}(aq) + 2Al(s) ⟶ 2Al^{3+}(aq) + 2Cu(s)$ (c) $Na(s) + LiNO_3(aq) ⟶ NaNO_3(aq) + Li(s)$ (d) $Ca(NO_3)_2(aq) + Ba(s) ⟶ Ba(NO_3)_2(aq) + Ca(s)$
### Question 17.25
17.25
25. Determine the overall reaction and its standard cell potential at 25 °C for this reaction. Is the reaction spontaneous at standard conditions? (a) $Cu(s)\ │\ Cu^{2+}(aq)\ ║\ Au^{3+}(aq)\ │\ Au(s)$
### Question 17.26
17.26
26. Determine the overall reaction and its standard cell potential at 25 °C for the reaction involving the galvanic cell made from a half-cell consisting of a silver electrode in 1 M silver nitrate solution and a half-cell consisting of a zinc electrode in 1 M zinc nitrate. Is the reaction spontaneous at standard conditions?
### Question 17.27
17.27
27. Determine the overall reaction and its standard cell potential at 25 °C for the reaction involving the galvanic cell in which cadmium metal is oxidized to 1 M cadmium(II) ion and a half-cell consisting of an aluminum electrode in 1 M aluminum nitrate solution. Is the reaction spontaneous at standard conditions?
### Question 17.28
17.28
28. Determine the overall reaction and its standard cell potential at 25 °C for these reactions. Is the reaction spontaneous at standard conditions? Assume the standard reduction for $Br_2(l)$ is the same as for $Br_2(aq).$ $Pt(s)\ │\ H_2(g) \│\ H^+(aq)\ ║\ Br_2(aq/) │\ Br^−(aq)\ │\ Pt(s)$
## 17.4 The Nernst Equation
By the end of this section, you will be able to:
• Relate cell potentials to free energy changes
• Use the Nernst equation to determine cell potentials at nonstandard conditions
• Perform calculations that involve converting between cell potentials, free energy changes, and equilibrium constants
We will now extend electrochemistry by determining the relationship between cell and the thermodynamics quantities such as Δ (Gibbs free energy) and K (the equilibrium constant). In galvanic cells, chemical energy is converted into electrical energy, which can do work. The electrical work is the product of the charge transferred multiplied by the potential difference (voltage):
The charge on 1 mole of electrons is given by Faraday’s constant (F)
total charge = (number of moles of e−) × F = nF
In this equation, n is the number of moles of electrons for the balanced oxidation-reduction reaction. The measured cell potential is the maximum potential the cell can produce and is related to the electrical work (wele) by
The negative sign for the work indicates that the electrical work is done by the system (the galvanic cell) on the surroundings. In an earlier chapter, the free energy was defined as the energy that was available to do work. In particular, the change in free energy was defined in terms of the maximum work (wmax), which, for electrochemical systems, is wele.
We can verify the signs are correct when we realize that n and F are positive constants and that galvanic cells, which have positive cell potentials, involve spontaneous reactions. Thus, spontaneous reactions, which have ΔG < 0, must have Ecell > 0. If all the reactants and products are in their standard states, this becomes
This provides a way to relate standard cell potentials to equilibrium constants, since
Most of the time, the electrochemical reactions are run at standard temperature (298.15 K). Collecting terms at this temperature yields
where n is the number of moles of electrons. For historical reasons, the logarithm in equations involving cell potentials is often expressed using base 10 logarithms (log), which changes the constant by a factor of 2.303:
Thus, if Δ, K, or cell is known or can be calculated, the other two quantities can be readily determined. The relationships are shown graphically in Figure 17.9.
Given any one of the quantities, the other two can be calculated.
### Example 17.5Equilibrium Constants, Standard Cell Potentials, and Standard Free Energy Changes
What is the standard free energy change and equilibrium constant for the following reaction at 25 °C?
Solution
The reaction involves an oxidation-reduction reaction, so the standard cell potential can be calculated using the data in Appendix L.
Remember that the cell potential for the cathode is not multiplied by two when determining the standard cell potential. With n = 2, the equilibrium constant is then
The two equilibrium constants differ slightly due to rounding in the constants 0.0257 V and 0.0592 V. The standard free energy is then
Check your answer: A positive standard cell potential means a spontaneous reaction, so the standard free energy change should be negative, and an equilibrium constant should be >1.
What is the standard free energy change and the equilibrium constant for the following reaction at room temperature? Is the reaction spontaneous?
Now that the connection has been made between the free energy and cell potentials, nonstandard concentrations follow. Recall that
where Q is the reaction quotient (see the chapter on equilibrium fundamentals). Converting to cell potentials:
This is the Nernst equation. At standard temperature (298.15 K), it is possible to write the above equations as
If the temperature is not 273.15 K, it is necessary to recalculate the value of the constant. With the Nernst equation, it is possible to calculate the cell potential at nonstandard conditions. This adjustment is necessary because potentials determined under different conditions will have different values.
### Example 17.6Cell Potentials at Nonstandard Conditions
Consider the following reaction at room temperature:
Is the process spontaneous?
Solution
There are two ways to solve the problem. If the thermodynamic information in Appendix G were available, you could calculate the free energy change. If the free energy change is negative, the process is spontaneous. The other approach, which we will use, requires information like that given in Appendix L. Using those data, the cell potential can be determined. If the cell potential is positive, the process is spontaneous. Collecting information from Appendix L and the problem,
The process is not spontaneous under standard conditions. Using the Nernst equation and the concentrations stated in the problem and n = 2,
The process is (still) nonspontaneous.
What is the cell potential for the following reaction at room temperature?
Al(s) │ Al3+(aq, 0.15 M) ║ Cu2+(aq, 0.025 M) │ Cu(s)
What are the values of n and Q for the overall reaction? Is the reaction spontaneous under these conditions?
Answer: n = 6; Q = 1440; Ecell = +1.97 V, spontaneous.
Finally, we will take a brief look at a special type of cell called a concentration cell. In a concentration cell, the electrodes are the same material and the half-cells differ only in concentration. Since one or both compartments is not standard, the cell potentials will be unequal; therefore, there will be a potential difference, which can be determined with the aid of the Nernst equation.
### Example 17.7Concentration Cells
What is the cell potential of the concentration cell described by
Zn(s) │ Zn2+(aq, 0.10 M) ║ Zn2+(aq, 0.50 M) │ Zn(s)
Solution
From the information given:
The standard cell potential is zero because the anode and cathode involve the same reaction; only the concentration of Zn2+ changes. Substituting into the Nernst equation,
and the process is spontaneous at these conditions.
Check your answer: In a concentration cell, the standard cell potential will always be zero. To get a positive cell potential (spontaneous process) the reaction quotient Q must be <1. Q < 1 in this case, so the process is spontaneous.
What value of Q for the previous concentration cell would result in a voltage of 0.10 V? If the concentration of zinc ion at the cathode was 0.50 M, what was the concentration at the anode?
Answer: Q = 0.00042; [Zn2+]cat = 2.1 × 10−4 M
## Exrecises
### Question 17.29
17.29
29. For the standard cell potentials given here, determine the ΔG° for the cell in kJ. (a) 0.000 V, n = 2 (b) +0.434 V, n = 2 (c) −2.439 V, n = 1
### Question 17.30
17.30
30. For the ΔG° values given here, determine the standard cell potential for the cell. (a) 12 kJ/mol, n = 3 (b) −45 kJ/mol, n = 1
### Question 17.31
17.31
31. Determine the standard cell potential and the cell potential under the stated conditions for the electrochemical reactions described here. State whether each is spontaneous or nonspontaneous under each set of conditions at 298.15 K. (a) $Hg(l) + S^{2−}(aq, 0.10 M) + 2Ag^+(aq, 0.25 M) ⟶ 2Ag(s) + HgS(s)$ (b) The galvanic cell made from a half-cell consisting of an aluminum electrode in 0.015 M aluminum nitrate solution and a half-cell consisting of a nickel electrode in 0.25 M nickel(II) nitrate solution. (c) The cell made of a half-cell in which 1.0 M aqueous bromine is oxidized to 0.11 M bromide ion and a half-cell in which aluminum ion at 0.023 M is reduced to aluminum metal. Assume the standard reduction potential for $Br_2(l)$ is the same as that of $Br_2(aq)$.
### Question 17.32
17.32
32. Determine ΔG and ΔG° for each of the reactions in the previous problem.
### Question 17.33
17.33
33. Use the data in Appendix L to determine the equilibrium constant for the following reactions. Assume 298.15K if no temperature is given. (a) $AgCl(s) ⇌ Ag^+(aq) + Cl^−(aq)$ (b)$CdS(s) ⇌ Cd^{2+}(aq) + S^{2−}(aq) at 377 K$ (c) $Hg^2+(aq) + 4Br^−(aq) ⇌ [HgBr_4]^{2−}(aq)$ (d) $H_2 O(l) ⇌ H^+(aq) + OH^−(aq) at 25 °C$
## 17.5 Batteries and Fuel Cells
By the end of this section, you will be able to:
• Classify batteries as primary or secondary
• List some of the characteristics and limitations of batteries
• Provide a general description of a fuel cell
A battery is an electrochemical cell or series of cells that produces an electric current. In principle, any galvanic cell could be used as a battery. An ideal battery would never run down, produce an unchanging voltage, and be capable of withstanding environmental extremes of heat and humidity. Real batteries strike a balance between ideal characteristics and practical limitations. For example, the mass of a car battery is about 18 kg or about 1% of the mass of an average car or light-duty truck. This type of battery would supply nearly unlimited energy if used in a smartphone, but would be rejected for this application because of its mass. Thus, no single battery is “best” and batteries are selected for a particular application, keeping things like the mass of the battery, its cost, reliability, and current capacity in mind. There are two basic types of batteries: primary and secondary. A few batteries of each type are described next.
### Primary Batteries
Primary batteries are single-use batteries because they cannot be recharged. A common primary battery is the dry cell (Figure 17.10). The dry cell is a zinc-carbon battery. The zinc can serves as both a container and the negative electrode. The positive electrode is a rod made of carbon that is surrounded by a paste of manganese(IV) oxide, zinc chloride, ammonium chloride, carbon powder, and a small amount of water. The reaction at the anode can be represented as the ordinary oxidation of zinc:
The reaction at the cathode is more complicated, in part because more than one reaction occurs. The series of reactions that occurs at the cathode is approximately
The overall reaction for the zinc–carbon battery can be represented as
with an overall cell potential which is initially about 1.5 V, but decreases as the battery is used. It is important to remember that the voltage delivered by a battery is the same regardless of the size of a battery. For this reason, D, C, A, AA, and AAA batteries all have the same voltage rating. However, larger batteries can deliver more moles of electrons. As the zinc container oxidizes, its contents eventually leak out, so this type of battery should not be left in any electrical device for extended periods.
Alkaline batteries (Figure 17.11) were developed in the 1950s partly to address some of the performance issues with zinc–carbon dry cells. They are manufactured to be exact replacements for zinc-carbon dry cells. As their name suggests, these types of batteries use alkaline electrolytes, often potassium hydroxide. The reactions are
An alkaline battery can deliver about three to five times the energy of a zinc-carbon dry cell of similar size. Alkaline batteries are prone to leaking potassium hydroxide, so these should also be removed from devices for long-term storage. While some alkaline batteries are rechargeable, most are not. Attempts to recharge an alkaline battery that is not rechargeable often leads to rupture of the battery and leakage of the potassium hydroxide electrolyte.
### Secondary Batteries
Secondary batteries are rechargeable. These are the types of batteries found in devices such as smartphones, electronic tablets, and automobiles.
Nickel-cadmium, or NiCd, batteries (Figure 17.12) consist of a nickel-plated cathode, cadmium-plated anode, and a potassium hydroxide electrode. The positive and negative plates, which are prevented from shorting by the separator, are rolled together and put into the case. This is a “jelly-roll” design and allows the NiCd cell to deliver much more current than a similar-sized alkaline battery. The reactions are
The voltage is about 1.2 V to 1.25 V as the battery discharges. When properly treated, a NiCd battery can be recharged about 1000 times. Cadmium is a toxic heavy metal so NiCd batteries should never be opened or put into the regular trash.
Lithium ion batteries (Figure 17.13) are among the most popular rechargeable batteries and are used in many portable electronic devices. The reactions are
With the coefficients representing moles, x is no more than about 0.5 moles. The battery voltage is about 3.7 V. Lithium batteries are popular because they can provide a large amount current, are lighter than comparable batteries of other types, produce a nearly constant voltage as they discharge, and only slowly lose their charge when stored.
The lead acid battery (Figure 17.14) is the type of secondary battery used in your automobile. It is inexpensive and capable of producing the high current required by automobile starter motors. The reactions for a lead acid battery are
Each cell produces 2 V, so six cells are connected in series to produce a 12-V car battery. Lead acid batteries are heavy and contain a caustic liquid electrolyte, but are often still the battery of choice because of their high current density. Since these batteries contain a significant amount of lead, they must always be disposed of properly.
### Fuel Cells
A fuel cell is a device that converts chemical energy into electrical energy. Fuel cells are similar to batteries but require a continuous source of fuel, often hydrogen. They will continue to produce electricity as long as fuel is available. Hydrogen fuel cells have been used to supply power for satellites, space capsules, automobiles, boats, and submarines (Figure 17.15).
In a hydrogen fuel cell, the reactions are
The voltage is about 0.9 V. The efficiency of fuel cells is typically about 40% to 60%, which is higher than the typical internal combustion engine (25% to 35%) and, in the case of the hydrogen fuel cell, produces only water as exhaust. Currently, fuel cells are rather expensive and contain features that cause them to fail after a relatively short time.
## Exercises
### Question 17.34
17.34
34. What are the desirable qualities of an electric battery?
### Question 17.35
17.25
35. List some things that are typically considered when selecting a battery for a new application.
### Question 17.36
17.36
36. Consider a battery made from one half-cell that consists of a copper electrode in 1 M $CuSO_4$ solution and another half-cell that consists of a lead electrode in 1 M $Pb(NO_3)_2$ solution. (a) What are the reactions at the anode, cathode, and the overall reaction? (b) What is the standard cell potential for the battery? (c) Most devices designed to use dry-cell batteries can operate between 1.0 and 1.5 V. Could this cell be used to make a battery that could replace a dry-cell battery? Why or why not. (d) Suppose sulfuric acid is added to the half-cell with the lead electrode and some $PbSO_4(s)$ forms. Would the cell potential increase, decrease, or remain the same?
### Question 17.37
17.37
37. Consider a battery with the overall reaction: $Cu(s) + 2Ag^+(aq) ⟶ 2Ag(s) + Cu^{2+}(aq)$. (a) What is the reaction at the anode and cathode? (b) A battery is “dead” when it has no cell potential. What is the value of Q when this battery is dead? (c) If a particular dead battery was found to have$[Cu^{2+}]$ = 0.11 M, what was the concentration of silver ion?
### Question 17.38
17.38
38. An inventor proposes using a SHE (standard hydrogen electrode) in a new battery for smartphones that also removes toxic carbon monoxide from the air:
### Question 17.39
17.39
39. Why do batteries go dead, but fuel cells do not?
### Question 17.40
17.40
40. Explain what happens to battery voltage as a battery is used, in terms of the Nernst equation.
### Question 17.41
17.41
41. Using the information thus far in this chapter, explain why battery-powered electronics perform poorly in low temperatures.
## 17.6 Corrosion
By the end of this section, you will be able to:
• Define corrosion
• List some of the methods used to prevent or slow corrosion
Corrosion is usually defined as the degradation of metals due to an electrochemical process. The formation of rust on iron, tarnish on silver, and the blue-green patina that develops on copper are all examples of corrosion. The total cost of corrosion in the United States is significant, with estimates in excess of half a trillion dollars a year.
### Chemistry in Everyday Life
Statue of Liberty: Changing Colors
The Statue of Liberty is a landmark every American recognizes. The Statue of Liberty is easily identified by its height, stance, and unique blue-green color (Figure 17.16). When this statue was first delivered from France, its appearance was not green. It was brown, the color of its copper “skin.” So how did the Statue of Liberty change colors? The change in appearance was a direct result of corrosion. The copper that is the primary component of the statue slowly underwent oxidation from the air. The oxidation-reduction reactions of copper metal in the environment occur in several steps. Copper metal is oxidized to copper(I) oxide (Cu2O), which is red, and then to copper(II) oxide, which is black
Coal, which was often high in sulfur, was burned extensively in the early part of the last century. As a result, sulfur trioxide, carbon dioxide, and water all reacted with the CuO
These three compounds are responsible for the characteristic blue-green patina seen today. Fortunately, formation of the patina created a protective layer on the surface, preventing further corrosion of the copper skin. The formation of the protective layer is a form of passivation, which is discussed further in a later chapter.
Perhaps the most familiar example of corrosion is the formation of rust on iron. Iron will rust when it is exposed to oxygen and water. The main steps in the rusting of iron appear to involve the following (Figure 17.17). Once exposed to the atmosphere, iron rapidly oxidizes.
The electrons reduce oxygen in the air in acidic solutions.
What we call rust is hydrated iron(III) oxide, which forms when iron(II) ions react further with oxygen.
The number of water molecules is variable, so it is represented by x. Unlike the patina on copper, the formation of rust does not create a protective layer and so corrosion of the iron continues as the rust flakes off and exposes fresh iron to the atmosphere.
One way to keep iron from corroding is to keep it painted. The layer of paint prevents the water and oxygen necessary for rust formation from coming into contact with the iron. As long as the paint remains intact, the iron is protected from corrosion.
Other strategies include alloying the iron with other metals. For example, stainless steel is mostly iron with a bit of chromium. The chromium tends to collect near the surface, where it forms an oxide layer that protects the iron.
Zinc-plated or galvanized iron uses a different strategy. Zinc is more easily oxidized than iron because zinc has a lower reduction potential. Since zinc has a lower reduction potential, it is a more active metal. Thus, even if the zinc coating is scratched, the zinc will still oxidize before the iron. This suggests that this approach should work with other active metals.
Another important way to protect metal is to make it the cathode in a galvanic cell. This is cathodic protection and can be used for metals other than just iron. For example, the rusting of underground iron storage tanks and pipes can be prevented or greatly reduced by connecting them to a more active metal such as zinc or magnesium (Figure 17.18). This is also used to protect the metal parts in water heaters. The more active metals (lower reduction potential) are called sacrificial anodes because as they get used up as they corrode (oxidize) at the anode. The metal being protected serves as the cathode, and so does not oxidize (corrode). When the anodes are properly monitored and periodically replaced, the useful lifetime of the iron storage tank can be greatly extended.
## Exercises
### Question 17.42
17.42
No correct answers: No correct answer has been set for this question
42. Which member of each pair of metals is more likely to corrode (oxidize)? (a) Mg or Ca (b) Au or Hg (c) Fe or Zn (d) Ag or Pt
A
a
B
b
C
c
D
d
### Question 17.43
17.43
43. Consider the following metals: Ag, Au, Mg, Ni, and Zn. Which of these metals could be used as a sacrificial anode in the cathodic protection of an underground steel storage tank? Steel is mostly iron, so use −0.447 V as thestandard reduction potential for steel.
### Question 17.44
17.44
44. Aluminum [/math] (E° {Al^{3+}/Al }\ =\ −2.07 V)[/math] is more easily oxidized than iron (E°{Fe^{3+}/Fe}\ =\ −0.477 V), and yet when both are exposed to the environment, untreated aluminum has very good corrosion resistance while the corrosion resistance of untreated iron is poor. Explain this observation.
### Question 17.45
17.45
45. If a sample of iron and a sample of zinc come into contact, the zinc corrodes but the iron does not. If a sample of iron comes into contact with a sample of copper, the iron corrodes but the copper does not. Explain this phenomenon
### Question 17.46
17.46
46. Suppose you have three different metals, A, B, and C. When metals A and B come into contact, B corrodes and A does not corrode. When metals A and C come into contact, A corrodes and C does not corrode. Based on this information, which metal corrodes and which metal does not corrode when B and C come into contact?
### Question 17.47
17.47
47. Why would a sacrificial anode made of lithium metal be a bad choice despite its $E°_{Li^+/Li}\ =\ −3.04 V$, which appears to be able to protect all the other metals listed in the standard reduction potential table?
## 17.7 Electrolysis
By the end of this section, you will be able to:
• Describe electrolytic cells and their relationship to galvanic cells
• Perform various calculations related to electrolysis
In galvanic cells, chemical energy is converted into electrical energy. The opposite is true for electrolytic cells. In electrolytic cells, electrical energy causes nonspontaneous reactions to occur in a process known as electrolysis. The charging electric car pictured in Figure 17.1 at the beginning of this chapter shows one such process. Electrical energy is converted into the chemical energy in the battery as it is charged. Once charged, the battery can be used to power the automobile.
The same principles are involved in electrolytic cells as in galvanic cells. We will look at three electrolytic cells and the quantitative aspects of electrolysis.
### The Electrolysis of Molten Sodium Chloride
In molten sodium chloride, the ions are free to migrate to the electrodes of an electrolytic cell. A simplified diagram of the cell commercially used to produce sodium metal and chlorine gas is shown in Figure 17.19. Sodium is a strong reducing agent and chlorine is used to purify water, and is used in antiseptics and in paper production. The reactions are
The power supply (battery) must supply a minimum of 4 V, but, in practice, the applied voltages are typically higher because of inefficiencies in the process itself.
### The Electrolysis of Water
It is possible to split water into hydrogen and oxygen gas by electrolysis. Acids are typically added to increase the concentration of hydrogen ion in solution (Figure 17.20). The reactions are
Note that the sulfuric acid is not consumed and that the volume of hydrogen gas produced is twice the volume of oxygen gas produced. The minimum applied voltage is 1.229 V.
### The Electrolysis of Aqueous Sodium Chloride
The electrolysis of aqueous sodium chloride is the more common example of electrolysis because more than one species can be oxidized and reduced. Considering the anode first, the possible reactions are
These values suggest that water should be oxidized at the anode because a smaller potential would be needed—using reaction (ii) for the oxidation would give a less-negative cell potential. When the experiment is run, it turns out chlorine, not oxygen, is produced at the anode. The unexpected process is so common in electrochemistry that it has been given the name overpotential. The overpotential is the difference between the theoretical cell voltage and the actual voltage that is necessary to cause electrolysis. It turns out that the overpotential for oxygen is rather high and effectively makes the reduction potential more positive. As a result, under normal conditions, chlorine gas is what actually forms at the anode.
Now consider the cathode. Three reductions could occur:
Reaction (v) is ruled out because it has such a negative reduction potential. Under standard state conditions, reaction (iii) would be preferred to reaction (iv). However, the pH of a sodium chloride solution is 7, so the concentration of hydrogen ions is only 1 × 10−7 M. At such low concentrations, reaction (iii) is unlikely and reaction (iv) occurs. The overall reaction is then
As the reaction proceeds, hydroxide ions replace chloride ions in solution. Thus, sodium hydroxide can be obtained by evaporating the water after the electrolysis is complete. Sodium hydroxide is valuable in its own right and is used for things like oven cleaner, drain opener, and in the production of paper, fabrics, and soap.
### Chemistry in Everyday Life
Electroplating
An important use for electrolytic cells is in electroplating. Electroplating results in a thin coating of one metal on top of a conducting surface. Reasons for electroplating include making the object more corrosion resistant, strengthening the surface, producing a more attractive finish, or for purifying metal. The metals commonly used in electroplating include cadmium, chromium, copper, gold, nickel, silver, and tin. Common consumer products include silver-plated or gold-plated tableware, chrome-plated automobile parts, and jewelry. We can get an idea of how this works by investigating how silver-plated tableware is produced (Figure 17.21).
In the figure, the anode consists of a silver electrode, shown on the left. The cathode is located on the right and is the spoon, which is made from inexpensive metal. Both electrodes are immersed in a solution of silver nitrate. As the potential is increased, current flows. Silver metal is lost at the anode as it goes into solution.
The mass of the cathode increases as silver ions from the solution are deposited onto the spoon
### Quantitative Aspects of Electrolysis
The amount of current that is allowed to flow in an electrolytic cell is related to the number of moles of electrons. The number of moles of electrons can be related to the reactants and products using stoichiometry. Recall that the SI unit for current (I) is the ampere (A), which is the equivalent of 1 coulomb per second (1 A = 1C/ s ). The total charge (Q, in coulombs) is given by
Where t is the time in seconds, n the number of moles of electrons, and F is the Faraday constant. Moles of electrons can be used in stoichiometry problems. The time required to deposit a specified amount of metal might also be requested, as in the second of the following examples.
### Example 17.8Converting Current to Moles of Electrons
In one process used for electroplating silver, a current of 10.23 A was passed through an electrolytic cell for exactly 1 hour. How many moles of electrons passed through the cell? What mass of silver was deposited at the cathode from the silver nitrate solution?
Solution
Faraday’s constant can be used to convert the charge (Q) into moles of electrons (n). The charge is the current (I) multiplied by the time
From the problem, the solution contains AgNO3, so the reaction at the cathode involves 1 mole of electrons for each mole of silver
The atomic mass of silver is 107.9 g/mol, so
Check your answer: From the stoichiometry, 1 mole of electrons would produce 1 mole of silver. Less than one-half a mole of electrons was involved and less than one-half a mole of silver was produced.
Aluminum metal can be made from aluminum ions by electrolysis. What is the half-reaction at the cathode? What mass of aluminum metal would be recovered if a current of 2.50 × 103 A passed through the solution for 15.0 minutes? Assume the yield is 100%.
### Example 17.9Time Required for Deposition
In one application, a 0.010-mm layer of chromium must be deposited on a part with a total surface area of 3.3 m2 from a solution of containing chromium(III) ions. How long would it take to deposit the layer of chromium if the current was 33.46 A? The density of chromium (metal) is 7.19 g/cm3 .
Solution
This problem brings in a number of topics covered earlier. An outline of what needs to be done is:
• If the total charge can be determined, the time required is just the charge divided by the current
• The total charge can be obtained from the amount of Cr needed and the stoichiometry
• The amount of Cr can be obtained using the density and the volume Cr required
• The volume Cr required is the thickness times the area
Solving in steps, and taking care with the units, the volume of Cr required is
Cubic centimeters were used because they match the volume unit used for the density. The amount of Cr is then
Since the solution contains chromium(III) ions, 3 moles of electrons are required per mole of Cr. The total charge is then
The time required is then
Check your answer: In a long problem like this, a single check is probably not enough. Each of the steps gives a reasonable number, so things are probably correct. Pay careful attention to unit conversions and the stoichiometry.
What mass of zinc is required to galvanize the top of a 3.00 m × 5.50 m sheet of iron to a thickness of 0.100 mm of zinc? If the zinc comes from a solution of Zn(NO3)2 and the current is 25.5 A, how long will it take to galvanize the top of the iron? The density of zinc is 7.140 g/cm3 .
Answer: 231 g Zn required 446 minutes
## Exercises
### Question 17.48
17.48
48. Identify the reaction at the anode, reaction at the cathode, the overall reaction, and the approximate potential required for the electrolysis of the following molten salts. Assume standard states and that the standard reduction potentials in Appendix L are the same as those at each of the melting points. Assume the efficiency is 100%. (a) $CaCl_2$ (b) $LiH$ (c) $AlCl_3$ (d) $CrBr_3$
### Question 17.49
17.49
49. What mass of each product is produced in each of the electrolytic cells of the previous problem if a total charge of $3.33 × 10^5\ C$ passes through each cell? Assume the voltage is sufficient to perform the reduction.
### Question 17.50
17.50
50. How long would it take to reduce 1 mole of each of the following ions using the current indicated? Assume the voltage is sufficient to perform the reduction. (a) $Al^{3+},\ 1.234\ A$ (b) $Ca{^2+},\ 22.2\ A$ (c) $Cr^{5+},\ 37.45\ A$ (d) $Au^{3+},\ 3.57\ A$
### Question 17.51
17.51
51. A current of 2.345 A passes through the cell shown in Figure for 45 minutes. What is the volume of the hydrogen collected at room temperature if the pressure is exactly 1 atm? Assume the voltage is sufficient to perform the reduction. (Hint: Is hydrogen the only gas present above the water?)
### Question 17.52
17.52
52. An irregularly shaped metal part made from a particular alloy was galvanized with zinc using a $Zn(NO_3)_2$ solution. When a current of 2.599 A was used, it took exactly 1 hour to deposit a 0.01123-mm layer of zinc on the part. What was the total surface area of the part? The density of zinc is 7.140 $g/cm^3$ . Assume the efficiency is 100%
## Key Terms
active electrode electrode that participates in the oxidation-reduction reaction of an electrochemical cell; the mass of an active electrode changes during the oxidation-reduction reaction
alkaline battery primary battery that uses an alkaline (often potassium hydroxide) electrolyte; designed to be an exact replacement for the dry cell, but with more energy storage and less electrolyte leakage than typical dry cell
anode electrode in an electrochemical cell at which oxidation occurs; information about the anode is recorded on the left side of the salt bridge in cell notation
battery galvanic cell or series of cells that produces a current; in theory, any galvanic cell
cathode electrode in an electrochemical cell at which reduction occurs; information about the cathode is recorded on the right side of the salt bridge in cell notation
cathodic protection method of protecting metal by using a sacrificial anode and effectively making the metal that needs protecting the cathode, thus preventing its oxidation
cell notation shorthand way to represent the reactions in an electrochemical cell
cell potential difference in electrical potential that arises when dissimilar metals are connected; the driving force for the flow of charge (current) in oxidation-reduction reactions
circuit path taken by a current as it flows because of an electrical potential difference
concentration cell galvanic cell in which the two half-cells are the same except for the concentration of the solutes; spontaneous when the overall reaction is the dilution of the solute
corrosion degradation of metal through an electrochemical process
current flow of electrical charge; the SI unit of charge is the coulomb (C) and current is measured in amperes
dry cell primary battery, also called a zinc-carbon battery; can be used in any orientation because it uses a paste as the electrolyte; tends to leak electrolyte when stored
electrical potential energy per charge; in electrochemical systems, it depends on the way the charges are distributed within the system; the SI unit of electrical potential is the volt (1 V = 1 J/s)
electrical work (welenegative of total charge times the cell potential; equal to wmax for the system, and so equals the free energy change (ΔG)
electrolysis process using electrical energy to cause a nonspontaneous process to occur
electrolytic cell electrochemical cell in which electrolysis is used; electrochemical cell with negative cell potentials
electroplating depositing a thin layer of one metal on top of a conducting surface
Faraday’s constant (F) charge on 1 mol of electrons; F = 96,485 C/mol e
fuel cell devices that produce an electrical current as long as fuel and oxidizer are continuously added; more efficient than internal combustion engines
galvanic cell electrochemical cell that involves a spontaneous oxidation-reduction reaction; electrochemical cells with positive cell potentials; also called a voltaic cell
galvanized iron method for protecting iron by covering it with zinc, which will oxidize before the iron; zinc-plated iron
half-reaction method method that produces a balanced overall oxidation-reduction reaction by splitting the reaction into an oxidation “half” and reduction “half,” balancing the two half-reactions, and then combining the oxidation half-reaction and reduction half-reaction in such a way that the number of electrons generated by the oxidation is exactly canceled by the number of electrons required by the reduction
inert electrode electrode that allows current to flow, but that does not otherwise participate in the oxidationreduction reaction in an electrochemical cell; the mass of an inert electrode does not change during the oxidationreduction reaction; inert electrodes are often made of platinum or gold because these metals are chemically unreactive.
lead acid battery secondary battery that consists of multiple cells; the lead acid battery found in automobiles has six cells and a voltage of 12 V
lithium ion battery very popular secondary battery; uses lithium ions to conduct current and is light, rechargeable, and produces a nearly constant potential as it discharges
Nernst equation equation that relates the logarithm of the reaction quotient (Q) to nonstandard cell potentials; can be used to relate equilibrium constants to standard cell potentials
nickel-cadmium battery (NiCd battery) secondary battery that uses cadmium, which is a toxic heavy metal; heavier than lithium ion batteries, but with similar performance characteristics
overpotential difference between the theoretical potential and actual potential in an electrolytic cell; the “extra” voltage required to make some nonspontaneous electrochemical reaction to occur
oxidation half-reaction the “half” of an oxidation-reduction reaction involving oxidation; the half-reaction in which electrons appear as products; balanced when each atom type, as well as the charge, is balanced
primary battery single-use nonrechargeable battery
reduction half-reaction the “half” of an oxidation-reduction reaction involving reduction; the half-reaction in which electrons appear as reactants; balanced when each atom type, as well as the charge, is balanced
sacrificial anode more active, inexpensive metal used as the anode in cathodic protection; frequently made from magnesium or zinc
secondary battery battery that can be recharged
standard cell potential (E°cellthe cell potential when all reactants and products are in their standard states (1 bar or 1 atm or gases; 1 M for solutes), usually at 298.15 K; can be calculated by subtracting the standard reduction potential for the half-reaction at the anode from the standard reduction potential for the half-reaction occurring at the cathode
standard hydrogen electrode (SHE) the electrode consists of hydrogen gas bubbling through hydrochloric acid over an inert platinum electrode whose reduction at standard conditions is assigned a value of 0 V; the reference point for standard reduction potentials
standard reduction potential (E°) the value of the reduction under standard conditions (1 bar or 1 atm for gases; 1 M for solutes) usually at 298.15 K; tabulated values used to calculate standard cell potentials
voltaic cell another name for a galvanic cell
## Summary
17.1 Balancing Oxidation-Reduction Reactions
An electric current consists of moving charge. The charge may be in the form of electrons or ions. Current flows through an unbroken or closed circular path called a circuit. The current flows through a conducting medium as a result of a difference in electrical potential between two points in a circuit. Electrical potential has the units of energy per charge. In SI units, charge is measured in coulombs (C), current in amperes (A = C/s) , and electrical potential in volts (V = J/C). Oxidation is the loss of electrons, and the species that is oxidized is also called the reducing agent. Reduction is the gain of electrons, and the species that is reduced is also called the oxidizing agent. Oxidation-reduction reactions can be balanced using the half-reaction method. In this method, the oxidation-reduction reaction is split into an oxidation half-reaction and a reduction half-reaction. The oxidation half-reaction and reduction half-reaction are then balanced separately. Each of the half-reactions must have the same number of each type of atom on both sides of the equation and show the same total charge on each side of the equation. Charge is balanced in oxidation half-reactions by adding electrons as products; in reduction half-reactions, charge is balanced by adding electrons as reactants. The total number of electrons gained by reduction must exactly equal the number of electrons lost by oxidation when combining the two half-reactions to give the overall balanced equation. Balancing oxidation-reduction reaction equations in aqueous solutions frequently requires that oxygen or hydrogen be added or removed from a reactant. In acidic solution, hydrogen is added by adding hydrogen ion (H+) and removed by producing hydrogen ion; oxygen is removed by adding hydrogen ion and producing water, and added by adding water and producing hydrogen ion. A balanced equation in basic solution can be obtained by first balancing the equation in acidic solution, and then adding hydroxide ion to each side of the balanced equation in such numbers that all the hydrogen ions are converted to water.
17.2 Galvanic Cells
Electrochemical cells typically consist of two half-cells. The half-cells separate the oxidation half-reaction from the reduction half-reaction and make it possible for current to flow through an external wire. One half-cell, normally depicted on the left side in a figure, contains the anode. Oxidation occurs at the anode. The anode is connected to the cathode in the other half-cell, often shown on the right side in a figure. Reduction occurs at the cathode. Adding a salt bridge completes the circuit allowing current to flow. Anions in the salt bridge flow toward the anode and cations in the salt bridge flow toward the cathode. The movement of these ions completes the circuit and keeps each halfcell electrically neutral. Electrochemical cells can be described using cell notation. In this notation, information about the reaction at the anode appears on the left and information about the reaction at the cathode on the right. The salt bridge is represented by a double line, ‖. The solid, liquid, or aqueous phases within a half-cell are separated by a single line, │. The phase and concentration of the various species is included after the species name. Electrodes that participate in the oxidation-reduction reaction are called active electrodes. Electrodes that do not participate in the oxidation-reduction reaction but are there to allow current to flow are inert electrodes. Inert electrodes are often made from platinum or gold, which are unchanged by many chemical reactions.
17.3 Standard Reduction Potentials
Assigning the potential of the standard hydrogen electrode (SHE) as zero volts allows the determination of standard reduction potentials, , for half-reactions in electrochemical cells. As the name implies, standard reduction potentials use standard states (1 bar or 1 atm for gases; 1 M for solutes, often at 298.15 K) and are written as reductions (where electrons appear on the left side of the equation). The reduction reactions are reversible, so standard cell potentials can be calculated by subtracting the standard reduction potential for the reaction at the anode from the standard reduction for the reaction at the cathode. When calculating the standard cell potential, the standard reduction potentials are not scaled by the stoichiometric coefficients in the balanced overall equation.
17.4 The Nernst Equation
Electrical work (wele) is the negative of the product of the total charge (Q) and the cell potential (Ecell). The total charge can be calculated as the number of moles of electrons (n) times the Faraday constant (F = 96,485 C/mol e ). Electrical work is the maximum work that the system can produce and so is equal to the change in free energy. Thus, anything that can be done with or to a free energy change can also be done to or with a cell potential. The Nernst equation relates the cell potential at nonstandard conditions to the logarithm of the reaction quotient. Concentration cells exploit this relationship and produce a positive cell potential using half-cells that differ only in the concentration of their solutes.
17.5 Batteries and Fuel Cells
Batteries are galvanic cells, or a series of cells, that produce an electric current. When cells are combined into batteries, the potential of the battery is an integer multiple of the potential of a single cell. There are two basic types of batteries: primary and secondary. Primary batteries are “single use” and cannot be recharged. Dry cells and (most) alkaline batteries are examples of primary batteries. The second type is rechargeable and is called a secondary battery. Examples of secondary batteries include nickel-cadmium (NiCd), lead acid, and lithium ion batteries. Fuel cells are similar to batteries in that they generate an electrical current, but require continuous addition of fuel and oxidizer. The hydrogen fuel cell uses hydrogen and oxygen from the air to produce water, and is generally more efficient than internal combustion engines.
17.6 Corrosion
Corrosion is the degradation of a metal caused by an electrochemical process. Large sums of money are spent each year repairing the effects of, or preventing, corrosion. Some metals, such as aluminum and copper, produce a protective layer when they corrode in air. The thin layer that forms on the surface of the metal prevents oxygen from coming into contact with more of the metal atoms and thus “protects” the remaining metal from further corrosion. Iron corrodes (forms rust) when exposed to water and oxygen. The rust that forms on iron metal flakes off, exposing fresh metal, which also corrodes. One way to prevent, or slow, corrosion is by coating the metal. Coating prevents water and oxygen from contacting the metal. Paint or other coatings will slow corrosion, but they are not effective once scratched. Zinc-plated or galvanized iron exploits the fact that zinc is more likely to oxidize than iron. As long as the coating remains, even if scratched, the zinc will oxidize before the iron. Another method for protecting metals is cathodic protection. In this method, an easily oxidized and inexpensive metal, often zinc or magnesium (the sacrificial anode), is electrically connected to the metal that must be protected. The more active metal is the sacrificial anode, and is the anode in a galvanic cell. The “protected” metal is the cathode, and remains unoxidized. One advantage of cathodic protection is that the sacrificial anode can be monitored and replaced if needed.
17.7 Electrolysis
Using electricity to force a nonspontaneous process to occur is electrolysis. Electrolytic cells are electrochemical cells with negative cell potentials (meaning a positive Gibbs free energy), and so are nonspontaneous. Electrolysis can occur in electrolytic cells by introducing a power supply, which supplies the energy to force the electrons to flow in the nonspontaneous direction. Electrolysis is done in solutions, which contain enough ions so current can flow. If the solution contains only one material, like the electrolysis of molten sodium chloride, it is a simple matter to determine what is oxidized and what is reduced. In more complicated systems, like the electrolysis of aqueous sodium chloride, more than one species can be oxidized or reduced and the standard reduction potentials are used to determine the most likely oxidation (the half-reaction with the largest [most positive] standard reduction potential) and reduction (the half-reaction with the smallest [least positive] standard reduction potential). Sometimes unexpected half-reactions occur because of overpotential. Overpotential is the difference between the theoretical half-reaction reduction potential and the actual voltage required. When present, the applied potential must be increased, making it possible for a different reaction to occur in the electrolytic cell. The total charge, Q, that passes through an electrolytic cell can be expressed as the current (I) multiplied by time (Q = It) or as the moles of electrons (n) multiplied by Faraday’s constant (Q = nF). These relationships can be used to determine things like the amount of material used or generated during electrolysis, how long the reaction must proceed, or what value of the current is required.
5300
(a) reduction; (b) oxidation; (c) oxidation; (d) reduction
Oxidized: (a) Ag; (b) Sn2+; (c) Hg; (d) Al; reduced: (a) Hg22+; (b) H2O2; (c) PbO2; (d) Cr2 O72−;
oxidizingagent: (a) Hg22+; (b) H2O2; (c) PbO2; (d) Cr2 O72−; reducing agent: (a) Ag; (b) Sn2+; (c)Hg; (d) Al
Oxidized = reducing agent: (a) SO32−; (b) Mn(OH)2; (c) H2; (d) Al; reduced = oxidizing agent: (a) Cu(OH)2; (b)O2; (c) NO3; (d) CrO42−
In basic solution, [OH] > 1 × 10−7 M > [H+]. Hydrogen ion cannot appear as a reactant because its concentration is essentially zero. If it were produced, it would instantly react with the excess hydroxide ion to produce water. Thus, hydrogen ion should not appear as a reactant or product in basic solution.
(a) Mg(s) │ Mg2+(aq) ║ Ni+(aq) │ Ni(s); (b) Cu(s) │ Cu2+(aq) ║ Ag+(aq) │ Ag(s); (c)
Mn(s) │ Mn2+(aq) ║ Sn2+(aq) │ Sn(s); (d) Pt(s) │ Cu+(aq), Cu2+(aq) ║ Au3+(aq) │ Au(s)
Without the salt bridge, the circuit would be open (or broken) and no current could flow. With a salt bridge, each half-cell remains electrically neutral and current can flow through the circuit.
Active electrodes participate in the oxidation-reduction reaction. Since metals form cations, the electrode would lose mass if metal atoms in the electrode were to oxidize and go into solution. Oxidation occurs at the anode.
(a) +2.115 V (spontaneous); (b) +0.4626 V (spontaneous); (c) +1.0589 V (spontaneous); (d) +0.727 V (spontaneous)
(a) 0 kJ/mol; (b) −83.7 kJ/mol; (c) +235.3 kJ/mol
(a) standard cell potential: 1.50 V, spontaneous; cell potential under stated conditions: 1.43 V, spontaneous; (b) standard cell potential: 1.405 V, spontaneous; cell potential under stated conditions: 1.423 V, spontaneous; (c) standard cell potential: −2.749 V, nonspontaneous; cell potential under stated conditions: −2.757 V, nonspontaneous
(a) 1.7 × 10−10; (b) 2.6 × 10−21; (c) 8.9 × 1019; (d) 1.0 × 10−14
Considerations include: cost of the materials used in the battery, toxicity of the various components (what constitutes proper disposal), should it be a primary or secondary battery, energy requirements (the “size” of the battery/how long should it last), will a particular battery leak when the new device is used according to directions, and its mass (the total mass of the new device).
Batteries are self-contained and have a limited supply of reagents to expend before going dead. Alternatively, battery reaction byproducts accumulate and interfere with the reaction. Because a fuel cell is constantly resupplied with reactants and products are expelled, it can continue to function as long as reagents are supplied.
Ecell, as described in the Nernst equation, has a term that is directly proportional to temperature. At low temperatures, this term is decreased, resulting in a lower cell voltage provided by the battery to the device—the same effect as a battery running dead.
Mg and Zn
Both examples involve cathodic protection. The (sacrificial) anode is the metal that corrodes (oxidizes or reacts). In the case of iron (−0.447 V) and zinc (−0.7618 V), zinc has a more negative standard reduction potential and so serves as the anode. In the case of iron and copper (0.34 V), iron has the smaller standard reduction potential and so corrodes (serves as the anode).
While the reduction potential of lithium would make it capable of protecting the other metals, this high potential is also indicative of how reactive lithium is; it would have a spontaneous reaction with most substances. This means that the lithium would react quickly with other substances, even those that would not oxidize the metal it is attempting to protect. Reactivity like this means the sacrificial anode would be depleted rapidly and need to be replaced frequently. (Optional additional reason: fire hazard in the presence of water.)
(a) mass Ca = 69.1 g mass Cl2 = 122 g ;
(b) mass Li = 23.9 g mass H2 = 3.48 g;
(c) mass Al = 31.0 g mass Cl2 = 122 g;
(d) mass Cr = 59.8 g mass Br2 = 276 g
0.79 L
|
{}
|
Question 1: Express each of the following as the product of sines and cosines:
i) $\sin 12 x + \sin 4x$ ii) $\sin 5x - \sin x$ iii) $\cos 12 x + \cos 8x$
iv) $\cos 12 x - \cos 4x$ v) $\sin 2x + \cos 4x$
i) $\sin 12 x + \sin 4x = 2 \sin \Big($ $\frac{12x + 4x}{2}$ $\Big) \cos \Big($ $\frac{12x - 4x}{2}$ $\Big) = 2 \sin 8x \cos 4x$
ii) $\sin 5x - \sin x = 2 \cos \Big($ $\frac{5x + x}{2}$ $\Big) \sin \Big($ $\frac{5x - x}{2}$ $\Big) = 2 \cos 3x \sin 2x$
iii) $\cos 12 x + \cos 8x = 2 \cos \Big($ $\frac{12x + 8x}{2}$ $\Big) \cos \Big($ $\frac{12x - 8x}{2}$ $\Big) = 2 \cos 10x \cos 2x$
iv) $\cos 12 x - \cos 4x = -2 \sin \Big($ $\frac{12x + 4x}{2}$ $\Big) \sin \Big($ $\frac{12x - 4x}{2}$ $\Big) = -2 \sin 8x \sin 4x$
v) $\sin 2x + \cos 4x = \sin 2x + \sin (90^o - 4x)$
$= 2 \sin \Big($ $\frac{2x + 90^o - 4x}{2}$ $\Big) \cos \Big($ $\frac{2x - 90^o + 4x}{2}$ $\Big)$
$= 2 \sin (45^o - x) \cos ( 3x - 45^o)$
$= 2 \sin (45^o - x) \cos ( 45^o- 3x)$
$\\$
Question 2: Prove that:
i) $\sin 38^o + \sin 22^o = \sin 82^o$ ii) $\cos 100^o + \cos 20^o = \cos 40^o$
iii) $\sin 50^o + \sin 10^o = \cos 20^o$ iv) $\sin 23^o + \sin 37^o = \cos 7^o$
v) $\sin 105^o + \cos 105^o = \cos 45^o$ vi) $\sin 40^o + \sin 20^o = \cos 10^o$
i) LHS $=\sin 38^o + \sin 22^o$
$=2 \sin \Big($ $\frac{38^o+22^o}{2}$ $\Big) \cos \Big($ $\frac{38^o-22^o}{2}$ $\Big)$
$= 2 \sin 30^o \cos 8^o$
$= 2 \times$ $\frac{1}{2}$ $\cos 8^o$
$= \cos (90^o-82^o) = \sin 82^o =$ RHS. Hence proved.
ii) LHS $=\cos 100^o + \cos 20^o$
$=2 \cos \Big($ $\frac{100^o+20^o}{2}$ $\Big) \cos \Big($ $\frac{100^o-20^o}{2}$ $\Big)$
$= 2 \cos 60^o \cos 40^o$
$= 2 \times$ $\frac{1}{2}$ $\cos 40^o$
$= \cos 40^o =$ RHS. Hence proved.
iii) LHS $=\sin 50^o + \sin 10^o$
$=2 \sin \Big($ $\frac{50^o+10^o}{2}$ $\Big) \cos \Big($ $\frac{50^o-10^o}{2}$ $\Big)$
$= 2 \sin 30^o \cos 20^o$
$= 2 \times$ $\frac{1}{2}$ $\cos 20^o$
$= \cos 20^o =$ RHS. Hence proved.
iv) LHS $=\sin 23^o + \sin 37^o$
$=2 \sin \Big($ $\frac{23^o+37^o}{2}$ $\Big) \cos \Big($ $\frac{23^o-37^o}{2}$ $\Big)$
$= 2 \sin 30^o \cos (-7^o)$
$= 2 \times$ $\frac{1}{2}$ $\cos 7^o$
$= \cos 7^o =$ RHS. Hence proved.
v) LHS $=\sin 105^o + \cos 105^o$
$=\sin 105^o + \cos (90^o+ 15^o)$
$=\sin 105^o - \sin 15^o$
$=2 \sin \Big($ $\frac{105^o-15^o}{2}$ $\Big) \cos \Big($ $\frac{105^o+15^o}{2}$ $\Big)$
$= 2 \sin 45^o \cos 60^o$
$= 2 \times$ $\frac{1}{\sqrt{2}}$ $\times$ $\frac{1}{2}$
$=$ $\frac{1}{\sqrt{2}}$ $= \cos 45^o =$ RHS. Hence proved.
vi) LHS $=\sin 40^o + \sin 20^o$
$=2 \sin \Big($ $\frac{40^o+20^o}{2}$ $\Big) \cos \Big($ $\frac{20^o-40^o}{2}$ $\Big)$
$= 2 \sin 30^o \cos (-10^o)$
$= 2 \times$ $\frac{1}{2}$ $\cos 10^o$
$= \cos 10^o =$ RHS. Hence proved.
$\\$
Question 3: Prove that:
i) $\cos 55^o + \cos 65^o + \cos 175^o = 0$ ii) $\sin 50^o - \sin 70^o + \sin 10^o = 0$
iii) $\cos 80^o + \cos 40^o - \cos 20^o = 0$ v) $\cos 20^o + \cos 100^o + \cos 140^o = 0$
v) $\sin$ $\frac{5\pi}{18}$ $- \cos$ $\frac{4\pi}{9}$ $= \sqrt{3} \sin$ $\frac{\pi}{9}$ vi) $\cos$ $\frac{\pi}{12}$ $- \sin$ $\frac{\pi}{12}$ $=$ $\frac{1}{\sqrt{2}}$
vii) $\sin 80^o - \cos 70^o = \cos 50^o$ viii) $\sin 51^o + \cos 81^o = \cos 21^o$
i) LHS $= \cos 55^o + \cos 65^o + \cos 175^o$
$= \cos 55^o + \cos 65^o + \cos ( 180^o - 5^o)$
$= \cos 55^o + \cos 65^o - \cos 5^o$
$= 2 \cos$ $\frac{55^o+65^o}{2}$ $\cos$ $\frac{55^o-65^o}{2}$ $- \cos 5^o$
$=2 \cos 60^o \cos 5^o - \cos 5^o$
$= 2 \times$ $\frac{1}{2}$ $\cos 5^o - \cos 5^o$
$= \cos 5^o - \cos 5^o = 0 =$ RHS. Hence proved.
ii) LHS $= \sin 50^o - \sin 70^o + \sin 10^o$
$= 2 \sin$ $\frac{50^o-70^o}{2}$ $\cos$ $\frac{50^o+70^o}{2}$ $+ \sin 10^o$
$=2 \sin (-10^o) \cos 60^o + \sin 10^o$
$= -2 \times$ $\frac{1}{2}$ $\sin 10^o + \sin 10^o$
$= -\sin 10^o + \sin 10^o = 0 =$ RHS. Hence proved.
iii) LHS $= \cos 80^o + \cos 40^o - \cos 20^o$
$= 2 \cos$ $\frac{80^o+40^o}{2}$ $\cos$ $\frac{80^o-40^o}{2}$ $- \cos 20^o$
$=2 \cos 60^o \cos 20^o - \cos 20^o$
$= 2 \times$ $\frac{1}{2}$ $\cos 20^o - \cos 20^o$
$= \cos 20^o - \cos 20^o = 0 =$ RHS. Hence proved.
iv) LHS $= \cos 20^o + \cos 100^o + \cos 140^o$
$= 2 \cos$ $\frac{20^o+100^o}{2}$ $\cos$ $\frac{20^o-100^o}{2}$ $+ \cos (180^o - 40^o)$
$=2 \cos 60^o \cos 40^o - \cos 40^o$
$= 2 \times$ $\frac{1}{2}$ $\cos 40^o - \cos 40^o$
$= \cos 40^o - \cos 40^o = 0 =$ RHS. Hence proved.
v) LHS $= \sin$ $\frac{5\pi}{18}$ $- \cos$ $\frac{4\pi}{9}$
$= \sin 50^o - \cos 80^o = \sin 50^o - \cos ( 90^o-10^o) = \sin 50^o - \sin 10^o$
$= 2 \sin$ $\frac{50^o-10^o}{2}$ $\cos$ $\frac{50^o+10^o}{2}$
$=2 \sin 20^o \cos 30^o$
$= 2 \times$ $\frac{\sqrt{3}}{2}$ $\sin 20^o$
$= \sqrt{3} \sin 20^o = \sqrt{3} \sin$ $\frac{\pi}{9}$ $=$ RHS. Hence proved.
vi) LHS $= \cos$ $\frac{\pi}{12}$ $- \sin$ $\frac{\pi}{12}$
$= \cos 15^o - \sin 15^o = \cos 15^o - \sin ( 90^o-75^o) = \cos 15^o - \cos 75^o$
$= 2 \sin$ $\frac{15^o+75^o}{2}$ $\sin$ $\frac{75^o-15^o}{2}$
$= 2 \sin 45^o \sin 30^o$
$= 2 \times$ $\frac{1}{\sqrt{2}}$ $\sin 30^o$
$= 2 \times$ $\frac{1}{\sqrt{2}}$ $\times$ $\frac{1}{2}$ $=$ $\frac{1}{\sqrt{2}}$ $=$ RHS. Hence proved.
vii) LHS $= \sin 80^o - \cos 70^o = \sin 80^o - \cos ( 90^o-20^o) = \sin 80^o - \sin 20^o$
$= 2 \sin$ $\frac{80^o-20^o}{2}$ $\cos$ $\frac{80^o+20^o}{2}$
$=2 \sin 30^o \cos 50^o$
$= 2 \times$ $\frac{1}{2}$ $\cos 50^o$
$= \cos 50^o =$ RHS. Hence proved.
viii) LHS $= \sin 51^o + \cos 81^o = \sin 51^o + \cos ( 90^o-9^o) = \sin 51^o + \sin 9^o$
$= 2 \sin$ $\frac{51^o+9^o}{2}$ $\cos$ $\frac{51^o-9^o}{2}$
$=2 \sin 30^o \cos 21^o$
$= 2 \times$ $\frac{1}{2}$ $\cos 21^o$
$= \cos 21^o =$ RHS. Hence proved.
$\\$
Question 4: Prove that:
i) $\cos \Big($ $\frac{3\pi}{4}$ $+x \Big) - \cos \Big($ $\frac{3\pi}{4}$ $-x \Big) = -\sqrt{2} \sin x$
ii) $\cos \Big($ $\frac{\pi}{4}$ $+x \Big) + \cos \Big($ $\frac{\pi}{4}$ $-x \Big) = -\sqrt{2} \cos x$
i) LHS $= \cos \Big($ $\frac{3\pi}{4}$ $+x \Big) - \cos \Big($ $\frac{3\pi}{4}$ $-x \Big)$
$= - \Big[ \cos \Big($ $\frac{3\pi}{4}$ $-x \Big) - \cos \Big($ $\frac{3\pi}{4}$ $+x \Big) \Big]$
$= - \Big[ -2 \sin \Big($ $\frac{\frac{3\pi}{4} - x + \frac{3\pi}{4} + x}{2}$ $\Big) \sin \Big($ $\frac{\frac{3\pi}{4} - x - \frac{3\pi}{4} - x}{2}$ $\Big) \Big]$
$= - \Big[ - 2 \sin$ $\frac{3\pi}{4}$ $\sin (-x) \Big]$
$= - 2 \sin \Big($ $\frac{\pi}{2}$ $+$ $\frac{\pi}{4}$ $\Big) \sin x$
$= -2 \cos$ $\frac{\pi}{4}$ $\sin x = - \sqrt{2} \sin x =$ RHS. Hence proved.
ii) LHS $= \cos \Big($ $\frac{\pi}{4}$ $+x \Big) + \cos \Big($ $\frac{\pi}{4}$ $-x \Big)$
$= 2 \cos \Big($ $\frac{\frac{\pi}{4}+ x + \frac{\pi}{4} - x}{2}$ $\Big) \cos \Big($ $\frac{\frac{\pi}{4}+ x - \frac{\pi}{4} + x}{2}$ $\Big)$
$= 2 \cos$ $\frac{\pi}{4}$ $\cos x$
$= \sqrt{2} \cos x =$ RHS. Hence proved.
$\\$
Question 5: Prove that:
i) $\sin 65^o + \cos 65^o = \sqrt{2} \cos 20^o$ ii) $\sin 47^o +\cos 77^o = \cos 17^o$
i) LHS $= \sin 65^o + \cos 65^o$
$= \sin 65^o + \cos ( 90^o-25^o)$
$= \sin 65^o + \sin 25^o$
$= 2 \sin \Big($ $\frac{65^o+ 25^o}{2}$ $\Big) \cos \Big($ $\frac{65^o- 25^o}{2}$ $\Big)$
$= 2 \sin 45^o \cos 20^o$
$= 2 \times$ $\frac{1}{\sqrt{2}}$ $\cos 20^o$
$= \sqrt{2} \cos 20^o =$ RHS. Hence proved.
ii) LHS $= \sin 47^o +\cos 77^o$
LHS $= \sin 47^o + \cos 77^o$
$= \sin 47^o + \cos ( 90^o-13^o)$
$= \sin 47^o + \sin 13^o$
$= 2 \sin \Big($ $\frac{47^o+ 13^o}{2}$ $\Big) \cos \Big($ $\frac{47^o- 13^o}{2}$ $\Big)$
$= 2 \sin 30^o \cos 17^o$
$= 2 \times$ $\frac{1}{2}$ $\cos 17^o$
$= \cos 17^o =$ RHS. Hence proved.
$\\$
Question 6: Prove that:
i) $\cos 3A + \cos 5A + \cos 7A + \cos 15A = 4 \cos 4A \cos 5A \cos 6A$
ii) $\cos A + \cos 3A + \cos 5A + \cos 7A = 4 \cos A \cos 2A \cos 4A$
iii) $\sin A + \sin 2A + \sin 4A + \sin 5A = 4 \cos$ $\frac{A}{2}$ $\cos$ $\frac{3A}{2}$ $\sin 3A$
iv) $\sin 3A + \sin 2A - \sin A = 4 \sin A \cos$ $\frac{A}{2}$ $\cos$ $\frac{3A}{2}$
v) $\cos 20^o \cos 100^o + \cos 100^o \cos 140^o - \cos 140^o \cos 200^o = -$ $\frac{3}{4}$
vi) $\sin$ $\frac{x}{2}$ $\sin$ $\frac{7x}{2}$ $+ \sin$ $\frac{3x}{2}$ $\sin$ $\frac{11x}{2}$ $= \sin 2x \sin 5x$
vii) $\cos x \cos$ $\frac{x}{2}$ $- \cos 3x \cos$ $\frac{9x}{2}$ $= \sin 4x \sin$ $\frac{7x}{2}$
i) LHS $= \cos 3A + \cos 5A + \cos 7A + \cos 15A$
$= [ \cos 5A + \cos 3A] + [ \cos 15A + \cos 7A ]$
$= 2 \cos \Big($ $\frac{5A + 3A}{2}$ $\Big) \cos \Big($ $\frac{5A - 3A}{2}$ $\Big) + 2 \cos \Big($ $\frac{15A + 7A}{2}$ $\Big) \cos \Big($ $\frac{15A - 7A}{2}$ $\Big)$
$= 2 \cos 4A \cos A + 2 \cos 11 A \cos 4A$
$= 2 \cos 4A [ \cos 11A + \cos A ]$
$= 2 \cos 4A \Big[ 2 \cos \Big($ $\frac{11A + A}{2}$ $\Big) \cos \Big($ $\frac{11A - A}{2}$ $\Big) \Big]$
$= 2 \cos 4A [ 2 \cos 6A \cos 5A ]$
$= 4 \cos 4A \cos 5A \cos 6A =$ RHS. Hence proved.
ii) LHS $= \cos A + \cos 3A + \cos 5A + \cos 7A$
$= [ \cos 3A + \cos A] + [ \cos 7A + \cos 5A ]$
$= 2 \cos \Big($ $\frac{3A + A}{2}$ $\Big) \cos \Big($ $\frac{3A - A}{2}$ $\Big) + 2 \cos \Big($ $\frac{7A + 5A}{2}$ $\Big) \cos \Big($ $\frac{7A - 5A}{2}$ $\Big)$
$= 2 \cos 2A \cos A + 2 \cos 6 A \cos A$
$= 2 \cos A [ \cos 6A + \cos 2A ]$
$= 2 \cos A \Big[ 2 \cos \Big($ $\frac{6A + 2A}{2}$ $\Big) \cos \Big($ $\frac{6A - 2A}{2}$ $\Big) \Big]$
$= 2 \cos A [ 2 \cos 4A \cos 2A ]$
$= 4 \cos A \cos 2A \cos 4A =$ RHS. Hence proved.
iii) LHS $= \sin A + \sin 2A + \sin 4A + \sin 5A$
$= [ \sin 2A + \sin A] + [ \sin 5A + \sin 4A ]$
$= 2 \sin \Big($ $\frac{2A + A}{2}$ $\Big) \cos \Big($ $\frac{2A - A}{2}$ $\Big) + 2 \sin \Big($ $\frac{5A + 4A}{2}$ $\Big) \cos \Big($ $\frac{5A - 4A}{2}$ $\Big)$
$= 2 \sin$ $\frac{3A}{2}$ $\cos$ $\frac{A}{2}$ $+ 2 \sin$ $\frac{9A}{2}$ $\cos$ $\frac{A}{2}$
$= 2 \cos$ $\frac{A}{2}$ $\Big[ \sin$ $\frac{3A}{2}$ $+ \sin$ $\frac{9A}{2}$ $\Big]$
$= 2 \cos$ $\frac{A}{2}$ $\Big[ 2 \sin \Big($ $\frac{\frac{9A}{2} + \frac{3A}{2}}{2}$ $\Big) \cos \Big($ $\frac{\frac{9A}{2} - \frac{3A}{2}}{2}$ $\Big) \Big]$
$= 2 \cos$ $\frac{A}{2}$ $[ 2 \sin 3A \sin$ $\frac{3A}{2}$ $]$
$= 4 \cos$ $\frac{A}{2}$ $\sin$ $\frac{3A}{2}$ $\sin 3A =$ RHS. Hence proved.
iv) LHS $= \sin 3A + \sin 2A - \sin A$
$= [ \sin 3A - \sin A ] + \sin 2A$
$= 2 \sin \Big($ $\frac{3A- A}{2}$ $\Big) \cos \Big($ $\frac{3A+ A}{2}$ $\Big) + \sin 2A$
$= 2 \sin A \cos 2A + \sin 2A$
$= 2 \sin A \cos 2A + 2\sin A \cos A$
$= 2 \sin A [ \cos 2A + \cos A ]$
$= 2 \sin A \Big[ 2 \cos \Big($ $\frac{2A+ A}{2}$ $\Big) \cos \Big($ $\frac{2A- A}{2}$ $\Big) \Big]$
$= 4 \sin A \cos$ $\frac{3A}{2}$ $\cos$ $\frac{A}{2}$ $=$ RHS. Hence proved.
v) LHS $= \cos 20^o \cos 100^o + \cos 100^o \cos 140^o - \cos 140^o \cos 200^o$
$=$ $\frac{1}{2}$ $\Big[ 2\cos 20^o \cos 100^o + 2\cos 100^o \cos 140^o - 2\cos 140^o \cos 200^o \Big]$
$=$ $\frac{1}{2}$ $\Big[ \cos (100^o+20^o)+ \cos (100^o-20^o) + \cos (140^o+100^o)+ \\ \cos (140^o-100^o) - \cos (200^o+140^o) -\cos (200^o-140^o) \Big]$
$=$ $\frac{1}{2}$ $\Big[ \cos 120^o+ \cos 80^o + \cos 240^o+ \cos 40^o - \cos 340^o -\cos 60^o \Big]$
$=$ $\frac{1}{2}$ $\Big[ \cos (90^o+30^o)+ \cos 80^o + \cos (180^o+60^o)+ \cos 40^o - \cos (360^o-20^o) -\cos 60^o \Big]$
$=$ $\frac{1}{2}$ $\Big[ -\sin 30^o + [ \cos 80^o + \cos 40^o ] - \cos 60^o - \cos 20^o -\cos 60^o \Big]$
$=$ $\frac{1}{2}$ $\Big[ -\sin 30^o + 2 \cos 60^o \cos 20^o - \cos 60^o - \cos 20^o -\cos 60^o \Big]$
$=$ $\frac{1}{2}$ $\Big[ -$ $\frac{1}{2}$ $+ 2 \times$ $\frac{1}{2}$ $\cos 20^o -$ $\frac{1}{2}$ $- \cos 20^o -$ $\frac{1}{2}$ $\Big]$
$=$ $\frac{1}{2}$ $\Big[ -$ $\frac{3}{2}$ $\Big] = -$ $\frac{3}{4}$ $=$ RHS. Hence proved.
vi) LHS $= \sin$ $\frac{x}{2}$ $\sin$ $\frac{7x}{2}$ $+ \sin$ $\frac{3x}{2}$ $\sin$ $\frac{11x}{2}$
$=$ $\frac{1}{2}$ $\Big[ 2\sin$ $\frac{x}{2}$ $\sin$ $\frac{7x}{2}$ $+ 2\sin$ $\frac{3x}{2}$ $\sin$ $\frac{11x}{2}$ $\Big]$
$=$ $\frac{1}{2}$ $\Big[ \cos \Big($ $\frac{7x}{2}$ $-$ $\frac{x}{2}$ $\Big) - \cos \Big($ $\frac{7x}{2}$ $+$ $\frac{x}{2}$ $\Big) +\cos \Big($ $\frac{11x}{2}$ $-$ $\frac{3x}{2}$ $\Big) - \cos \Big($ $\frac{11x}{2}$ $+$ $\frac{3x}{2}$ $\Big) \Big]$
$=$ $\frac{1}{2}$ $\Big[ \cos 3x - \cos 4x + \cos 4x - \cos 7x \Big]$
$=$ $\frac{1}{2}$ $\Big[ \cos 3x - \cos 7x \Big]$
$= -$ $\frac{1}{2}$ $\Big[ \cos 7x - \cos 3x \Big]$
$= -$ $\frac{1}{2}$ $\Big[ -2 \sin \Big($ $\frac{7x+3x}{2}$ $\Big) \sin \Big($ $\frac{7x-3x}{2}$ $\Big) \Big]$
$= \sin 5x \sin 2x =$ RHS. Hence proved.
vii) LHS $= \cos x \cos$ $\frac{x}{2}$ $- \cos 3x \cos$ $\frac{9x}{2}$
$=$ $\frac{1}{2}$ $\Big[ 2\cos x \cos$ $\frac{x}{2}$ $- 2\cos 3x \cos$ $\frac{9x}{2}$ $\Big]$
$=$ $\frac{1}{2}$ $\Big[ \cos \Big( x +$ $\frac{x}{2}$ $\Big) + \cos \Big( x -$ $\frac{x}{2}$ $\Big) - \cos \Big($ $\frac{9x}{2}$ $+ 3x \Big) - \cos \Big($ $\frac{9x}{2}$ $- 3x \Big) \Big]$
$=$ $\frac{1}{2}$ $\Big[ \cos$ $\frac{3x}{2}$ $+ \cos$ $\frac{x}{2}$ $- \cos$ $\frac{15x}{2}$ $- \cos$ $\frac{3x}{2}$ $\Big]$
$=$ $\frac{1}{2}$ $\Big[ \cos$ $\frac{x}{2}$ $- \cos$ $\frac{15x}{2}$ $\Big]$
$=$ $\frac{1}{2}$ $\Big[ - 2 \sin \Big($ $\frac{\frac{x}{2}+\frac{15x}{2} }{2}$ $\Big) \sin \Big($ $\frac{\frac{x}{2}-\frac{15x}{2} }{2}$ $\Big) \Big]$
$= \sin 4x \sin$ $\frac{7x}{2}$ $=$ RHS. Hence proved.
$\\$
Question 7: Prove that:
i) $\frac{\sin A + \sin 3A}{\cos A - \cos 3A}$ $= \cot A$ ii) $\frac{\sin 9A - \sin 7A}{\cos 7A - \cos 9A}$ $= \cot 8A$
iii) $\frac{\sin A - \sin B}{\cos A + \cos B}$ $= \tan \Big($ $\frac{A-B}{2}$ $\Big)$ iv) $\frac{\sin A + \sin B}{\sin A - \sin B}$ $= \tan \Big($ $\frac{A+B}{2}$ $\Big) \cot \Big($ $\frac{A-B}{2}$ $\Big)$
v) $\frac{\cos A + \cos B}{\cos B - \cos A}$ $= \cot \Big($ $\frac{A+B}{2}$ $\Big) \cot \Big($ $\frac{A-B}{2}$ $\Big)$
i) LHS $= \frac{\sin A + \sin 3A}{\cos A - \cos 3A}$
$=$ $\frac{2 \sin \Big( \frac{A+3A}{2} \Big) \cos \Big( \frac{A-3A}{2} \Big) }{-2 \sin \Big( \frac{A+3A}{2} \Big) \sin \Big( \frac{A-3A}{2} \Big)}$
$=$ $\frac{-2 \sin 2A \cos (-A) }{2 \sin 2A \sin (-A) }$ $=$ $\frac{- \cos ( -A)}{\sin ( -A)}$ $=$ $\frac{\cos A}{\sin A}$ $= \cot A =$ RHS. Hence proved.
ii) LHS $=$ $\frac{\sin 9A - \sin 7A}{\cos 7A - \cos 9A}$
$=$ $\frac{2 \sin \Big( \frac{9A-7A}{2} \Big) \cos \Big( \frac{9A+7A}{2} \Big) }{-2 \sin \Big( \frac{7A+9A}{2} \Big) \sin \Big( \frac{7A-9A}{2} \Big)}$
$=$ $\frac{-2 \sin A \cos 8A }{2 \sin 8A \sin (-A) }$ $=$ $\frac{- \sin A \cos 8A}{- \sin A \sin 8A }$ $=$ $\frac{\cos 8A}{\sin 8A}$ $= \cot 8A =$ RHS. Hence proved.
iii) LHS $=$ $\frac{\sin A - \sin B}{\cos A + \cos B}$
$=$ $\frac{2 \cos \Big( \frac{A+B}{2} \Big) \sin \Big( \frac{A-B}{2} \Big) }{2 \cos \Big( \frac{A+B}{2} \Big) \cos \Big( \frac{A-B}{2} \Big)}$
$=$ $\frac{ \sin \Big( \frac{A-B}{2} \Big) }{ \cos \Big( \frac{A-B}{2} \Big)}$ $= \tan \Big( \frac{A-B}{2} \Big) =$ RHS. Hence proved.
iv) LHS $=$ $\frac{\sin A + \sin B}{\sin A - \sin B}$
$=$ $\frac{2 \sin \Big( \frac{A+B}{2} \Big) \cos \Big( \frac{A-B}{2} \Big) }{2 \sin \Big( \frac{A-B}{2} \Big) \cos \Big( \frac{A+B}{2} \Big)}$
$= \tan \Big( \frac{A+B}{2} \Big) \cot \Big( \frac{A-B}{2} \Big) =$ RHS. Hence proved.
v) LHS $=$ $\frac{\cos A + \cos B}{\cos B - \cos A}$
$=$ $\frac{2 \cos \Big( \frac{A+B}{2} \Big) \cos \Big( \frac{A-B}{2} \Big) }{-2 \sin \Big( \frac{B+A}{2} \Big) \sin \Big( \frac{B-A}{2} \Big)}$
$=$ $\frac{2 \cos \Big( \frac{A+B}{2} \Big) \cos \Big( \frac{A-B}{2} \Big) }{2 \sin \Big( \frac{A+B}{2} \Big) \sin \Big( \frac{A-B}{2} \Big)}$
$= \cot \Big( \frac{A+B}{2} \Big) \cot \Big( \frac{A-B}{2} \Big) =$ RHS. Hence proved.
$\\$
Question 8: Prove that:
i) $\frac{\sin A + \sin 3A + \sin 5A}{\cos A + \cos 3A + \cos 5A}$ $= \tan 3A$
ii) $\frac{\cos 3A + 2\cos 5A + \cos 7A}{\cos A + 2\cos 3A + \cos 5A}$ $=$ $\frac{\cos 5A}{\cos 3A}$
iii) $\frac{\cos 4A + \cos 3A + \cos 2A}{\sin 4A + \sin 3A + \sin 2A}$ $= \cot 3A$
iv) $\frac{\sin 3A + \sin 5A + \sin 7A + \sin 9A}{\cos 3A + \cos 5A + \cos 7A + \cos 9A}$ $= \tan 6A$
v) $\frac{\sin 5A - \sin 7A + \sin 8A - \sin 4A}{\cos 4A + \cos 7A - \cos 5A - \cos 8A}$ $= \cot 6A$
vi) $\frac{\sin 5A \cos 2A - \sin 6A \cos A}{\sin A \sin 2A - \cos 2A \cos 3A}$ $= \tan A$
vii) $\frac{\sin 11A \sin A + \sin 7A \sin 3A}{\cos 11A \sin A + \cos 7A \sin 3A}$ $= \tan 8A$
viii) $\frac{\sin 3A \cos 4A - \sin A \cos 2A}{\sin 4A \sin A + \cos 6A \cos A}$ $= \tan 2A$
ix) $\frac{\sin A \sin 2A + \sin 3A \sin 6A}{\sin A \cos 2A + \sin 3A \cos 6A}$ $= \tan 5A$
x) $\frac{\sin A + 2\sin 3A + \sin 5A}{\sin 3A + 2\sin 5A + \sin 7A}$ $=$ $\frac{\sin 3A}{\sin 5A}$
xi) $\frac{\sin (\theta + \phi) -2 \sin \theta + \sin (\theta - \phi)}{\cos (\theta + \phi) -2 \cos \theta + \cos (\theta - \phi)}$ $= \tan \theta$
i) LHS $=$ $\frac{\sin A + \sin 3A + \sin 5A}{\cos A + \cos 3A + \cos 5A}$
$=$ $\frac{(\sin 5A + \sin A) + \sin 3A}{(\cos 5A + \cos A) + \cos 3A}$
$=$ $\frac{2 \sin \Big( \frac{5A+A}{2} \Big) \cos \Big( \frac{5A-A}{2} \Big) + \sin 3A } {2 \cos \Big( \frac{5A+A}{2} \Big) \cos \Big( \frac{5A-A}{2} \Big) + \cos 3A }$
$=$ $\frac{ 2 \sin 3A \cos 2A + \sin 3A}{ 2 \cos 3A \cos 2A+ \cos 3A}$
$=$ $\frac{ \sin 3A ( 2 \cos 2A + 1)}{\cos 3A ( 2 \cos 2A + 1)}$
$= \tan 3A =$ RHS. Hence proved.
ii) LHS $=$ $\frac{\cos 3A + 2\cos 5A + \cos 7A}{\cos A + 2\cos 3A + \cos 5A}$
$=$ $\frac{(\cos 3A + \cos 7A) + 2\cos 5A}{(\cos A + \cos 5A) + 2\cos 3A}$
$=$ $\frac{2 \cos \Big( \frac{7A+3A}{2} \Big) \cos \Big( \frac{7A-3A}{2} \Big) + 2\cos 5A } {2 \cos \Big( \frac{5A+A}{2} \Big) \cos \Big( \frac{5A-A}{2} \Big) + 2\cos 3A }$
$=$ $\frac{ 2 \cos 5A \cos 2A + 2\cos 5A}{ 2 \cos 3A \cos 2A+ 2\cos 3A}$
$=$ $\frac{ 2\cos 5A ( \cos 2A + 1)}{2\cos 3A ( \cos 2A + 1)}$
$=$ $\frac{\cos 5A}{\cos 3A}$ $=$ RHS. Hence proved.
iii) LHS $=$ $\frac{\cos 4A + \cos 3A + \cos 2A}{\sin 4A + \sin 3A + \sin 5A}$
$=$ $\frac{(\cos 4A + \cos 2A) + \cos 3A}{(\sin 4A + \sin 2A) + \sin 3A}$
$=$ $\frac{2 \cos \Big( \frac{4A+2A}{2} \Big) \cos \Big( \frac{4A-2A}{2} \Big) + \cos 3A } {2 \sin \Big( \frac{4A+2A}{2} \Big) \cos \Big( \frac{4A-2A}{2} \Big) + \sin 3A }$
$=$ $\frac{ 2 \cos 3A \cos A + \cos 3A}{ 2 \sin 3A \cos A+ \sin 3A}$
$=$ $\frac{ \cos 3A ( 2\cos A + 1)}{\sin 3A ( 2\cos A + 1)}$
$= \frac{\cos 3A}{\cos 3A}$ $= \cot 3A$ RHS. Hence proved.
iv) LHS $=$ $\frac{\sin 3A + \sin 5A + \sin 7A + \sin 9A}{\cos 3A + \cos 5A + \cos 7A + \cos 9A}$
$=$ $\frac{ (\sin 9A + \sin 3A) + (\sin 7A + \sin 5A) }{ (\cos 9A + \cos 3A) + (\cos 7A + \cos 5A) }$
$=$ $\frac{ 2 \sin \Big( \frac{9A+3A}{2} \Big) \cos \Big( \frac{9A-3A}{2} \Big) + 2 \sin \Big( \frac{7A+5A}{2} \Big) \cos \Big( \frac{7A-5A}{2} \Big) }{ 2 \cos \Big( \frac{9A+3A}{2} \Big) \cos \Big( \frac{9A-3A}{2} \Big) + 2 \cos \Big( \frac{7A+5A}{2} \Big) \cos \Big( \frac{7A-5A}{2} \Big) }$
$=$ $\frac{2 \sin 6A \cos 3A + 2 \sin 6A \cos 2A}{2 \cos 6A \cos 3A + 2 \cos 6A \cos 2A}$
$=$ $\frac{2 \sin 6A (\cos 3A + \cos 2A) }{2 \cos 6A (\cos 3A + \cos 2A)}$
$= \tan 6A =$ RHS. Hence proved.
v) LHS $=$ $\frac{\sin 5A - \sin 7A + \sin 8A - \sin 4A}{ \cos 4A + \cos 7A - \cos 5A - \cos 8A}$
$=$ $\frac{ - (\sin 7A - \sin 5A) + (\sin 8A - \sin 4A) }{ - (\cos 7A - \cos 5A) - (\cos 8A - \cos 4A) }$
$=$ $\frac{ -2 \sin \Big( \frac{7A-5A}{2} \Big) \cos \Big( \frac{7A+5A}{2} \Big) + 2 \sin \Big( \frac{8A-4A}{2} \Big) \cos \Big( \frac{8A+4A}{2} \Big) }{- 2 \sin \Big( \frac{7A+5A}{2} \Big) \sin \Big( \frac{7A-5A}{2} \Big) +2 \sin \Big( \frac{8A+4A}{2} \Big) \sin \Big( \frac{8A-4A}{2} \Big) }$
$=$ $\frac{-2 \sin A \cos 6A + 2 \sin 2A \cos 6A}{-2 \sin 6A \sin A + 2 \sin 6A \sin 2A}$
$=$ $\frac{2 \cos 6A (-\sin A + \sin 2A) }{2 \sin 6A (-\sin A + \sin 2A) }$
$= \cot 6A =$ RHS. Hence proved.
vi) LHS $=$ $\frac{\sin 5A \cos 2A - \sin 6A \cos A}{\sin A \sin 2A - \cos 2A \cos 3A}$
$=$ $\frac{2\sin 5A \cos 2A - 2\sin 6A \cos A}{2\sin A \sin 2A - 2\cos 2A \cos 3A}$
$=$ $\frac{ \sin ( 5A +2A) + \sin (5A - 2A) - [ \sin ( 6A +A) + \sin (6A - A) ] }{ \cos ( 2A - A) - \cos ( 2A + A) - [ \cos ( 3A + 2A) + \cos ( 3A - 2 A) ] }$
$=$ $\frac{ \sin 7A + \sin 3A - [ \sin 7A + \sin 5A ] }{ \cos A - \cos 3A - [ \cos 5A + \cos A ] }$
$=$ $\frac{ \sin 3A - \sin 5A }{ - \cos 3A - \cos 5A }$
$=$ $\frac{ -( \sin 5A - \sin 3A) }{ - (\cos 5A + \cos 3A) }$
$=$ $\frac{2 \sin \Big( \frac{5A-3A}{2} \Big) \cos \Big( \frac{5A+3A}{2} \Big)}{2 \cos \Big( \frac{5A+3A}{2} \Big) \cos \Big( \frac{5A-3A}{2} \Big)}$
$=$ $\frac{\sin A \cos 4A}{\cos 4A \cos A}$
$=$ $\frac{\sin A }{\cos A}$ $= \tan A =$ RHS
vii) LHS $=$ $\frac{\sin 11A \sin A + \sin 7A \sin 3A}{\cos 11A \sin A + \cos 7A \sin 3A}$
$=$ $\frac{2\sin 11A \sin A + 2\sin 7A \sin 3A}{2\cos 11A \sin A + 2\cos 7A \sin 3A}$
$=$ $\frac{ \cos ( 11A -A) - \cos (11A + A) + \cos ( 7A -3A) - \cos (7A + 3A) }{ \sin ( 11A + A) - \sin ( 11A - A) + \sin ( 7A + 3A) - \sin ( 7A - 3 A) }$
$=$ $\frac{ \cos 10A - \cos 12A + \cos 4A - \cos 10A }{ \sin 12A - \sin 10A + \sin 10A - \sin 4A }$
$=$ $\frac{ - ( \cos 12A - \cos 4A) }{ \sin 12A - \sin 4A }$
$=$ $\frac{ - \Big[ -2 \sin \Big( \frac{12A+4A}{2} \Big) \cos \Big( \frac{12A-4A}{2} \Big) \Big] }{2 \sin \Big( \frac{12A-4A}{2} \Big) \cos \Big( \frac{12A+4A}{2} \Big)}$
$=$ $\frac{2\sin 8A \sin 4A}{2\sin 4A \cos 8A}$
$=$ $\frac{\sin 8A }{\cos 8A}$ $= \tan 8A =$ RHS
viii) LHS $=$ $\frac{\sin 3A \cos 4A - \sin A \cos 2A}{\sin 4A \sin A + \cos 6A \cos A}$
$=$ $\frac{2\sin 3A \cos 4A - 2\sin A \cos 2A}{2\sin 4A \sin A + 2\cos 6A \cos A}$
$=$ $\frac{ \sin ( 4A +3A) - \sin (4A -3A) - [ \sin ( 2A+A) - \sin (2A -A) ] }{ \cos ( 4A - A) - \cos ( 4A + A) + \cos ( 6A + A) + \cos ( 6A - A) }$
$=$ $\frac{ \sin 7A - \sin A - \sin 3A + \sin A }{ \cos 3A - \cos 5A + \cos 7A + \cos 5A }$
$=$ $\frac{ \sin 7A - \sin 3A }{ \cos 3A + \cos 7A }$
$=$ $\frac{2 \sin \Big( \frac{7A-3A}{2} \Big) \cos \Big( \frac{7A+3A}{2} \Big)}{2 \cos \Big( \frac{7A+3A}{2} \Big) \cos \Big( \frac{7A-3A}{2} \Big)}$
$=$ $\frac{2\sin 2A \cos 5A}{2\cos 5A \cos 2A}$
$=$ $\frac{\sin 2A }{\cos 2A}$ $= \tan 2A =$ RHS
ix) LHS $=$ $\frac{\sin A \sin 2A + \sin 3A \sin 6A}{\sin A \cos 2A + \sin 3A \cos 6A}$
$=$ $\frac{2\sin A \sin 2A + 2\sin 3A \sin 6A}{2\sin A \cos 2A + 2\sin 3A \cos 6A}$
$=$ $\frac{ \cos ( 2A -A) - \cos (2A +A) + \cos ( 6A-3A) - \cos (6A+3A) }{ \sin ( 2A + A) - \sin ( 2A - A) + \sin ( 6A + 3A) - \sin ( 6A - 3A) }$
$=$ $\frac{ \cos A - \cos 3A + \cos 3A - \cos 9A }{ \sin 3A - \sin A + \sin 9A - \sin 3A }$
$=$ $\frac{ \cos A - \cos 9A }{ - \sin A + \sin 9A }$
$=$ $\frac{ - (\cos 9A - \cos A) }{ ( \sin 9A - \sin A) }$
$=$ $\frac{- \Big[ -2 \sin \Big( \frac{9A+A}{2} \Big) \sin \Big( \frac{9A-A}{2} \Big) \Big] }{2 \sin \Big( \frac{9A-A}{2} \Big) \cos \Big( \frac{9A+A}{2} \Big)}$
$=$ $\frac{2\sin 5A \sin 4A}{2\sin 4A \cos 5A}$
$=$ $\frac{\sin 5A }{\cos 5A}$ $= \tan 5A =$ RHS
x) LHS $=$ $\frac{\sin A + 2\sin 3A + \sin 5A}{\sin 3A + 2\sin 5A + \sin 7A}$
$=$ $\frac{ (\sin A + \sin 5A )+ 2\sin 3A }{ (\sin 3A + \sin 7A) + 2\sin 5A }$
$=$ $\frac{2 \sin \Big( \frac{5A+A}{2} \Big) \cos \Big( \frac{5A-A}{2} \Big) + 2\sin 3A } {2 \sin \Big( \frac{7A+3A}{2} \Big) \cos \Big( \frac{7A-3A}{2} \Big) + 2\sin 5A }$
$=$ $\frac{ 2 \sin 3A \cos 2A + 2 \sin 3A}{ 2 \sin 5A \cos 2A+ 2\sin 5A}$
$=$ $\frac{ \sin 3A ( 1+ 2\cos A )}{\sin 5A ( 1+ 2\cos A )}$
$= \frac{\sin 3A}{\sin 5A}$ $= \cot 3A$ RHS. Hence proved.
xi) LHS $=$ $\frac{\sin (\theta + \phi) -2 \sin \theta + \sin (\theta - \phi)}{\cos (\theta + \phi) -2 \cos \theta + \cos (\theta - \phi)}$ $= \tan \theta$
$=$ $\frac{ [ \sin (\theta + \phi) + \sin (\theta - \phi) ] -2 \sin \theta}{ [ \cos (\theta + \phi) + \cos (\theta - \phi) ] -2 \cos \theta }$
$=$ $\frac{ 2 \sin \Big( \frac{\theta + \phi+\theta - \phi}{2} \Big) \cos \Big( \frac{\theta + \phi-\theta + \phi}{2} \Big) -2 \sin \theta}{ 2 \cos \Big( \frac{\theta + \phi+\theta - \phi}{2} \Big) \cos \Big( \frac{\theta + \phi-\theta + \phi}{2} \Big) -2 \cos \theta }$
$=$ $\frac{2\sin \theta \cos \phi - 2 \sin \theta}{2\cos \theta \cos \phi - 2 \cos \theta}$
$=$ $\frac{2\sin \theta ( \cos \phi - 1) }{2\cos \theta ( \cos \phi - 1)}$
$=$ $\frac{\sin \theta }{\cos \theta}$
$= \tan \theta =$ RHS. Hence proved.
$\\$
Question 9: Prove that:
i) $\sin \alpha + \sin \beta + \sin \gamma - \sin (\alpha + \beta + \gamma) = 4 \sin \Big($ $\frac{\alpha + \beta}{2}$ $\Big) \sin \Big($ $\frac{\beta + \gamma}{2}$ $\Big) \sin \Big($ $\frac{\gamma+ \alpha}{2}$ $\Big)$
ii) $\cos ( A+B+C) + \cos (A-B+C) + \cos ( A + B - C) + \cos (-A+B+C) = 4 \cos A \cos B \cos C$
i) LHS $= \sin \alpha + \sin \beta + \sin \gamma - \sin (\alpha + \beta + \gamma)$
$= [ \sin \alpha + \sin \beta ] + [ \sin \gamma - \sin (\alpha + \beta + \gamma) ]$
$= 2 \sin \Big($ $\frac{\alpha + \beta}{2}$ $\Big) \cos \Big($ $\frac{\alpha - \beta}{2}$ $\Big) + 2 \sin \Big($ $\frac{\gamma - (\alpha + \beta + \gamma)}{2}$ $\Big) \cos \Big($ $\frac{\gamma + (\alpha + \beta + \gamma)}{2}$ $\Big)$
$= 2 \sin \Big($ $\frac{\alpha + \beta}{2}$ $\Big) \cos \Big($ $\frac{\alpha - \beta}{2}$ $\Big) + 2 \sin \Big($ $\frac{ - (\alpha + \beta )}{2}$ $\Big) \cos \Big($ $\frac{\alpha + \beta + 2\gamma}{2}$ $\Big)$
$= 2 \sin \Big($ $\frac{\alpha + \beta}{2}$ $\Big) \Big[ \cos \Big($ $\frac{\alpha - \beta}{2}$ $\Big) - \cos \Big($ $\frac{\alpha + \beta + 2\gamma)}{2}$ $\Big) \Big]$
$= 2 \sin \Big($ $\frac{\alpha + \beta}{2}$ $\Big) \Big[ - 2 \sin \Big($ $\frac{\frac{\alpha - \beta}{2} +\frac{\alpha + \beta + 2\gamma}{2}}{2}$ $\Big) \sin \Big($ $\frac{\frac{\alpha - \beta}{2} - \frac{\alpha + \beta + 2\gamma}{2}}{2}$ $\Big) \Big]$
$= 2 \sin \Big($ $\frac{\alpha + \beta}{2}$ $\Big)\Big[ -2 \sin \Big($ $\frac{\alpha + \gamma}{2}$ $\Big) \sin \Big($ $\frac{-(\beta + \gamma)}{2}$ $\Big) \Big]$
$= 4 \sin \Big($ $\frac{\alpha + \beta}{2}$ $\Big)\Big[ \sin \Big($ $\frac{\alpha + \gamma}{2}$ $\Big) \sin \Big($ $\frac{\beta + \gamma}{2}$ $\Big) \Big]$
$= 4 \sin \Big($ $\frac{\alpha + \beta}{2}$ $\Big)\Big[ \sin \Big($ $\frac{\beta + \gamma}{2}$ $\Big) \sin \Big($ $\frac{\alpha + \gamma}{2}$ $\Big) \Big] =$ RHS. Hence proved.
ii) LHS $= \cos ( A+B+C) + \cos (A-B+C) + \cos ( A + B - C) + \cos (-A+B+C)$
$= [ \cos ( A+B+C) + \cos (A-B+C) ] + [ \cos ( A + B - C) + \cos (-A+B+C) ]$
$= 2 \cos \Big( \frac{A+B+C + A - B + C}{2} \cos \frac{A + B + C - A + B - C}{2} \Big) + 2 \cos \Big( \frac{A+B-C - A + B + C}{2} \cos \frac{A + B - C + A - B - C}{2} \Big)$
$= 2 \cos ( A + C) \cos B + 2 \cos B \cos ( A -C)$
$= 2 \cos B [ \cos ( A + C) + \cos ( A - C) ]$
$= 2 \cos B \Big[ 2 \cos \Big($ $\frac{A + C + A - C}{2}$ $\Big) \cos \Big($ $\frac{A + C - A +C}{2}$ $\Big) \Big]$
$= 4 \cos B \cos A \cos C$
$= 4 \cos A \cos B \cos C =$ RHS. Hence proved.
$\\$
Question 10: If $\cos A + \cos B =$ $\frac{1}{2}$ and $\sin A + \sin B =$ $\frac{1}{4}$, prove that:
$\tan \Big($ $\frac{A+B}{2}$ $\Big) =$ $\frac{1}{2}$
Given: $\cos A + \cos B =$ $\frac{1}{2}$ and $\sin A + \sin B =$ $\frac{1}{4}$
Dividing one by another
$\Rightarrow$ $\frac{\sin A + \sin B}{\cos A + \cos B}$ $=$ $\frac{\frac{1}{4}}{\frac{1}{2}}$
$\Rightarrow$ $\frac{2 \sin \Big( \frac{A+B}{2} \Big) \cos \Big( \frac{A-B}{2} \Big) }{2 \cos \Big( \frac{A+B}{2} \Big) \cos \Big( \frac{A-B}{2} \Big)}$ $=$ $\frac{1}{2}$
$\Rightarrow$ $\frac{ \sin \Big( \frac{A+B}{2} \Big) }{ \cos \Big( \frac{A+B}{2} \Big) }$ $=$ $\frac{1}{2}$
$\Rightarrow \tan \Big($ $\frac{A+B}{2}$ $\Big) =$ $\frac{1}{2}$
$\\$
Question 11: If $\mathrm{cosec} A + \sec A = \mathrm{cosec} B + \sec B$, then prove that $\tan A \tan B = \cot \Big($ $\frac{A+B}{2}$ $\Big)$
Given: $\mathrm{cosec} A + \sec A = \mathrm{cosec} B + \sec B$
$\Rightarrow \sec B - \sec A = \mathrm{cosec} B - \mathrm{cosec} A$
$\Rightarrow$ $\frac{1}{\cos A}$ $-$ $\frac{1}{\cos B}$ $=$ $\frac{1}{\sin B}$ $- \frac{1}{\sin A}$
$\Rightarrow$ $\frac{\cos B - \cos A}{\cos A \cos B}$ $=$ $\frac{\sin A - \sin B}{\sin A \sin B}$
$\Rightarrow$ $\frac{\sin A \sin B}{\cos A \cos B}$ $=$ $\frac{\sin A - \sin B}{\cos B - \cos A}$
$\Rightarrow \tan A \tan B =$ $\frac{2 \sin \frac{A-B}{2} \cos \frac{A+B}{2}}{2 \sin \frac{B-A}{2} \sin \frac{B+A}{2}}$
$\Rightarrow \tan A \tan B = \cot$ $\frac{A+B}{2}$
$\\$
Question 12: If $\sin 2A = \lambda \sin 2B$, prove that $\frac{\tan (A+B)}{\tan (A-B)}$ $=$ $\frac{\lambda+1}{\lambda - 1}$
Given $\sin 2A = \lambda \sin 2B$
$\Rightarrow \lambda =$ $\frac{\sin 2A}{\sin 2B}$
Applying componendo and dividendo
$\frac{\lambda+1}{\lambda-1}$ $=$ $\frac{\sin 2A + \sin 2B}{\sin 2A - \sin 2B}$
$\Rightarrow$ $\frac{\lambda+1}{\lambda-1}$ $=$ $\frac{2 \sin ( A+B) \cos ( A - B) }{2 \sin ( A-B) \cos ( A + B) }$
$\Rightarrow$ $\frac{\lambda+1}{\lambda-1}$ $=$ $\frac{\tan (A+B) }{\tan (A-B) }$
Hence proved.
$\\$
Question 13: Prove that:
i) $\frac{\cos (A+B+C) + \cos (-A + B + C) +\cos (A-B+C) + \cos (A+B-C) }{\sin (A+B+C) + \sin (-A + B + C) +\sin (A-B+C) + \sin (A+B-C)}$ $= \cot C$
ii) $\sin (B-C) \cos (A-D) + \sin (C-A) \cos (B-D) + \sin (A-B) \cos (C-D) = 0$
i) LHS $= \frac{\cos (A+B+C) + \cos (-A + B + C) +\cos (A-B+C) + \cos (A+B-C) }{\sin (A+B+C) + \sin (-A + B + C) +\sin (A-B+C) + \sin (A+B-C)}$
$=$ $\frac{2 \cos \Big( \frac{A+B+C-A+B+C}{2} \Big) \cos \Big( \frac{A+B+C+A-B-C}{2} \Big) + 2 \cos \Big( \frac{A-B+C+A+B-C}{2} \Big) \cos \Big( \frac{A-B+C-A-B+C}{2} \Big) }{2 \sin \Big( \frac{A+B+C-A+B+C}{2} \Big) \cos \Big( \frac{A+B+C+A-B-C}{2} \Big) + 2 \sin \Big( \frac{A-B+C+A+B-C}{2} \Big) \cos \Big( \frac{A-B+C-A-B+C}{2} \Big) }$
$=$ $\frac{2 \cos (B+C) \cos A + 2 \cos A \cos (C-B) }{2 \sin (B+C) \cos A + 2 \sin (C-B) \cos A}$
$=$ $\frac{2 \cos A [ \cos ( B+C) + \cos (C-B) ]}{2 \cos A [ \sin ( B + C) + \sin (C-B) ]}$
$=$ $\frac{ \cos ( B+C) + \cos (C-B) }{ \sin ( B + C) + \sin (C-B) }$
$=$ $\frac{ 2 \cos \Big( \frac{B+C+C-B}{2} \Big) \cos \Big( \frac{B+C-C+B}{2} \Big) }{ 2 \sin \Big( \frac{B+C+C-B}{2} \Big) \cos \Big( \frac{B+C-C+B}{2} \Big) }$
$=$ $\frac{2 \cos C \cos B}{ 2 \sin C \cos B}$
$=$ $\frac{\cos C}{\sin C}$ $= \cot C$
ii) LHS $= \sin (B-C) \cos (A-D) + \sin (C-A) \cos (B-D) + \sin (A-B) \cos (C-D)$
$=$ $\frac{1}{2}$ $[ 2\sin (B-C) \cos (A-D) + 2\sin (C-A) \cos (B-D) + 2\sin (A-B) \cos (C-D) ]$
$=$ $\frac{1}{2}$ $[ \sin ( B-C+A-D ) + \sin ( B - C - A + D) + \sin ( C - A + B - D) + \\ \sin ( C - A - B + D) + \sin (A - B + C - D) + \sin ( A - B - C + D) ]$
$=$ $\frac{1}{2}$ $[ \sin ( A + B - C - D ) + \sin ( B+D - A - C) + \sin ( B+C - A - D) \\ - \sin ( A + B - C - D ) - \sin (B + D - A - C) - \sin ( B+C-A-D) ]$
$=$ $\frac{1}{2}$ $[ 0 ] =$ RHS. Hence proved.
$\\$
Question 14: If $\frac{\cos (A-B) }{\cos (A+B)}$ $+$ $\frac{\cos (C+D)}{\cos (C-D)}$ $= 0$, prove that $\tan A \tan B \tan C \tan D = -1$
Given $\frac{\cos (A-B) }{\cos (A+B)}$ $+$ $\frac{\cos (C+D)}{\cos (C-D)}$ $= 0$
$\Rightarrow$ $\frac{\cos ( A - B) \cos ( C - D) + \cos ( C+D) \cos (A+B) }{\cos ( A+ B) \cos ( C+D)}$ $= 0$
$\Rightarrow \cos ( A - B) \cos ( C - D) + \cos ( C+D) \cos (A+B) = 0$
$\Rightarrow \cos ( A - B) \cos ( C - D) = - \cos ( C+D) \cos (A+B)$
$\Rightarrow (\cos A \cos B + \sin A \sin B) (\cos C \cos D + \sin C \sin D) = - ( \cos C \cos D - \sin C \sin D)(\cos A \cos B - \sin A \sin B)$
Dividing both sides by $\cos A \cos B \cos C \cos D$
$\Rightarrow \frac{ (\cos A \cos B + \sin A \sin B) (\cos C \cos D + \sin C \sin D)}{\cos A \cos B \cos C \cos D} = - \frac{ ( \cos C \cos D - \sin C \sin D)(\cos A \cos B - \sin A \sin B)}{\cos A \cos B \cos C \cos D}$
$\Rightarrow \frac{ (\cos A \cos B + \sin A \sin B)}{\cos A \cos B } \frac{ (\cos C \cos D + \sin C \sin D)}{ \cos C \cos D} = - \frac{ ( \cos C \cos D - \sin C \sin D)}{\cos A \cos B } \frac{ (\cos A \cos B - \sin A \sin B)}{ \cos C \cos D}$
$\Rightarrow ( 1 + \tan A \tan B) ( 1 + \tan C \tan D) = - ( 1- \tan C \tan D) ( 1 - \tan A \tan B)$
$\Rightarrow ( 1 + \tan A \tan B) ( 1 + \tan C \tan D) = ( \tan C \tan D - 1) ( 1 - \tan A \tan B)$
$\Rightarrow 1 + \tan A \tan B + \tan C \tan D + \tan A \tan B \tan C \tan D = \tan C \tan D - \tan A \tan B \tan C \tan D - 1 + \tan A \tan B$
$\Rightarrow 2 \tan A \tan B \tan C \tan D = -2$
$\Rightarrow \tan A \tan B \tan C \tan D = -1$
Hence proved.
$\\$
Question 15: If $\cos (\alpha + \beta) \sin (\gamma + \delta) = \cos (\alpha - \beta) \sin (\gamma - \delta)$, then prove that $\cot \alpha \cot \beta \cot \gamma = \cot \delta$
Given $\cos (\alpha + \beta) \sin (\gamma + \delta) = \cos (\alpha - \beta) \sin (\gamma - \delta)$
$\Rightarrow [ \cos \alpha \cos \beta - \sin \alpha \sin \beta ] [ \sin \gamma \cos \delta + \cos \gamma \sin \delta ] = [ \cos \alpha \cos \beta + \sin \alpha \sin \beta ] [ \sin \gamma \cos \delta - \cos \gamma \sin \delta ]$
Dividing both sides by $\sin \alpha \sin \beta \sin \gamma \sin \delta$
$\Rightarrow$ $\frac{ [ \cos \alpha \cos \beta - \sin \alpha \sin \beta ] [ \sin \gamma \cos \delta + \cos \gamma \sin \delta ]}{\sin \alpha \sin \beta \sin \gamma \sin \delta}$ $=$ $\frac{ [ \cos \alpha \cos \beta + \sin \alpha \sin \beta ] [ \sin \gamma \cos \delta - \cos \gamma \sin \delta ]}{\sin \alpha \sin \beta \sin \gamma \sin \delta}$
$\Rightarrow (\cot \alpha \cot \beta - 1 ) ( \cot \delta + \cot \gamma) = (\cot \alpha \cot \beta + 1 ) ( \cot \delta - \cot \gamma)$
$\Rightarrow \cot \alpha \cot \beta \cot \delta - \cot \delta + \cot \alpha \cot \beta \cot \gamma - \cot \gamma = \cot \alpha \cot \beta \cot \delta + \cot \delta - \cot \alpha \cot \beta \cot \gamma - \cot \gamma$
$\Rightarrow 2 \cot \alpha \cot \beta \cot \gamma = 2 \cot \delta$
$\Rightarrow \cot \alpha \cot \beta \cot \gamma = \cot \delta$
Hence proved.
$\\$
Question 16: If $y \sin \phi = x \sin (2 \theta + \phi)$, prove that $(x+y) \cot (\theta + \phi)= (y-x) \cot \theta$
Given $y \sin \phi = x \sin (2 \theta + \phi)$
$\Rightarrow$ $\frac{y}{x}$ $=$ $\frac{\sin (2 \theta + \phi)}{\sin \phi}$
Applying componendo and dividendo
$\frac{y-x}{y+x}$ $=$ $\frac{\sin (2 \theta + \phi) - \sin \phi}{ \sin (2 \theta + \phi) + \sin \phi}$
$\Rightarrow$ $\frac{y-x}{y+x}$ $=$ $\frac{2 \sin \Big( \frac{2 \theta + \phi - \phi}{2} \Big) \cos \Big( \frac{2 \theta + \phi + \phi}{2} \Big)}{ 2 \sin \Big( \frac{2 \theta + \phi + \phi}{2} \Big) \cos \Big( \frac{2 \theta + \phi - \phi}{2} \Big)}$
$\Rightarrow$ $\frac{y-x}{y+x}$ $=$ $\frac{\sin \theta \cos ( \theta + \phi) }{\sin ( \theta + \phi) \cos \theta}$
$\Rightarrow$ $\frac{y-x}{y+x}$ $=$ $\frac{\cot ( \theta + \phi) }{\cos \theta}$
$\Rightarrow (y-x) \cos \theta = (y+x) \cot ( \theta + \phi)$
Hence proved.
$\\$
Question 17: If $\cos (A+B) \sin (C-D) = \cos (A-B) \sin (C+D)$, prove that $\tan A \tan B \tan C + \tan D = 0$
Given $\cos (A+B) \sin (C-D) = \cos (A-B) \sin (C+D)$
$\Rightarrow (\cos A \cos B - \sin A \sin B) ( \sin C \cos D - \cos C \sin D) = (\cos A \cos B + \sin A \sin B) ( \sin C \cos D + \cos C \sin D)$
Divide both sides by $\cos A \cos B \cos C\cos D$ we get
$\frac{(\cos A \cos B - \sin A \sin B) ( \sin C \cos D - \cos C \sin D)}{\cos A \cos B \cos C\cos D} = \frac{(\cos A \cos B + \sin A \sin B) ( \sin C \cos D + \cos C \sin D)}{\cos A \cos B \cos C\cos D}$
$\Rightarrow ( 1 - \tan A \tan B)(\tan C - \tan D) = (1 + \tan A \tan B) (\tan C + \tan D)$
$\Rightarrow \tan C - \tan D - \tan A \tan B \tan C + \tan A \tan B \tan D = \tan C + \tan D + \tan A \tan B \tan C + \tan A \tan B \tan D$
$\Rightarrow -2 \tan D = 2 \tan A \tan B \tan C$
$\Rightarrow \tan A \tan B \tan C + \tan D = 0$
Hence proved.
$\\$
Question 18: If $x \cos \theta = y \cos \Big( \theta+$ $\frac{2\pi}{3}$ $\Big) = z \cos \Big( \theta +$ $\frac{4\pi}{3}$ $\Big)$, prove that $xy + yz + zx = 0$
Given $x \cos \theta = y \cos \Big( \theta+$ $\frac{2\pi}{3}$ $\Big) = z \cos \Big( \theta +$ $\frac{4\pi}{3}$ $\Big) = k$
$\Rightarrow x =$ $\frac{k}{\cos \theta}$
$\Rightarrow y =$ $\frac{k}{\cos \Big( \theta + \frac{2\pi}{3} \Big) }$
$\Rightarrow z =$ $\frac{k}{\cos \Big( \theta + \frac{4\pi}{3} \Big) }$
$\therefore xy + yz + zx$
$= \Big[$ $\frac{k}{\cos \theta} \frac{k}{\cos \big( \theta + \frac{2\pi}{3} \big) } + \frac{k}{\cos \big( \theta + \frac{2\pi}{3} \big) } \frac{k}{\cos \big( \theta + \frac{4\pi}{3} \big) } + \frac{k}{\cos \big( \theta + \frac{4\pi}{3} \big) } \frac{k}{\cos \theta}$ $\Big]$
$= k^2 \Big[$ $\frac{1}{\cos \theta} \frac{1}{\cos \big( \theta + \frac{2\pi}{3} \big) } + \frac{1}{\cos \big( \theta + \frac{2\pi}{3} \big) } \frac{1}{\cos \big( \theta + \frac{4\pi}{3} \big) } + \frac{1}{\cos \big( \theta + \frac{4\pi}{3} \big) } \frac{1}{\cos \theta}$ $\Big]$
$= k^2 \Big[$ $\frac{\cos \big( \theta + \frac{4\pi}{3} \big) + \cos \theta + \cos \big( \theta + \frac{2\pi}{3} \big)}{\cos \big( \theta+ \frac{4\pi}{3} \big) \cos \theta \cos \big( \theta + \frac{2\pi}{3} \big)}$ $\Big]$
$= k^2 \Big[$ $\frac{ \cos \theta \cos \frac{4\pi}{3} - \sin \theta \sin \frac{4\pi}{3} + \cos \theta + \cos \theta \cos \frac{2\pi}{3} - \sin \theta \sin \frac{2\pi}{3}}{\cos \big( \theta + \frac{4\pi}{3} \big) \cos \theta \cos \big( \theta + \frac{2\pi}{3} \big)}$ $\Big]$
$= k^2 \Big[$ $\frac{ \cos \theta (-\frac{1}{2}) - \sin \theta (-\frac{\sqrt{3}}{2}) + \cos \theta + \cos \theta (-\frac{1}{2}) - \sin \theta (\frac{\sqrt{3}}{2}) }{\cos \big( \theta + \frac{4\pi}{3} \big) \cos \theta \cos \big( \theta + \frac{2\pi}{3} \big)}$ $\Big]$
$= k^2 \Big[$ $\frac{ -\cos \theta + (\frac{\sqrt{3}}{2}) \sin \theta + \cos \theta - (\frac{\sqrt{3}}{2}) \sin \theta }{\cos \big( \theta + \frac{4\pi}{3} \big) \cos \theta \cos \big( \theta + \frac{2\pi}{3} \big)}$ $\Big]$
$= k^2 \Big[$ $\frac{ 0 }{\cos \big( \theta + \frac{4\pi}{3} \big) \cos \theta \cos \big( \theta + \frac{2\pi}{3} \big)}$ $\Big]$
$= 0 =$ RHS. Hence proved.
$\\$
Question 19: If $m \sin \theta = n \sin (\theta + 2 \alpha)$, prove that $\tan (\theta + \alpha) \cot \alpha =$ $\frac{m+n}{m-n}$
Given: $m \sin \theta = n \sin (\theta + 2 \alpha)$
$\frac{m}{n}$ $=$ $\frac{\sin (\theta + 2 \alpha)}{\sin \theta}$
Applying componendo and dividendo
$\frac{m+n}{m-n}$ $=$ $\frac{\sin (\theta + 2 \alpha) +\sin \theta }{\sin (\theta + 2 \alpha) -\sin \theta}$
$\frac{m+n}{m-n}$ $=$ $\frac{2 \sin \big( \frac{\theta + 2 \alpha + \theta}{2} \big) \cos \big( \frac{\theta + 2 \alpha - \theta}{2} \big) }{2 \cos \big( \frac{\theta + 2 \alpha + \theta}{2} \big) \sin \big( \frac{\theta + 2 \alpha - \theta}{2} \big)}$
$\frac{m+n}{m-n}$ $=$ $\frac{2 \sin ( \theta + \alpha) \cos \alpha}{2 \cos ( \theta + \alpha) \sin \alpha}$
$\frac{m+n}{m-n}$ $=$ $\frac{\tan ( \theta + \alpha) }{\tan \alpha}$
$\tan ( \theta + \alpha) =$ $\frac{m+n}{m-n}$ $\tan \alpha$
Hence proved.
|
{}
|
# Finding the symbolic inverse of a function
Is there a way of inverting this function to obtain $$r(\rho)$$?
rho[r_, b0_, q_] :=
r (1 + (Sqrt[π]Gamma[1/(q - 1)])/((1 - q) Gamma[1/2 ((q + 1)/(q - 1))]) b0 /r + (1 + q)/(2 q) (b0/r)^(1 - q))
Note that $$q<0$$ and $$b0$$ is some positive constant.
The typical way to do this, is to use Solve or Reduce get r to one side of the equality. It seems like Mathematica cannot solve the equation, unfortunately:
Reduce[
{
rho == r (1 + (Sqrt[\[Pi]] Gamma[1/(q - 1)])/((1 - q) Gamma[1/2 ((q + 1)/(q - 1))]) b0/r + (1 + q)/(2 q) (b0/r)^(1 - q)),
q < 0,
b0 > 0
},
r
]
During evaluation of In[2]:= Reduce::nsmet: This system cannot be solved with the methods available to Reduce.
Out[2]= Reduce[{rho == r (1 + ((1 + q) (b0/r)^(1 - q))/(2 q) + ( b0 Sqrt[[Pi]] Gamma[1/(-1 + q)])/((1 - q) r Gamma[(1 + q)/(2 (-1 + q))])), q < 0, b0 > 0}, r]
• What if I specify the value for $q$ and $b0$? I tried it for $q=-2$ and $b0=1$ but I don't understand what Mathematica spit out. – user583893 Feb 6 at 14:56
• It spits out Root objects, which are symbolic representations of the exact roots of polynomials. In this case, it gives roots of 3rd-degree polynomials, so you can use the option Cubics -> True in Reduce to expand them. – Sjoerd Smit Feb 6 at 15:38
• Both Solve and Reduce should give answers in terms of Root for integer values of q. Typically, there are no solutions otherwise. – bbgodfrey Feb 6 at 18:38
|
{}
|
# Secure Multiplication and Match-Making
To do general secure computation, we will of course need to do more than secure addition. It turns out that the secret sharing scheme from the previous subsection already allows us to do more: we can also do secure multiplication.
Suppose two numbers $a,b \in \mathbb{Z}_p$ have been secret shared as described above, so that $a= a_1+a_2+a_3\bmod p$ and $b= b_1+b_2+b_3\bmod p$, and we wish to compute the product $ab\bmod p$ securely. We obviously have
$ab= a_1b_1 + a_1b_2+ a_1b_3 +a_2b_1 + a_2b_2+ a_2b_3 + a_3b_1 + a_3b_2+ a_3b_3\bmod p$.
It is now easy to see that if the $a_i$‘s and $b_i$‘s have been distributed as described above, it is the case that for each product $a_i b_j$, there is at least one player among the three who knows $a_i$ and $b_j$ and therefore can compute $a_i b_j$. For instance, $\mathsf{P}_1$ has been given $a_2,a_3,b_2,b_3$ and can therefore compute $a_2b_2, a_2b_3, a_3b_2$ and $a_3b_3$. The situation is, therefore, that the desired result $ab$ is the sum of some numbers where each summand can be computed by at least one of the players. But now we are essentially done, since from Secure Addition, we already know how to add securely!
The protocol resulting from these observations is shown below. To argue why it works, one first notes that correctness, namely $ab= u_1+u_2+u_3 \bmod p$, follows trivially from the above. To show that nothing except $ab\bmod p$ is revealed, one notes that nothing new about $a,b$ is revealed in the first step, and because Secure Addition is private, nothing except the sum of the inputs is revealed in the last step, and this sum always equals $ab\bmod p$.
Secure Multiplication
Participants are $\mathsf{P}_1, \mathsf{P}_2, \mathsf{P}_3$, input for $\mathsf{P}_1$ is $a\in \mathbb{Z}_p$, input for $\mathsf{P}_2$ is $b\in \mathbb{Z}_p$, where $p$ is a fixed prime agreed upon in advance. $\mathsf{P}_3$ has no input.
1. $\mathsf{P}_1$ distributes shares $a_1,a_2,a_3$ of $a$, while $\mathsf{P}_2$ distributes shares $b_1,b_2,b_3$ of $b$.
2. $\mathsf{P}_1$ locally computes $u_1= a_2b_2+ a_2b_3+ a_3b_2\bmod p$, $\mathsf{P}_2$ computes $u_2= a_3b_3+a_1b_3+ a_3b_1\bmod p$, and $\mathsf{P}_3$ computes $u_3= a_1b_1+a_1b_2+a_2b_1\bmod p$.
3. The players use Secure Addition to compute the sum $u_1+u_2+u_3\bmod p$ securely, where $\mathsf{P}_i$ uses $u_i$ as input.
It is interesting to note that even in a very simple case where both $a$ and $b$ are either $0$ or $1$, secure multiplication has a meaningful application: consider two parties Alice and Bob. Suppose Alice is wondering whether Bob wants to go out with her, and also Bob is asking himself if Alice is interested in him. They would very much like to find out if there is mutual interest, but without running the risk of the embarrassment that would result if for instance Bob just tells Alice that he is interested, only to find that Alice turns him down. The problem can be solved if we let Alice choose $a\in \mathbb{Z}_p$ where $a=1$ if she is interested in Bob and $a=0$ otherwise. In the same way, Bob chooses $b$ to be $0$ or $1$. Then we compute the function $f(a,b)= ab\bmod p$ securely. It is clear that the result is $1$ if and only if there is mutual interest. But on the other hand if, for instance, Alice is not interested, she will choose $a=0$ and in this case she learns nothing new from the protocol. To see why, notice that security of the protocol implies that the only (possibly) new information Alice will learn is the result $ab\bmod p$. But she already knows that result will be $0$! In particular, she does not learn whether Bob was interested or not, so Bob is safe from embarrassment. By a symmetric argument, this is of course also the case for Alice.
This argument assumes, of course, that both players choose their inputs honestly according to their real interests. In the following section we discuss what happens if players do not follow the instructions and what we can do about the problems resulting from this.
From Secure Multiplication, we see that if Alice and Bob play the roles of $\mathsf{P}_1$ and $\mathsf{P}_2$, respectively, they just need to find a third party to help them, to do the multiplication securely. Note that this third party is not a completely trusted third party of the kind we discussed before: he does not learn anything about $a$ or $b$ other than $ab\bmod p$. Alice and Bob do have to trust, however, that the third party does not share his information with Bob or with Alice.
It is an obvious question whether one can do secure multiplication such that only Alice and Bob have to be involved? The answer turns out to be yes, but then information theoretic security. Instead one has to use solutions based on cryptography. Such solutions can always be broken if one party has enough computing power, but on the other hand, this is an issue with virtually all the cryptographic techniques we use in practice.
For completeness, we remark that Alice and Bob’s problem is a special case of the so-called match-making problem which has somewhat more serious applications than secure dating. Consider a set of companies where each company has a set of other companies it would prefer to do business with. Now we want that each pair of companies finds out whether there is mutual interest, but without forcing companies to reveal their strategy by announcing their interests in public.
Next we will investigate how to design protocols which remain secure even if some parties try to cheat in the protocols, i.e., we ask the question What if Players Do Not Follow Instructions?
|
{}
|
Mathematics of zombies
What do you do if there is a Zombie attack? Can mathematics help? This page is (humorously) dedicated to collecting links to papers or blog posted related to the mathematical models of Zombies.
George Romero’s 1968 Night of the Living Dead, now in the public domain, introduced reanimated ghouls, otherwise known as zombies, which craved live human flesh. Romero’s script was insired on Richard Matheson’s I Am Legend. In Romero’s version, the zombies could be killed by destroying the zombie’s brain. A dead human could, in some cases be “reanimated,” turning into a zombie. These conditions are modeled mathematically in several papers, given below.
1. When Zombies Attack! Mathematical Modelling of a Zombie Outbreak!, paper by Mathematicians at the University of Ottawa, Philip Munz, Ioan Hudea, Joe Imad and Robert J. Smith? (yes, his last name is spelled “Smith?”).
2. youtube video 28 Minutes Later – The Maths of Zombies , made by Dr James Grime (aka, “siningbanana”), which references the above paper.
3. Epidemics in the presence of social attraction and repulsion, Oct 2010 Zombie paper by Evelyn Sander and Chad M. Topaz.
4. Statistical Inference in a Zombie Outbreak Model, slides for a talk given by Des Higman, May 2010.
5. Mathematics kills zombies dead!, 08/17/2009 blog post by “Zombie Research Society Staff”.
6. The Mathematics of Zombies, August 18, 2009 blog post by Mike Elliot.
7. Love, War and Zombies – Systems of Differential Equations using Sage, April 2011 slides by David Joyner. Sage commands for Love, War and Zombies talk. This was given as a Project Mosaic/M-cast broadcast.
8. Public domain 1968 film Night of the Living Dead by George Romero.
Paley graphs in Sage
Let $q$ be a prime power such that $q\equiv 1 \pmod 4$. Note that this implies that the unique finite field of order $q$, $GF(q)$, has a square root of $-1$. Now let $V=GF(q)$ and
$E = \{(a,b)\in V\times V\ |\ a-b\in GF(q)^2\}.$
By hypothesis, $(a,b)\in E$ if and only if $(b,a)\in E$. By definition $G = (V, E)$ is the Paley graph of order $q$.
Paley was a brilliant mathematician who died tragically at the age of 26. Paley graphs are one of the many spin-offs of his work. The following facts are known about them.
1. The eigenvalues of Paley graphs are $\frac{q-1}{2}$ (with multiplicity $1$) and
$\frac{-1 \pm \sqrt{q}}{2}$ (both with multiplicity $\frac{q-1}{2}$).
2. It is known that a Paley graph is a Ramanujan graph.
3. It is known that the family of Paley graphs of prime order is a vertex expander graph family.
4. If $q=p^r$, where $p$ is prime, then $Aut(G)$ has order $rq(q-1)/2$.
Here is Sage code for the Paley graph (thanks to Chris Godsil, see [GB]):
def Paley(q):
K = GF(q)
return Graph([K, lambda i,j: i != j and (i-j).is_square()])
(Replace “K” by “$K.\langle a\rangle$” above; I was having trouble rendering it in html.) Below is an example.
sage: X = Paley(13)
sage: X.vertices()
[0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12]
sage: X.is_vertex_transitive()
True
sage: X.degree_sequence()
[6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6]
sage: X.spectrum()
[6, 1.302775637731995?, 1.302775637731995?, 1.302775637731995?,
1.302775637731995?, 1.302775637731995?, 1.302775637731995?,
-2.302775637731995?, -2.302775637731995?, -2.302775637731995?,
-2.302775637731995?, -2.302775637731995?, -2.302775637731995?]
sage: G = X.automorphism_group()
sage: G.cardinality()
78
We see that this Paley graph is regular of degree $6$, it has only three distinct eigenvalues, and its automorphism group is order $13\cdot 12/2 = 78$.
Here is an animation of this Paley graph:
The frames in this animation were constructed one-at-a-time by deleting an edge and plotting the new graph.
Here is an animation of the Paley graph of order $17$:
The frames in this animation were constructed using a Python script:
X = Paley(17)
E = X.edges()
N = len(E)
EC = X.eulerian_circuit()
for i in range(N):
X.plot(layout="circular", graph_border=True, dpi=150).save(filename="paley-graph_"+str(int("1000")+int("%s"%i))+".png")
X.delete_edge(EC[i])
X.plot(layout="circular", graph_border=True, dpi=150).save(filename="paley-graph_"+str(int("1000")+int("%s"%N))+".png")
Instead of removing the frames “by hand” they are removed according to their occurrence in a Eulerian circuit of the graph.
Here is an animation of the Paley graph of order $29$:
[GB] Chris Godsil and Rob Beezer, Explorations in Algebraic Graph Theory with Sage, 2012, in preparation.
Sestinas and Sage
According to [1], a sestina is a highly structured poem consisting of six six-line stanzas followed by a tercet (called its envoy or tornada), for a total of thirty-nine lines. The same set of six words ends the lines of each of the six-line stanzas, but in a shuffled order each time. The shuffle used is very similar to the Mongean shuffle.
Define $f_n(k) = 2k$, if k <= n/2 and $f_n(k) = 2n+1-2k$, if $k > n/2.$ Let $p = (p_1,...,p_n) \in S_n$, where $p_j = f_n(p_{j-1})$ and $S_n$ is the symmetric group of order $n$. From [2], we have the following result.
Theorem: If p is an n-cycle then 2n+1 is a prime.
Call such a prime a “sestina prime”. Which primes are sestina primes?
Here is Python/Sage code for this permutation:
def sestina(n):
"""
Computes the element of the symmetric group S_n associated to the shuffle above.
EXAMPLES:
sage: sestina(4)
(1,2,4)
sage: sestina(6)
(1,2,4,5,3,6)
sage: sestina(8)
(1,2,4,8)(3,6,5,7)
sage: sestina(10)
(1,2,4,8,5,10)(3,6,9)
sage: sestina(12)
(1,2,4,8,9,7,11,3,6,12)(5,10)
sage: sestina(14)
(1,2,4,8,13,3,6,12,5,10,9,11,7,14)
sage: sestina(16)
(1,2,4,8,16)(3,6,12,9,15)(5,10,13,7,14)
sage: sestina(18)
(1,2,4,8,16,5,10,17,3,6,12,13,11,15,7,14,9,18)
sage: sestina(20) (1,2,4,8,16,9,18,5,10,20)(3,6,12,17,7,14,13,15,11,19)
sage: sestina(22) (1,2,4,8,16,13,19,7,14,17,11,22)(3,6,12,21)(5,10,20)(9,18)
"""
def fcn(k, n):
if k<=int(n/2):
return 2*k
else:
return 2*n+1-2*k
L = [fcn(k,n) for k in range(1,n+1)]
G = SymmetricGroup(n)
return G(L)
And here is an example due to Ezra Pound [3]:
I
Damn it all! all this our South stinks peace.
You whoreson dog, Papiols, come! Let’s to music!
I have no life save when the swords clash.
But ah! when I see the standards gold, vair, purple, opposing
And the broad fields beneath them turn crimson,
Then howl I my heart nigh mad with rejoicing.
II
In hot summer have I great rejoicing
When the tempests kill the earth’s foul peace,
And the light’nings from black heav’n flash crimson,
And the fierce thunders roar me their music
And the winds shriek through the clouds mad, opposing,
And through all the riven skies God’s swords clash.
III
Hell grant soon we hear again the swords clash!
And the shrill neighs of destriers in battle rejoicing,
Spiked breast to spiked breast opposing!
Better one hour’s stour than a year’s peace
With fat boards, bawds, wine and frail music!
Bah! there’s no wine like the blood’s crimson!
IV
And I love to see the sun rise blood-crimson.
And I watch his spears through the dark clash
And it fills all my heart with rejoicing
And prys wide my mouth with fast music
When I see him so scorn and defy peace,
His lone might ’gainst all darkness opposing.
V
The man who fears war and squats opposing
My words for stour, hath no blood of crimson
But is fit only to rot in womanish peace
Far from where worth’s won and the swords clash
For the death of such sluts I go rejoicing;
Yea, I fill all the air with my music.
VI
Papiols, Papiols, to the music!
There’s no sound like to swords swords opposing,
No cry like the battle’s rejoicing
When our elbows and swords drip the crimson
And our charges ’gainst “The Leopard’s” rush clash.
May God damn for ever all who cry “Peace!”
VII
And let the music of the swords make them crimson
Hell grant soon we hear again the swords clash!
Hell blot black for always the thought “Peace”!
References:
[1] http://en.wikipedia.org/wiki/Sestina
[2] Richard Dore and Anton Geraschenko,”Sestinas and Primes” posted to http://stacky.net/wiki/index.php?title=Course_notes, and http://math.berkeley.edu/~anton/written/sestina.pdf
[3] Ezra Pound, “Sestina: Altaforte” (1909), (originally published int the English Review, 1909)
[4] John Bullitt, N. J. A. Sloane and J. H. Conway , http://oeis.org/A019567
|
{}
|
V8 engine Cessna 172
Help Support Homebuilt Aircraft & Kit Plane Forum:
robertl
Well-Known Member
I'm not a mechanic, or an engineer but I have stayed at a Holiday Inn before so I feel that I'm qualified to make this statement. How about you take the auto engine that you are working with, make whatever changes you want to make and just manufacture your own, under whatever criteria is standard for the aviation industry and call it an aircraft engine and not an auto engine. I know it would still have to be proven and all the FAA BS and this and that and bla, bla, bla stuff would still have to be done. But, that way, you're not trying to use something that was meant for one thing, an automobile, and using it for something else, an airplane. Do Lycoming and Continental manufacture their own engines and sub assemblies, "In House", or, are the engine blocks, cases, etc. manufactured by 3rd parties? If you're using a Chevy 350 ci engine for example, whoever is making if for Chevy could be your 3rd party supplier, if they were willing to do whatever extra testing is required, but it seems you are doing that extra testing already.
Bob
rv6ejguy
Well-Known Member
HBA Supporter
The problem is, as soon as you certify the engine, you add liability concerns and nobody in the manufacturing chain wants any part of that exposure. Look at how lawyers sometimes use the "shotgun approach" after an accident- they name the airframer, prop, engine, avionics, fuel pump manufacturer etc. Any component maker who had parts on that plane can be named in a lawsuit. It simply isn't worth the trouble for many companies to provide parts for certified aircraft.
Also doesn't matter what you call something to the FAA. You'll have to vett and satisfy the process control and batching of each manufactured part going into an engine which is a huge task and greatly increases the costs of the finished product.
Is it really any wonder why almost no small company produces new certified engines today? Most people asking how a Lyconental or new certified diesel can cost $40-$100,000 dollars don't have a clue of what is involved to get there and make money at it. Look at all the failed attempts- Orenda, Mistral, EPS, Honda, Toyota just to name a few.
The reality is there are way easier ways to make a living in this world than producing certified aircraft engines.
Last edited:
slociviccoupe
Well-Known Member
Ls7 comes with a factory dry sump pan and tank. Only rewuires changing the oio pump for the dual inline pump pressure and scavenge pump, the oil pan, the tank and hoses. How much better can you get than oem gm parts for a factory dry sump.
dwalker
Well-Known Member
HBA Supporter
Ls7 comes with a factory dry sump pan and tank. Only rewuires changing the oio pump for the dual inline pump pressure and scavenge pump, the oil pan, the tank and hoses. How much better can you get than oem gm parts for a factory dry sump.
The Dailey Engineering dry sump is better than OEM.
wrmiles
Member
Many years ago when I was at Cessna, there was a preliminary approach to GM to supply aircraft engines. I don't know many details except it didn't go anywhere. I imagine product liability, configuration control/QC requirements, and small volume were the cause of it not going anywhere.
rv7charlie
Well-Known Member
When I was a lot younger and dumber, and Toyota went public with their aviation related development work on their V8, I brought it up with my stockbroker friend, wanting to buy some Toyota stock. He basically laughed at me, pointing out that the engine project was basically coffee break conversation for a company the size of Toyota. Of course, he was right. They could handle the liability issues (far more exposure in autos than any general aviation aircraft), but there was no real money to be made for them in aviation, compared to their revenue in more ground-bound endeavors.
speedracer
Well-Known Member
When I was a lot younger and dumber, and Toyota went public with their aviation related development work on their V8, I brought it up with my stockbroker friend, wanting to buy some Toyota stock. He basically laughed at me, pointing out that the engine project was basically coffee break conversation for a company the size of Toyota. Of course, he was right. They could handle the liability issues (far more exposure in autos than any general aviation aircraft), but there was no real money to be made for them in aviation, compared to their revenue in more ground-bound endeavors.
Many years ago Toyota contracted Shirl Dickey to install a Toyota engine in his Eracer (a side by side LongEZ spin off which he designed and marketed plans for). He did one runway flight for the camera and got paid.
PMD
Well-Known Member
Thielert and then Austro managed reasonably well by sourcing automotive engines and components to make aviation engines. You might, of course, notice that those did NOT come from the USA. The problem is how the US deals with the LLL - making people and companies completely unprotected from ridiculous lawsuits and settlements. Until the US (and to a lesser extent Canada) gets the legal and insurance runaway problem under control, we can simply watch the Genav and many, many other businesses simply move production to safe places outside of the USA. Do you think it is an co-incidence that the largest volume aviation engine manufacturer started, grew and is sustained by Canadian companies with European production?
BBerson
Light Plane Philosopher
HBA Supporter
Do you think it is an co-incidence that the largest volume aviation engine manufacturer started, grew and is sustained by Canadian companies with European production?
The Special Light Sport rule and strict standards have prevented any new small USA engine manufacture. And the previous industry has disappeared.
TarDevil
Well-Known Member
Any component maker who had parts on that plane can be named in a lawsuit.
TarDevil
Come on; do you really believe that Slick (almost certainly have a lawyer on retainer) would write a $300k check to avoid a 10 minute session in front of a judge to show that they weren't involved with the plane? You can take the up with the Slick guys. I personally enjoyed working with that guy and can't imagine him lying to me. He was a no-nonsense type. For what it's worth, I asked the same question. I don't remember his exact reply - it was 38 years ago. thjakits Well-Known Member Come on; do you really believe that Slick (almost certainly have a lawyer on retainer) would write a$300k check to avoid a 10 minute session in front of a judge to show that they weren't involved with the plane?
.....especially if they could just sue that lawyer in return for accusing them of the impossible...
slociviccoupe
Well-Known Member
The Dailey Engineering dry sump is better than OEM.
You are correct. They make really nice dry sumps. My only concern is the aluminum rotors and longetivity. And worst case scenario the belt comes off. The oem ls7 pumps are mounted on nose of the crank.
BJC
Well-Known Member
HBA Supporter
Until the US (and to a lesser extent Canada) gets the legal and insurance runaway problem under control,
Even though I agree with you, I don’t expect to see it happen.
A rep from Slick was traveling with me and said they were sued in a crash and the airplane had Bendix equipment. When they informed the plaintiff's lawyers, they were just told, "Come to court and prove it."
It is common practice to make settlements based on minimizing costs rather that resolving right from wrong. BTDT with corporate money rather than mine. At times, I wonder if things would be different if defendants asserted innocence and spent the money to prove it, even if it cost more.
BJC
TarDevil
Well-Known Member
At times, I wonder if things would be different if defendants asserted innocence and spent the money to prove it, even if it cost more.
Oh, just once!
HBA Supporter
PMD
Well-Known Member
Even though I agree with you, I don’t expect to see it happen.
My concern is that this and several other factors (mostly the switch from Capitalism to Casino Capitalism) plus mindless drift into globalism that will take the US economy down. It is already well on its way. Sadly, genav is a lot more fragile and less resilient than other business sectors and happens to be the one we are caught up in.
On the US lawsuit and settlement issue: I have unfortunately been there, done that and lost the T shirt. I have seen first hand the strategy where litigants literally calculate what they think you can spend and make darn sure to defend oneself/company against their malfiesence they clearly demonstrate that it will cost more than you can afford. That strategy (a common one) means you just bend over and stroke the settlement cheque. A legal system that not only allows, but encourages and rewards such things is iindeed broken.
Last edited:
tspear
Well-Known Member
Come on; do you really believe that Slick (almost certainly have a lawyer on retainer) would write a $300k check to avoid a 10 minute session in front of a judge to show that they weren't involved with the plane? I was a subject matter expert on an IT related suite about three years ago. The individual being sued was accused of stealing a program written in Java; because the new application he developed with no functional overlap or data model overlap and developed in JavaScript. Total bill$450K spent before the ten minute hearing where the judge threw out the case. That was money spent just on all the motions, and preliminary hearings.
Tim
lelievre12
Well-Known Member
HBA Supporter
I was a subject matter expert on an IT related suite about three years ago. The individual being sued was accused of stealing a program written in Java; because the new application he developed with no functional overlap or data model overlap and developed in JavaScript.
Total bill $450K spent before the ten minute hearing where the judge threw out the case. That was money spent just on all the motions, and preliminary hearings. Tim Not just motions and hearings but the costs for each party to prepare the cases. Any reasonably complex affidavit is going to be >$50K and usually more than one is needed for both defendant and the plaintiff. In my own business I was around \$50K down this road when I realized that all this money on both sides could be used to settle instead of fight.
I instead called the plaintiffs, (two ex employees) arranged for a nice lunch and then settled out of court directly over a nice steak and glass of wine. My lawyers of course were furious that I had settled 'without them' but they knew that I knew that had it been left up to them, we never would have settled and would have fought until the very expensive 'end'. Lawyers billing by the hour have zero interest to settle. My advice always is to swallow your pride, and mediate directly with the plaintiff no matter how 'right' you are.
|
{}
|
# Mighty Max C-Axis and Igus
I mentioned in Accelstepper that I had been working on a 5 Axis platform for a laser cutting project. Well I’ve sent the drawings in and parts are slowly starting to come! We’ve already ordered and have many of the smaller parts…motors, drivers, (not yet controller….that’ll be another post) and the rotary parts for our rotational axis Namely the C-axis…Remember the ABC are the ones the rotate around our XYZ.
Buuuut of course. Nothing ever works the way a person wants it to and here is a little tale…or maybe review of our purchase.
We decided to buy Igus slew rings¹ instead of designing a complicated set of axel+axialbearings to rotate a table that barely weighs 300 grams. They’re pretty expensive per part, and have all the fancy Igus PTFE² or whatever plastics they use (it’s teflon). Usually When one decides to go high precision, you can guess by the price of something an expectation of quality, one can also expect (wrongly) a certain kind of tolerance. If you want to rotate a thing…it should not move eccentric around any of the XYZs…Our expectation was exactly (and falsely) that.
Once the parts actually showed up, a person could literally shift the outer HTD ring around it’s sliding ‘elements’ Infact, as I just see now in the picture. The black core was ø37mm and the ptfe elements were ø37.25mm! That may not seem like a lot. But for a high precision machine like the one I’m building. It’s nearly 250 times what I can even tolerate.
Fortunately we arn’t working in the micrometres anymore, just in the 10s of micrometres! So it’s only 25 times larger than what I can deal with. Thats something I can work with!
How you ask?
Guage band!
This is a exactly defined (also with a tolerance but whatevs) strip of metal. The ones I have are 0.1mm 0.02 and 0.05. When a person measures things using something like a venier calipre you can expect that your measurements might be off…especially if it’s a low quality (again the quality here) something called Abbe-error³ shows up or you moved the things after trying to remove it from the thing you were measuring!
So it’s kinda trial and error. And thats why I had different thickness to play around with. I immediately tried to use the o.1mm band and found it was way too thick to use. Once reassembled the ring would barely move. Waaaay too much friction. So I next tried the 0.05 and the 0.02. (spoiler, they worked)
From my old work I learned that when you’re going to use something like these you should cut them at an angle instead of straight. The moving parts inside are less likely to catch the ends and maybe start ripping things up. Of course we have a part that is a little jagged and sharp, eventually the parts will wear. You can reduce this wearing by doing this simple trick! Not to mention it looks nicer.
So in the end we closed up at slop of 0.14mm which means (assuming the 250 micrometre was accurate) we’re down to 0.055mm eccentricity. Thats one I can deal with. So whats my lesson learned? Maybe just use the axel+bearings instead of going with Igus slew rings…
$e^{\sqrt{2}}$
|
{}
|
# Generic Group Analyzer Unbounded "Do it unbounded"
## Introduction
Generic Group Analyzer Unbounded is a tool to automatically prove computational security statements in the Generic Group Model. It is especially designed to deal with security experiments where the attacker is allowed to interact with an Oracle an unbounded number of times and this interaction may be adaptive, i.e., new queries can depend on earlier Oracle responses.
This overcomes previous tools like the Generic Group Analyzer (Crypto 2014), paper available at http://eprint.iacr.org/2014/458, which cannot deal with adaptive adversaries; or its extension (PKC 2015), whose paper is available at http://eprint.iacr.org/2015/019 and it only deals with experiments where the number of queries is fixed.
The tool focuses on pairing-based cryptographic constructions, in particular, on schemes defined over bilinear groups.
A bilinear group is a tupple $(p, \mathbb{G}_{1}, \mathbb{G}_{2}, \mathbb{G}_{T}, g_{1}, g_{2}, e)$ where,
• $p$ is a prime number,
• $\mathbb{G}_{1}, \mathbb{G}_{2}, \mathbb{G}_{T}$ are cyclic groups of prime order $p$,
• $g_{1}$ is a generator of $\mathbb{G}_{1}$ and $g_{2}$ is a generator of $\mathbb{G}_{2}$,
• $e$, known as the pairing, is an efficiently computable map $e: \mathbb{G}_{1} \times \mathbb{G}_{2} \rightarrow \mathbb{G}_{T}$, that satisfies the bilinear property: $$\forall a,b \in \mathbb{Z}_{p}, \ e(g_{1}^{a}, g_{2}^{b}) = e(g_{1},g_{2})^{ab}$$ and it is non-degenerate, i.e, $e(g_{1},g_{2})$ is a generator of $\mathbb{G}_{T}$.
Therefore, the tool can be used to analyze the generic security of:
• Structure-Preserving Signatures (signature schemes defined over bilinear groups where the messages and signatures consist of group elements and the verification of signatures consists of evaluation pairing product equations).
• Algebraic Message-Authentication Codes.
• Assumptions.
and more cryptographic construction defined over bilinear groups whose security is defined as a computational experiment.
### Example
This is an example of a cryptographic construction that can be analyzed with the tool. It is taken from the AsiaCrypt 2015 paper, https://eprint.iacr.org/2014/635, by S. Chatterjee and A. Menezes:
#### Re-randomizable Structure-Preserving Signature scheme in Type III
• Setup ${\cal P}(1^{\lambda})$:
Return $\mathit{PP} = (p, \mathbb{G}_{1}, \mathbb{G}_{2}, \mathbb{G}_{T}, g_{1}, g_{2}, e) \leftarrow {\cal G}(1^{\lambda})$, where ${\cal G}$ is an efficient algorithm that on input $1^{\lambda}$ outputs a description of a bilinear group in Type III, with groups of order $p$ for a $\lambda$-bit prime $p$.
• Key generation ${\cal K}(\mathit{PP})$:
Choose $v,w \stackrel{\$}{\leftarrow} \mathbb{Z}_{p}^{*}$and compute$\mathit{VK} = (\mathit{PP}, V, W)$and$\mathit{SK} = (\mathit{PP}, v, w)$as $$V \leftarrow g_{1}^{v} \ \text{ and } \ W \leftarrow g_{1}^{w}$$ • Signing${\cal S}_{\mathit{SK}}(M)$: For$M \in \mathbb{G}_{2}$, choose$r \stackrel{\$}{\leftarrow} \mathbb{Z}_{p}^{*}$ and compute the signature $(T_{1},T_{2},S) \in \mathbb{G}_{1}\times \mathbb{G}_{2}^{2}$ as $$T_{1} \leftarrow g_{1}^{r}, \ \ \ \ \ T_{2} \leftarrow g_{2}^{r}, \ \ \ \text{ and } \ \ \ S \leftarrow M^{v} \cdot g_{2}^{w} \cdot g_{2}^{r^2}$$
• Verification ${\cal V}_{\mathit{VK}}(M, (T_{1},T_{2},S))$:
Accept if and only if $T_{1} \in \mathbb{G}_{1}$, $M,T_{2},S \in \mathbb{G}_{2}$ and $$e(g_{1}, S) = e(V,M) \cdot e(W,g_{2}) \cdot e(T_{1}, T_{2}) \ \ \land \ \ \ e(T_{1}, g_{2}) = e(g_{1}, T_{2})$$
A possible notion of security for a signature scheme is the Existential Unforgeability Against Chosen Message Attacks. We say that a signature scheme $({\cal P}, {\cal K}, {\cal S}, {\cal V})$ is EUF-CMA secure if for all polynomial time algorithms ${\cal A}$, $$\mathit{Pr}\left[ \begin{array}{ll} \mathit{PP} \leftarrow {\cal P}(1^{\lambda}) & \\ (\mathit{VK}, \mathit{SK}) \leftarrow {\cal K}(\mathit{PP}) & : M \not \in Q \ \land \ {\cal V}_{\mathit{VK}}(M,\Sigma) = 1 \\ (M,\Sigma) \leftarrow {\cal A}^{{\cal S}_{\mathit{SK}}(\cdot)}(\mathit{VK}) & \\ \end{array} \right] \approx \mathit{negl}(\lambda)$$ where $Q$ is the set of queries made by ${\cal A}$ to the signing oracle. Note that, since the adversary is polynomial, $|Q| \approx \mathit{poly}(\lambda)$, but this cardinality is not bounded.
Intuitively, this notion says that:
There is no polynomial time adversary that can produce a valid signature (with significant probability) on a message without the knowledge of the secret key, even if she has access to a signing Oracle that produces valid signatures for messages of her choice.
Note that the adversary must find a valid signature on a message that was not sent to the Oracle.
## Installation
1. Install Opam.
In Ubuntu,
apt-get install -y ocaml ocaml-native-compilers opam libtool libtool-bin libgmp-dev libffi-dev m4 libz-dev libssl-dev camlp4-extra
In OS X, use homebrew,
brew install opam
2. Install the right compiler and the right libraries:
opam pin add gga-unbounded GGA_UNBOUNDED_DIR -n
opam install gga-unbounded --deps-only
3. The tool uses Sage and Z3 as backend.
• For Sage, you should be able to start sage -python. (We used SageMath, Version 6.8).
• We use a Z3 wrapper written in Python. Visit the Z3 GitHub project.
4. Set the path variable:
export UBT_PATH=GGA_UNBOUNDED_DIR
5. To compile the tool use make. This will produce two executables:
• ubt.native to perform automated analysis.
• wsubt.native to communicate via web sockets with the interactive tool.
6. (Interactive Mode) The interactive mode uses MathJax. You can install it, by changing to the web directory, i.e., cd web/ and running make get-mathjax.
## Usage
### Input files
The description of the cryptographic construction and the security game is provided to the tool as an input file. As an example, we present the input file Chatterjee-Menezes.ubt expressing the Structure-Preserving Scheme described above and the EUC-CMA security experiment:
group_setting 3.
sample V,W.
input [V,W] in G1.
oracle o1(M:G2) =
sample R;
return [ R ] in G1,
[ R, M*V + R^2 + W] in G2.
win (wM:G2, wR1:G1, wR2:G2, wS:G2) =
(forall i: wM <> M_i /\ wR1 = wR2 /\ wS = V*wM + wR1*wR2 + W).
### Automated mode
You can use the automated algorithm by running, e.g.:
./ubt.native examples/Automatic/Chatterjee-Menezes.ubt
This will produce the following output,
simplify.
extract_coeffs.
simplify.
simplify.
case_distinction p9_j'1.
simplify.
simplify.
divide_by_param p9_i'1.
simplify.
Proven!
Time 3.49 seconds
which corresponds to the sequence of commands that the automatic algorithm executed to prove the security of the cryptographic scheme. Visit the Documentation section for details about these commands.
We note that there may be several ways to prove the statement, i.e., several sequences of commands representing a valid proof. The automatic algorithm follows a heuristic to apply the available rules, therefore, the produced result may not be the shortest or simplest existing proof.
### Interactive mode
We provide an interactive web interface where you can execute commands while watching the current state of the proof. This can be used to analyze given proofs or to produce customized ones.
To start the web interface, execute:
./wsubt.native examples/Interactive/Chatterjee-Menezes.ubt
and open the shown URL in your browser,
Open the following URL in your browser (websocket support required):
file:///GGA_UNBOUNDED_DIR/web/index.html
Files: examples/Interactive/Chatterjee-Menezes.ubt
This is a screenshot of the interactive mode (click on the image to enlarge it):
On the left side there is the code window, were you must define the security experiment and you can execute commands. On the right you can see the equations associated with the current state of the proof.
Visit the Documentation section for details about how use the interactive mode.
|
{}
|
# Setting text color in TikZ without changing line color; conflict with double
I have a certain TikZ style that I would like to use over and over again (so I'd like to use it as a scope). Namely, I want the color of the curves to be red, with white borders (to give a "crossing over" effect when the curves go over each other), and I want the color of the text to be red too. Currently I am using the "double" effect to accomplish this, like so:
\tikzset{curveinscope/.style={every path/.style={draw=white, double distance=1pt, line width=2pt, double=red, color=red}}}
However this clashes with the "double" effect: setting "color=red" to make the text color red sets the drawing color of the first line in the "double" to also be red, which messes up the effect (it needs to be white).
How can I fix this?
-
You should set the color using text=red instead of color=red:
\documentclass{article}
\usepackage{tikz}
\begin{document}
\begin{tikzpicture}
\tikzset{curveinscope/.style={every path/.style={draw=white, double distance=1pt, line width=2pt, double=red, text=red}}}
\begin{scope}[curveinscope]
\node at (1,0) {X};
\draw (0,0) -- (2,2);
\draw (2,1) -- (0,1);
\end{scope}
\end{tikzpicture}
\end{document}
-
I knew there was a text-specific colour option but tried text color first and got an error. Next time, I'll check the manual before posting. – Loop Space Jun 15 '11 at 11:30
@Andrew: Hehe, I did exactly the same thing (trying text color first, that is). Maybe there should be an alias for the option. – Jake Jun 15 '11 at 11:31
I have a memory of a text color option. Maybe it was an older version of PGF that was dropped in favour of the more concise text=colour. – Loop Space Jun 15 '11 at 11:47
Thanks a lot! I couldn't find that text=red color option in the manual. This is definitely a simple solution :-) – Bruce Bartlett Jun 15 '11 at 14:05
@Bruce: Section 16.4.1 in the manual for PGF2.10. – Loop Space Jun 15 '11 at 16:38
Separate your path-specific options from your node-specific ones. Put the node-specific ones in an every node/.style:
\documentclass{standalone}
\usepackage{tikz}
\tikzset{curve in scope/.style={
every path/.style={
draw=white,
double distance=1pt,
line width=2pt,
double=red,
},
every node/.style={
color=red
}
},
curve in scope with bad colours/.style={
every path/.style={
draw=white,
double distance=1pt,
line width=2pt,
double=red,
color=red
}
}
}
\begin{document}
\begin{tikzpicture}
\begin{scope}[curve in scope]
\path (0,0) -- node[auto] {hello world} (3,0);
\end{scope}
\begin{scope}[curve in scope with bad colours,yshift=-1cm]
\path (0,0) -- node[auto] {hello world} (3,0);
\end{scope}
\end{tikzpicture}
\end{document}
Result:
-
Thanks Andrew, this teaches me some good practice here. I didn't know about the every node option. – Bruce Bartlett Jun 15 '11 at 14:03
@Bruce: Yes, whilst Jake's is the more concise of the solutions, I think that the principle of separation is worth pointing out too. – Loop Space Jun 15 '11 at 16:37
|
{}
|
# An Efficient Soft-Margin Kernel SVM Implementation In Python
Published:
This short tutorial aims at introducing support vector machine (SVM) methods from its mathematical formulation along with an efficient implementation in a few lines of Python! Do play with the full code hosted on my github page. I strongly recommend reading Support Vector Machine Solvers (from L. Bottou & C-J. Lin) for an in-depth cover of the topic, along with the LIBSVM library. The present post naturally follows this introduction on SVMs.
## Support Vector Machines - The model
##### Figure 1: An optimal hyperplane.
Support vector machines (SVMs) are supervised learning models for classification (or regression) defined by supporting hyperplanes. SVM is one of the most widely used algorithms since it relies on strong theoretical foundations and has good performance in practice.
As illustrated on Figure 1, SVMs represent examples as points in space, mapped so that the examples of the different categories are divided by a clear gap that is as wide as possible. New examples are then mapped into that same space and assigned a category based on which side of the gap they fall.
Let's consider a dataset $D = { (\mathbf{x}_{i}, y_{i}), \mathbf{x} \in \mathbb{R}^d, y \in \{ -1, 1 \}}$ , and $\phi$ a feature mapping - a possibly non-linear function used to get features from the datapoints $\{x_i\}$.
The main idea is to find an hyperplane $\w^*$ separating our dataset and maximising the margin of this hyperplane:
$$\label{eq:hard_objective} \min _{\w, b} \mathcal{P}(\w, b) = \cfrac{1}{2} \w^2$$ $$\label{eq:hard_conditions} \text{subject to} \ \forall i \ \ y_i(\w^T \phi(\x_i)+b) \ge 1$$ The conditions (\ref{eq:hard_conditions}) enforce that datapoints (after being mapped through $\phi$) from the first (respectively second) category lie below (respectively above) the hyperplane $\w^*$, while the objective (\ref{eq:hard_objective}) maximises the margin (the distance between the hyperplane $\w^*$ and the closest example).
### Soft margin
We implicitly made the assumption that the dataset $D$ was separable: that there exists an hyperplane $\w^* \in \mathbb{R}^d$ such that all red points (i.e. $y=-1)$ lie on one side of the hyperplane (i.e. $\w^T \phi(\x_i)+b \le 0$) and blue points (y=$+1)$ lie on the other side of the hyperplane (i.e. $\w^T \phi(\x_i)+b \ge 0$). This formulation is called hard margin, since the margin cannot let some datapoints go through (all datapoints are well classified).
This assumption can be relaxed by introducing positive slack variables $\mathbf{\xi}=(\xi_1, \dots, \xi_n)$ allowing some examples to violate the margin constraints (\ref{eq:hard_conditions}). $\xi_i$ are non-zero only if $\x_i$ sits on the wrong side of the hyperplane, and is equal to the distance between $\x_i$ and the hyperplane $\w$. Then an hyperparameter $C$ controls the compromise between large margins and small margin violations.
$$\label{eq:hard_primal} \min _{\w, b, \mathbf{\xi}} \mathcal{P}(\w, b, \mathbf{\xi}) = \cfrac{1}{2} \w^2 + C \sum_{i=1}^n \xi_i \\ \text{subject to} \begin{cases} \forall i \quad y_i(\w^T \phi(\x_i)+b) \ge 1 - \xi_i \\ \forall i \quad \xi_i \ge 0 \end{cases}$$
## How to fit the model ?
Once one have such a mathematical representation of the model through an optimisation problem, the next natural question arising is how should we solve this problem (once we know it is well-posed) ? The SVM solution is the optimum of a well defined convex optimisation problem (\ref{eq:hard_primal}). Since the optimum does not depend on the manner it has been calculated, the choice of a particular optimisation algorithm can be made on the sole basis of its computational requirements.
### Dual formulation
Directly solving (\ref{eq:hard_primal}) is difficult because the constraints are quite complex. A classic move is then to simplify this problem via Lagrangian duality (see L. Bottou et al for more details), yielding the dual optimisation problem: $$\label{eq:soft_dual} \max _{\alpha} \mathcal{D}(\alpha) = \sum_{i=1}^n \alpha_i - \cfrac{1}{2} \sum_{i,j=1}^n y_i \alpha_i y_j \alpha_j \mathbf{K}(\x_i, \x_j) \\$$ $$\label{eq:soft_dual_cons} \text{subject to} \begin{cases} \forall i \quad 0 \le \alpha_i \le C \\ \sum_i y_i\alpha_i = 0 \end{cases}$$ with $\{\alpha_i\}_{i=1,\dots,n}$ being the dual coefficients to solve and $\mathbf{K}$ being the kernel associated with $\phi$: $\forall i,j \ \ \mathbf{K}(\x_i, \x_j)=\left< \phi(\x_i) , \phi(\x_j)\right>$. That problem is much easier to solve since the constraints are much simpler. Then, the direction $\w^*$ of the optimal hyperplane is recovered from a solution $\alpha^*$ of the dual optimisation problem (\ref{eq:soft_dual}-\ref{eq:soft_dual_cons}) (by forming the Lagragian and taking its minimum w.r.t. $\w$ - which is a strongly convex function): $$\w^* = \sum_{i} \alpha^*_i y_i \phi(\x_i)$$ The optimal hyperplane is therefore a weighted combination over the datapoints with non-zero dual coefficient $\alpha^*_i$. Those datapoints are therefore called support vectors, hence «support vector machines». This property is quite elegant and really useful since in practice only a few $\alpha^*_i$ are non-zeros. Hence, a new datapoint prediction only requires to evaluate: $$\text{sign}\left(\w^{*T} \phi(\x)+b\right) = \text{sign}\left(\sum_{i} \alpha^*_i y_i \phi(\x_i)^T\phi(\mathbf{x}) +b\right) = \text{sign}\left(\sum_{i} \alpha^*_i y_i \mathbf{K}(\mathbf{x}_i, \mathbf{x}) +b \right)$$
The SVM optimisation problem (\ref{eq:soft_dual}) is a Quadratic Problem (QP), a well studied class of optimisation problems for which good libraries has been developed for. This is the approach taken in this intro on SVM, relying on the Python's quadratic program solver cvxopt.
Yet this approach can be inefficient since such packages were often designed to take advantage of sparsity in the quadratic part of the objective function. Unfortunately, the SVM kernel matrix $\mathbf{K}$ is rarely sparse but sparsity occurs in the solution of the SVM problem. Moreover, the specification of a SVM problem rarely fits in memory and generic optimisation packages sometimes make extra work to locate the optimum with high accuracy which is often useless. Let's then described an algorithm tailored to efficiently solve that optimisation problem.
### The Sequential Minimal Optimisation (SMO) algorithm
One way to avoid the inconveniences above-mentioned is to rely on the decomposition method. The idea is to decompose the optimisation problem in a sequence of subproblems where only a subset of coefficients $\alpha_i$, $i \in \mathcal{B}$ needs to be optimised, while leaving the remaining coefficients $\alpha_j$, $j \notin \mathcal{B}$ unchanged: $$\label{eq:smo} \max _{\alpha'} \mathcal{D}(\alpha') = \sum_{i=1}^n \alpha'_i - \cfrac{1}{2} \sum_{i,j=1}^n y_i \alpha'_i y_j \alpha'_j \mathbf{K}(\x_i, \x_j) \\ \text{subject to} \begin{cases} \forall i \notin \mathcal{B} \quad \alpha'_i=\alpha_i \\ \forall i \in \mathcal{B} \quad 0 \le \alpha'_i \le C \\ \sum_i y_i\alpha'_i = 0 \end{cases}$$ One need to decide how to choose the working set $\mathcal{B}$ for each subproblem. The simplest is to always use the smallest possible working set, that is, two elements (such as the maximum violating pair scheme, which is discussed in Section 7.2 in Support Vector Machine Solvers ). The equality constraint $\sum_i y_i \alpha'_i = 0$ then makes this a one dimensional optimisation problem.
##### Figure 2: Direction search - from L. Bottou & C-J. Lin.
The subproblem optimisation can then be achieved by performing successive direction searches along well chosen successive directions. Such a method seeks to maximizes an optimisation problem restricted to the half line ${\mathbf{\alpha} + \lambda \mathbf{u}, \lambda \in \Lambda}$, with $\mathbf{u} = (u_1,\dots,u_n)$ a feasible direction (i.e. can slightly move the point $\mathbf{\alpha}$ along direction $\mathbf{u}$ without violating the constraints).
The equality constraint (\ref{eq:smo}) restricts $\mathbf{u}$ to the linear subspace $\sum_i y_i u_i = 0$. Each subproblem is therefore solved by performing a search along a direction $\mathbf{u}$ containing only two non zero coefficients: ${u}_i = y_i$ and ${u}_j = −y_j$.
The set $\Lambda$ of all coefficients $\lambda \ge 0$ is defined such that the point $\mathbf{\alpha} + \lambda \mathbf{u}$ satisfies the constraints. Since the feasible polytope is convex and bounded $\Lambda = [0, \lambda^{\max}]$. Direction search is expressed by the simple optimisation problem $$\lambda^* = \arg\max_{\lambda \in \Lambda}{\mathcal{D}(\mathbf{\alpha }+ \lambda \mathbf{u})}$$ Since the dual objective function is quadratic, $\mathcal{D}(\mathbf{\alpha }+ \lambda \mathbf{u})$ is shaped like a parabola. The location of its maximum $\lambda^+$ is easily computed using Newton’s formula: $$\lambda^+ = \cfrac{ \partial \mathcal{D}(\mathbf{\alpha }+ \lambda \mathbf{u}) / \partial \lambda \ |_{\lambda=0} } {\partial^2 \mathcal{D}(\mathbf{\alpha }+ \lambda \mathbf{u}) / \partial \lambda^2 \ |_{\lambda=0}} = \cfrac{\mathbf{g}^T \mathbf{u}}{\mathbf{u}^T \mathbf{H} \mathbf{u}}$$
where vector $\mathbf{g}$ and matrix $\mathbf{H}$ are the gradient and the Hessian of the dual objective function $\mathcal{D}(\mathbf{\alpha})$: $$g_i = 1 - y_i \sum_j{y_j \alpha_j K_{ij}} \quad \text{and} \quad H_{ij} = y_i y_j K_{ij}$$ Hence $\lambda^* = \max \left(0, \min \left(\lambda^{\max}, \lambda^+ \right)\right) = \max \left(0, \min \left(\lambda^{\max}, \cfrac{\mathbf{g}^T \mathbf{u}}{\mathbf{u}^T \mathbf{H} \mathbf{u}} \right)\right)$.
### Implementation
From a Python’s class point of view, an SVM model can be represented via the following attributes and methods:
Then the _compute_weights method is implemented using the SMO algorithm described above:
## Demonstration
We demonstrate this algorithm on a synthetic dataset drawn from a two dimensional standard normal distribution. Running the example script will generate the synthetic dataset, then train a kernel SVM via the SMO algorithm and eventually plot the predicted categories.
##### Optimal hyperplane with predicted labels for radial basis (left) and linear (right) kernel SVMs.
The material and code is available on my github page. I hope you enjoyed that tutorial !
### Acknowledgments
I’m grateful to Thomas Pesneau for his comments.
Tags:
|
{}
|
# M2: Milestones second funding period
• Analysis of the various data sets of BESIII above 4.1 GeV/c2 center of mass energy by one simultaneous fit to the data. This will be performed for the two reaction channels Y → J/ψπ+π- and Y → DD̅*. In order to achieve this goal, the Common Partial Wave Analysis Framework ComPWA, developed and maintained by the Helmholtz Institute Mainz, will be extended to allow for these simultaneous fits.
• A search for the isospin singlet states decaying to J/ψη and ηcη will be performed.
• A detailed investigation of the structures in the D recoil spectrum will be performed.
• The full BESIII data sets taken on the \jpsi and \psip resonances will be analyzed to derive input data for an amplitude analysis of the decays to \pions. Starting from a simple isobar analysis, we will test more advanced models involving e.g. rescattering effects.
• Analysis of ω → \piz \epem, ω → ηγ with A2 data.
• Dalitz plot analysis of $\etapr \to 3 \piz$ with A2 data.
• Consistent quantum-field theoretical description of the PVγ system. By extending Dirac's Hamiltonian method to also include interactions with pseudoscalars and by combining it with renormalizability in the sense of effective field theory, we will derive constraints for the coupling constants of the most general Lagrangian. The unstable nature of vector mesons will be treated in the framework of the complex-mass scheme.
• Quantum corrections to the chiral anomaly in the processes η' → π+π-γ and η'→ ππγ, making use of the recently developed ηη' mixing scheme.
• We will organize a workshop together with project S3 in which common partial wave analysis techniques will be discussed.
|
{}
|
It has been a puzzling question why several organisms reproduce sexually. Fisher and Muller hypothesized that reproducing by sex can speed up the evolution. They explained that in the sexual reproduction, recombination can combine beneficial alleles that lie on different chromosomes, which speeds up the time that those beneficial alleles spread to the entire population. We consider a population model of fixed size N, in which we will focus on two loci on a chromosome. Each allele at each locus can mutate into a beneficial allele at rate \mu_N. The individuals with 0, 1, and 2 beneficial alleles die at rates 1, 1-s_N and 1-2s_N respectively. When an individual dies, with probability 1-r_N, the new individual inherits both alleles from one parent, chosen at random from the population, while with probability r_N, recombination occurs, and the new individual receives its two alleles from different parents. Under certain assumptions on the parameters N, \mu_N, s_N$and$r_N, we obtain an asymptotic approximation for the time that both beneficial alleles spread to the entire population. When the recombination probability is small, we show that recombination does not speed up the time that the two beneficial alleles spread to the entire population, while when the recombination probability is large, we show that recombination decreases the time, which agrees with Fisher-Muller hypothesis, and confirms the advantage of sexual reproduction.
|
{}
|
# Logarithms Tutorial
The logarithm is perhaps the single, most useful arithmetic concept in all the sciences; and an understanding of them is essential to an understanding of many scientific ideas. Logarithms may be defined and introduced in several different ways. But for our purposes, let's adopt a simple approach. This approach originally arose out of a desire to simplify multiplication and division to the level of addition and subtraction. Of course, in this era of the cheap hand calculator, this is not necessary anymore but it still serves as a useful way to introduce logarithms. The question is, therefore:
Is there any operation in mathematics which produces a multiplication by the performance of an addition?
With not too much thought, the answer should come to you.
What is $2^3 \times 2^4$.
The answer is $2^7$ which is obtained by adding the powers 3 and 4. This is correct, of course, since $2^3 \times 2^4$ is just seven 2s multiplied together. Note that this addition trick does not work for the case of $3^3 \times 2^4$. The base numbers must be the same, as in the first case, where we used 2.
In general, this addition trick can be written as $p^a \times p^b = p^{a+b}$. This expression will do our job of multiplying any two numbers, say $1.3$ and $6.9$, if we can only express $1.3$ as $p^a$ and $6.9$ as $p^b$.
What number will we use for the base $p$? Any number will do, but traditionally, only two are in common use:
Ten ($10$) and the transcendental number $e (= 2.71828...)$, giving logarithms to the base $10$ or common logarithms ($\log$), and logarithms to the base $e$ or natural logarithms ($ln$).
#### If you would like to know why this strange number $e$ is used click here.
The two bases for logarithms in common use are a) $10$ and b) the transcendental number $e = 2.71828---$
• a) For ordinary computations, logarithms to the base $10$ are most common. This is the common or Briggsian system first devised by Henry Briggs (1560-1631) with the assistance of Napier.
• b) Logarithms to the base e are called Naperian or natural logarithms for their inventor John Napier (1550-1617). For general numerical computation they have no advantage over common logarithms. Their origin and usefulness must be sought in more sophisticated mathematics than just multiplication and division of numbers and is connected to the special properties of the number e. Some of these properties are:
1. The exponential function ex is the only one that has itself as a derivative i.e.,
$de^x/dx = e^x$
2. The simple but very important differential equation expressing constant percentage growth with time $t$
$dx/x = (const)dt$
has the solution
$x = x_0e^{(const)t}$
or taking natural logarithms
$ln(x/x_0) = (const)t$
3. The numerical values of logarithms are determined by summing the terms of an infinite series. The series for $e$ is
$e = 1 + 1/1! + 1/2! + 1/3! + 1/4! + ----$
[Remember $n!$ is n-factorial or $(n)X(n-1)X(n-2)X---(1)$]
For e raised to the power $x$ it is
$e^x = 1 + x + x^2/2! + x^3/3! + ---$
Evaluating numbers in this way is labour intensive so series that converge rapidly are the best. The most rapidly converging series is that for $ln$ (i.e., to the base $e$). Any other base logarithm (e.g., to base $b$) can be found from the simple relation
$\log _ba = ln\; a/ln \; b$
Therefore once the most efficient set of logarithms have been found any others follow from a simple ratio.
Let's first talk about logarithms to the base $10$ or common logs. We thus choose to let our number $1.3$ be equal to $10a$.
$1.3 = 10a$
$a$' is called "the logarithm of $1.3$". How large is $a$'? Well, it's not $0$ since $100 = 1$ and it's less than $1$ since $10^1 = 10$. Therefore, we see that all numbers between $1$ and $10$ have logarithms between $0$ and $1.$ If you look at the table below you'll see a summary of this.
Number Range Logarithm Range
$1 - 10$ or $10^0 - 10^1$ $0 - 1$
$10 - 100$ or $10^1 - 10^2$ $1 - 2$
$100 -1000$ or $10^2 - 10^3$ $2 - 3$
etc. etc.
You see, we have the number range listed on the left and the logarithm range listed on the right. For numbers between $1$ and $10$, that is between $10^0$ and $10^1$, the logarithm lies in the range $0$ to $1.$ For numbers between $10$ and $100$, that is between $10^1$ and $10^2$, the logarithm lies in the range $1$ to $2$, and so on. Now in the bad old days before calculators, you would have to learn to use a set of logarithm tables to find the logarithm of our number, $1.3$, that we asked for earlier. But nowadays, you can get it at the press of a button on your calculator.
I'm going to take a moment to discuss your calculator. If you don't have a calculator with scientific functions on it, you should get one before proceeding in this tutorial since the rest depends on it.
Most calculators are very straightforward in obtaining the logarithm. They either have two logarithm keys or a dual function key. In any case, the labels will be $\log$' and $ln$' which is often pronounced lon'.
Log is the key for logs to the base $10$ and $ln$ is for natural logs. We want logs to the base $10$ in our example so we use $\log$'. Enter $1.3$ on your calculator, and then press the log key.
Do you have $0.113943$? You should have. This number then is $a$' back in our previous expression and therefore the logarithm of $1.3.$ Pause now and determine $b$' in that expression, the logarithm of $6.9.$
You should have $0.838849$ for the log of $6.9.$ If not, review what we have done and try again. Now we are going to do something silly in view of the fact that you have a calculator. We're going to use the two logarithms you have evaluated to find the product of $1.3$ and $6.9.$ Of course, you can do it quickly with your calculator, but this will show that logarithms do what they are supposed to do. According to our original idea, the sum of the two logarithms was supposed to be the logarithm of the answer.
Now add the two logarithms. The sum is $0.952792.$ This is the logarithm of the answer. If we only knew what number had $0.952792$ as its logarithm, we would know the value of $1.3 \times 6.9.$
The problem of finding a number when you know its logarithm is called finding the "antilogarithm" or sometimes "exponentiation". Again, lets look at your calculator. Here is where calculators differ a lot and I hope I mention one that is something like yours.
You should look for a key on your calculator that says something like $10^x$ or $10^y$. If so, then pressing that key will take the antilog of the number in the display. Alternatively, your calculator may have an "inverse" key. If so, then pressing inverse and then log will take the antilog of the number in the display. Enter $0.952792$ into your calculator and find the antilog.
Did you get $8.97$? If not, try again. Of course, you might have got something like $8.96999$ but of course that really is $8.97.$ Now, multiply $1.3 \times 6.9$ on your calculator and you'll see that $8.97$ is indeed the correct answer.
The whole operation could be done with natural logarithms as well as shown below.
$1.3 \times 6.9 = ?$
$ln 1.3 = 0.262364$ i.e. $1.3 = e^{0.262364}$
$ln 6.9 = 1.931521$ i.e. $6.9 = e^{1.931521}$
$total = 2.193885$ i.e. $1.3 \times 6.9 = e^{2.193885}$
$antiln 2.193885 = 8.97$
If the sum of logarithms gives the product of two numbers, then the difference gives the quotient.
In the instance below, I've taken the difference between the ln of $1.3$ and the ln of $6.9.$ Check it on your calculator.
$1.3/6.9 = ?$
$ln 1.3 = 0.262364$
$ln 6.9 = 1.931521$
$ln 1.3/6.9 = -1.669157$
$antiln (-1.669157) = 0.1884$
Don't be afraid of the negative sign. It simply means that the answer is less than $1.$ Enter $-1.669157$ on your calculator, then find its antiln. Note we are working with natural logs in this example.
If you didn't get $0.1884$, try again. Of course, this is just $1.3$ divided by $6.9.$ In the instance below, I have done the whole problem over again using common logs. Pause here and check it.
$\log 1.3 = 0.113943$
$\log 6.9 = 0.838849$
$\log(1.3/6.9) = -0.724906$
$antilog (-0.724906) = 0.1884$
The logarithmic and exponential functions are very important since many physical and biological processes can be described by them.
For example, suppose you have a certain number of radioactive atoms at time $t = 0.$ Let's let this number be N0. Radioactivity behaves in such a way that the number $N$ of radioactive atoms remaining at a later time $t$ is given by a linear variation of the logarithm of $N$ with $t.$
That is, a graph of $ln$ $N$ vs $t$ is a straight line. You know that the equation of such a straight line is given by $y = mx+b$ where $m$ is the slope and $b$ is the $y$ intercept. Therefore, the equation of radioactivity is $ln N = -kt + ln N_0$ where $ln N_0$ is the y intercept and the slope of the line is $-k$.
Let's now examine the equation we derived for radioactivity, $ln$ $N = ln N_0 -kt$. Here is where it is important to be able to do algebra with logarithms. Lets get the logarithms on one side so that we get
$ln N - ln N_0 = -kt.$
But we know that the difference of logarithms is the logarithm of the quotient so the left-hand side becomes ln $N/N_0$. Now let's take antilogarithms and do the right side first. The antiln of any quantity is the number $e$ to the power of that quantity so the right-hand side becomes $e^{-kt}$. The left-hand side is the antiln of the ln and so it just becomes $N$ divided by $N_0$. Finally, we can rearrange to put the final equation in the form
$N = N_0 e^{-kt}$
which is called the equation of "exponential decay", so you can see why taking an antiln is often called "exponentiation".
## Quiz
1. $ln (7.42) = ?$
2. $\log (7.42) = ?$
3. $ln (ekt) = ?$
4. $\log (ekt) = ?$
5. $\text{antilog} \; 0.8704 = ?$
6. $\text{antiln} \; 2.0042 = ?$
7. $100.8704 = ?$
8. $e2.0042 = ?$
3. $kt$
4. $kt \log e = 0.4343kt$
|
{}
|
WIKISKY.ORG
Home Getting Started To Survive in the Universe News@Sky Astro Photo The Collection Forum Blog New! FAQ Press Login
# NGC 6886
Contents
### Images
DSS Images Other Images
### Related articles
An Atlas of [N II] and [O III] Images and Spectra of Planetary NebulaeWe present an atlas of Hubble Space Telescope images and ground-based,long-slit, narrowband spectra centered on the 6584 Å line of [NII] and the 5007 Å line of [O III]. The spectra were obtained fora variety of slit positions across each target (as shown on the images)in an effort to account for nonspherical nebular geometries in a robustmanner. We have extended the prolate ellipsoidal shell model originallydevised by Aaquist, Zhang, and Kwok to generate synthetic images, aswell as long-slit spectra. Using this model, we have derived basicparameters for the subsample of PNe that present ellipsoidal appearancesand regular kinematic patterns. We find differences between ourparameters for the target PNe as compared to those of previous studies,which we attribute to increased spatial resolution for our image dataand the inclusion of kinematic data in the model fits. The data andanalysis presented in this paper can be combined with detections ofnebular angular expansion rates to determine precise distances to the PNtargets. The Abundances of Light Neutron-Capture Elements in Planetary Nebulae. I. Photoionization Modeling and Ionization CorrectionsWe have conducted a large-scale survey of 120 planetary nebulae (PNe) tosearch for the near-infrared emission lines [Kr III] 2.199 μm and [SeIV] 2.287 μm. The neutron (n)-capture elements Se and Kr may beenriched in a PN if its progenitor star experienced s-processnucleosynthesis and third dredge-up. In order to determine Se and Krabundances, we have added these elements to the atomic databases of thephotoionization codes Cloudy and XSTAR, which we use to deriveionization correction factors (ICFs) to account for the abundances ofunobserved Se and Kr ions. However, much of the atomic data governingthe ionization balance of these two elements are unknown, and have beenapproximated from general principles. We find that uncertainties in theatomic data can lead to errors approaching 0.3 dex in the derived Seabundances and up to 0.2-0.25 dex for Kr. To reduce the uncertainties inthe Kr ionization balance stemming from the approximate atomic data, wehave modeled 10 bright PNe in our sample, selected because they exhibitemission lines from multiple Kr ions in their optical and near-infraredspectra. We have empirically adjusted the uncertain Kr atomic data untilthe observed line intensities of the various Kr ions are adequatelyreproduced by our models. Using the adjusted Kr atomic data, we havecomputed a grid of models over a wide range of physical parameters(central star temperature, nebular density, and ionization parameter)and derived formulae that can be used to compute Se and Kr ICFs. In thesecond paper of this series, we will apply these ICFs to our full sampleof 120 PNe, which comprises the first large-scale survey of n-captureelements in PNe.This paper includes data taken at the McDonald Observatory of theUniversity of Texas at Austin. Planetary nebulae abundances and stellar evolutionA summary is given of planetary nebulae abundances from ISOmeasurements. It is shown that these nebulae show abundance gradients(with galactocentric distance), which in the case of neon, argon, sulfurand oxygen (with four exceptions) are the same as HII regions and earlytype star abundance gradients. The abundance of these elements predictedfrom these gradients at the distance of the Sun from the center areexactly the solar abundance. Sulfur is the exception to this; the reasonfor this is discussed. The higher solar neon abundance is confirmed;this is discussed in terms of the results of helioseismology. Evidenceis presented for oxygen destruction via ON cycling having occurred inthe progenitors of four planetary nebulae with bilobal structure. Theseprogenitor stars had a high mass, probably greater than 5 Mȯ. Thisis deduced from the high values of He/H and N/H found in these nebulae.Formation of nitrogen, helium and carbon are discussed. The high massprogenitors which showed oxygen destruction are shown to have probablydestroyed carbon as well. This is probably the result of hot bottomburning. Bubbles in planetary nebulae and clusters of galaxies: jet bendingWe study the bending of jets in binary stellar systems. A compactcompanion accretes mass from the slow wind of the mass-losing primarystar, forms an accretion disc and blows two opposite jets. These fastjets are bent by the slow wind. Disregarding the orbital motion, we findthe dependence of the bending angle on the properties of the slow windand the jets. Bending of jets is observed in planetary nebulae which arethought to be the descendants of interacting binary stars. For example,in some of these planetary nebulae, the two bubbles (lobes) which areinflated by the two opposite jets are displaced to the same side of thesymmetry axis of the nebula. Similar displacements are observed inbubble pairs in the centre of some clusters and groups of galaxies. Wecompare the bending of jets in binary stellar systems with that inclusters of galaxies. The structure of planetary nebulae: theory vs. practiceContext.This paper is the first in a short series dedicated to thelong-standing astronomical problem of de-projecting the bi-dimensional,apparent morphology of a three-dimensional mass of gas. Aims.Wefocus on the density distribution in real planetary nebulae (and alltypes of expanding nebulae). Methods. We introduce some basictheoretical notions, discuss the observational methodology, and developan accurate procedure for determining the matter radial profile withinthe sharp portion of nebula in the plane of the sky identified by thezero-velocity-pixel-column (zvpc) of high-resolution spectral images.Results. The general and specific applications of the method (andsome caveats) are discussed. Moreover, we present a series of evolutivesnapshots, combining illustrative examples of both model and trueplanetary nebulae. Conclusions. The zvpc radial-densityreconstruction - added to tomography and 3D recovery developed at theAstronomical Observatory of Padua (Italy) - constitutes a very usefultool for looking more closely at the spatio-kinematics, physicalconditions, ionic structure, and evolution of expanding nebulae. The importance of soft X-rays for the excitation of H2 emission in planetary nebulaeWe note that H2 emitting planetary nebulae tend to haveZanstra temperatures TZ(HeII) > 90 kK. This is shown to beconsistent with a large evolutionary lifetime, and the kinematic ages ofthe envelopes. Non-local thermodynamic equilibrium stellar atmosphericmodelling also shows that levels of soft X-ray emission increase morerapidly than has previously been assumed, and are preferentially largein H2 emitting sources. It is suggested that this may holdthe key to explaining the strengths of the H2 transitions. The Use of K_S Band Photometric Excesses to Investigate H(2) Emission in Planetary NebulaeWe have determined the distribution of H(2) emission in 14 planetarynebulae (PNe), using imaging and photometry published by the 2MASSinfrared survey. This technique is only applicable under certainstringent conditions, and requires precise broad band photometry, andaccurate spatial registration between the K_S and H band images. It is,in addition, only applicable to certain sources, and excludes outflowsin which central star and grain thermal excesses are appreciable. Ourresults for NGC 3132, NGC 6720, IC 4406 and M 2-9 are closely similar tothose of previous narrow band imaging, and confirm that H(2) emissionis confined to narrow, highly fragmented shells. Similar results areobtained for M 1-7, M 1-8, and M 3-5. Our spatial profiles also confirmthat the emission extends outside of the primary ionised shells. Whereenvelopes are large, and the PNe are more evolved, then the fractionalextensions Deltatheta/$theta appear to be at their smallest. They arealso similar to the radial widths predicted for H(2) abundanceprofiles, and to the values DeltaR/R determined throughmagnetohydrodynamic modelling of shocks. There appears, finally, to beevidence for an evolution in this parameter, such that Deltatheta/$thetavaries with increasing envelope size d(H) as Deltatheta/\$theta ~d(H)(-2.2) . Morpho-Kinematic Modeling of Gaseous Nebulae with SHAPEWe present a powerful new tool to analyse and disentangle the 3-Dgeometry and kinematic structure of gaseous nebulae. The method consistsin combining commercially available digital animation software tosimulate the 3-D structure and expansion pattern of the nebula with adedicated, purpose-built rendering software that produces the finalimages and long slit spectra which are compared to the real data. Weshow results for the complex planetary nebulae NGC 6369 and Abell 30based on long slit spectra obtained at the San Pedro MártirObservatory. Morpho-kinematic modelling of gaseous nebulae with ShapeWe present a powerful new tool to disentangle the 3-D geometry andkinematic structure of gaseous nebulae. The method consists of combiningcommercially available digital animation software to simulate the 3-Dstructure and expansion pattern of the nebula with a dedicated, purposebuilt rendering software that produces the final images and long slitspectra which are compared to the real data. In this contribution weshow results for the complex planetary nebula NGC369 based on long slitspectra obtained at the San Pedro Mártir observatory. Polycyclic aromatic hydrocarbon emission bands in selected planetary nebulae: a study of the behaviour with gas phase C/O ratioAirborne and space-based low-resolution spectroscopy in the 1980sdiscovered tantalizing quantitative relationships between the gas phaseC/O abundance ratio in planetary nebulae (PNe) and the fractions oftotal far-infrared (FIR) luminosity radiated by the 7.7- and 11.3-μmbands (the C = C stretch and C-H bend, respectively), of polycyclicaromatic hydrocarbons (PAHs). Only a very small sample of nebulae wasstudied in this context, limited by airborne observations of the7.7-μm band, or the existence of adequate IRAS Low ResolutionSpectrometer data for the 11.3-μm band. To investigate these trendsfurther, we have expanded the sample of planetaries available for thisstudy using Infrared Space Observatory (ISO) low-resolution spectrasecured with the Short Wavelength Spectrometer and the Long WavelengthSpectrometer. The new sample of 43 PNe, of which 17 are detected in PAHemission, addresses the range from C/O = 0.2-13 with the objective oftrying to delineate the pathways by which carbon dust grains might haveformed in planetaries. For the 7.7-μm and 11.3-μm bands, weconfirm that the ratio of band strength to total infrared (IR)luminosity is correlated with the nebular C/O ratio. Expressed inequivalent width terms, the cut-on C/O ratio for the 7.7-μm band isfound to be 0.6+0.2-0.4, in good accord with thatfound from sensitive ground-based measurements of the 3.3-μ band. The 3-D shaping of NGC 6741: A massive, fast-evolving Planetary Nebula at the recombination-reionization edgeWe infer the gas kinematics, diagnostics and ionic radial profiles,distance and central star parameters, nebular photo-ionization model,spatial structure and evolutionary phase of the Planetary Nebula NGC6741 by means of long-slit ESO NTT+EMMI high-resolution spectra at nineposition angles, reduced and analysed according to the tomographic and3-D methodologies developed at the Astronomical Observatory of Padua(Italy). NGC 6741 (distance≃2.0 kpc, age≃ 1400 yr, ionizedmass Mion≃ 0.06 Mȯ) is a dense(electron density up to 12 000 cm-3), high-excitation,almost-prolate ellipsoid (0.036 pc × 0.020 pc × 0.018 pc,major, intermediate and minor semi-axes, respectively), surrounded by asharp low-excitation skin (the ionization front), and embedded in aspherical (radius≃ 0.080 pc), almost-neutral, high-density (n(HI)≃ 7 ×103 atoms cm-3) halo containinga large fraction of the nebular mass (Mhalo≥ 0.20Mȯ). The kinematics, physical conditions and ionicstructure indicate that NGC 6741 is in a deep recombination phase,started about 200 years ago, and caused by the rapid luminosity drop ofthe massive (M*=0.66{-}0.68 Mȯ), hot (logT* ≃ 5.23) and faint (logL*/Lȯ ≃ 2.75) post-AGB star, which hasexhausted the hydrogen-shell nuclear burning and is moving along thewhite dwarf cooling sequence. The general expansion law of the ionizedgas in NGC 6741, Vexp(km s-1)=13 × R arcsec,fails in the innermost, highest-excitation layers, which move slowerthan expected. The observed deceleration is ascribable to the luminositydrop of the central star (the decreasing pressure of the hot-bubble nolonger balances the pressure of the ionized gas), and appears instriking contrast to recent reports inferring that acceleration is acommon property of the Planetary Nebulae innermost layers. A detailedcomparative analysis proves that the "U"-shaped expansion velocity fieldis a spurious, incorrect result due to a combination of: (a) simplisticassumptions (spherical shell hypothesis for the nebula); (b) unfitreduction method (emission profiles integrated along the slit); and (c)inappropriate diagnostic choice (λ4686 Å of He II, i.e. athirteen fine-structure components recombination line). Some generalimplications for the shaping mechanisms of Planetary Nebulae arediscussed. The Chemical Composition of Galactic Planetary Nebulae with Regard to Inhomogeneity in the Gas Density in Their EnvelopesThe results of a study of the chemical compositions of Galacticplanetary nebulae taking into account two types of inhomogeneity in thenebular gas density in their envelopes are reported. New analyticalexpressions for the ionization correction factors have been derived andare used to determine the chemical compositions of the nebular gas inGalactic planetary nebulae. The abundances of He, N, O, Ne, S, and Arhave been found for 193 objects. The Y Z diagrams for various Heabundances are analyzed for type II planetary nebulae separately andjointly with HII regions. The primordial helium abundance Y p andenrichment ratio dY/dZ are determined, and the resulting values arecompared with the data of other authors. Radial abundance gradients inthe Galactic disk are studied using type II planetary nebulae. Abundances in planetary nebulae: NGC 6886ISO and IUE spectra of the round-shaped planetary nebula NGC6886 are combined with spectra in the visual wavelength regiontaken from the literature to obtain a complete, extinction correctedspectrum from ultraviolet to infrared wavelengths. The characteristicsof the nebula and its central star are determined by various methodsincluding photoionization modeling using Cloudy. The results of themodeling are checked against the observational data and compared tothose derived from a more classical abundance determination approach.The abundances determined are found to differ substantially from earlierresults although the observations used are essentially the same, exceptfor the inclusion of the ISO results. The reasons for this differenceare discussed. Finally, the main results are interpreted in terms of theevolutionary stage of NGC 6886 and its central star.Based on observations with ISO, an ESA project with instruments fundedby ESA Member States (especially the PI countries: France, Germany, TheNetherlands and the UK) and with the participation of ISAS and NASA.This work has also used IUE and HST archival data. A reexamination of electron density diagnostics for ionized gaseous nebulaeWe present a comparison of electron densities derived from opticalforbidden line diagnostic ratios for a sample of over a hundred nebulae.We consider four density indicators, the [O II]λ3729/λ3726, [S II] λ6716/λ6731, [Cl III]λ5517/λ5537 and [Ar IV] λ4711/λ4740 doubletratios. Except for a few H II regions for which data from the literaturewere used, diagnostic line ratios were derived from our own high qualityspectra. For the [O II] λ3729/λ3726 doublet ratio, we findthat our default atomic data set, consisting of transition probabilitiesfrom Zeippen (\cite{zeippen1982}) and collision strengths from Pradhan(\cite{pradhan}), fit the observations well, although at high electrondensities, the [O II] doublet ratio yields densities systematicallylower than those given by the [S II] λ6716/λ6731 doubletratio, suggesting that the ratio of transition probabilities of the [OII] doublet, A(λ3729)/A(λ3726), given by Zeippen(\cite{zeippen1982}) may need to be revised upwards by approximately 6per cent. Our analysis also shows that the more recent calculations of[O II] transition probabilities by Zeippen (\cite{zeippen1987a}) andcollision strengths by McLaughlin & Bell (\cite{mclaughlin}) areinconsistent with the observations at the high and low density limits,respectively, and can therefore be ruled out. We confirm the earlierresult of Copetti & Writzl (\cite{copetti2002}) that the [O II]transition probabilities calculated by Wiese et al. (\cite{wiese}) yieldelectron densities systematically lower than those deduced from the [SII] λ6716/λ6731 doublet ratio and that the discrepancy ismost likely caused by errors in the transition probabilities calculatedby Wiese et al. (\cite{wiese}). Using our default atomic data set for [OII], we find that Ne([O II]) Ne([S II]) ≈Ne([Cl III])< Ne([Ar IV]). Planetary nebula distances re-examined: an improved statistical scaleThe distances of planetary nebulae (PNe) are still quite uncertain.Although observational estimates are available for a small proportion ofPNe, based on statistical parallax and the like, such distances are verypoorly determined for the majority of galactic PNe. In particular,estimates of so-called statistical' distance appear to differ byfactors of ~2.7.We point out that there is a well-defined correlation between the 5-GHzluminosity of the sources, L5, and their brightnesstemperatures, TB. This represents a different trend to thoseinvestigated in previous statistical analyses, and permits us todetermine independent distances to a further 449 outflows. Thesedistances are shown to be closely comparable to those determined using aTB-R correlation, providing that the latter trend is taken tobe non-linear.This non-linearity in the TB-R plane has not been noted inprevious analyses, and is likely responsible for the broad (andconflicting) ranges of distance that have previously been published.Finally, we point out that there is a close accord between observedtrends within the L5-TB and TB-Rplanes, and the variation predicted through nebular evolutionarymodelling. This is used to suggest that observational biases areprobably modest, and that our revised distance scale is reasonablytrustworthy. On the O II Ground Configuration Energy LevelsThe most accurate way to measure the energy levels for the O II2p3 ground configuration has been from the forbidden lines inplanetary nebulae. We present an analysis of modern planetary nebuladata that nicely constrain the splitting within the 2D termand the separation of this term from the ground4S3/2 level. We extend this method to H II regionsusing high-resolution spectroscopy of the Orion Nebula, covering all sixvisible transitions within the ground configuration. These data confirmthe splitting of the 2D term while additionally constrainingthe splitting of the 2P term. The energies of the2P and 2D terms relative to the ground(4S) term are constrained by requiring that all six linesgive the same radial velocity, consistent with independent limits placedon the motion of the O+ gas and the planetary nebula data. Wind accretion by a binary stellar system and disc formationI calculate the specific angular momentum of mass accreted by a binarysystem embedded in the dense wind of a mass-losing asymptotic giantbranch star. The accretion flow is of the Bondi-Hoyle-Lyttleton type.For most of the space of the relevant parameters the flow is basicallyan isothermal high Mach number accretion flow. I find that when theorbital plane of the accreting binary system and the orbital plane ofthe triple system are not parallel to each other, the accreted mass onto one or two of the binary system components has high specific angularmomentum. For a large fraction of triple-star systems, accretion discswill be formed around one or two of the stars in the binary system,provided that the mass ratio of the two stars in the accreting binarysystem is >~0.5. Such discs may blow jets which shape the descendantplanetary nebula (PN). The axis of jets will be almost parallel to theorbital plane of the triple-star system. One jet is blown outwardrelative to the wind, while the other jet passes near the mass-losingstar, and is more likely to be slowed down or deflected. I find thatduring the final asymptotic giant branch phase, when the mass-loss rateis very high, an accretion disc may form for orbital separation betweenthe accreting binary systems and the mass-losing star of up to ~400-800au. I discuss the implications for the shape of the descendant PN, andlist several PN which may have been shaped by an accreting binary-starsystem, i.e. by a triple-star system. Flux Ratio [Nev] 14.3/24.3 as a Test of Collision StrengthsFrom ISO [Nev] 14.3/24.3 μm line flux ratios, we find that 10 out of20 planetary nebulae (PNs) have measured ratios below the low-electrondensity (Ne) theoretical predicted limit. Such astronomicaldata serve to provide important tests of atomic data, collisionstrengths in this case. In principle, well-calibrated measurements ofthe [Nev] 14.3/24.3 flux ratio could improve upon the existing atomicdata. A reanalysis of chemical abundances in galactic PNe and comparison with theoretical predictions New determinations of chemical abundances for He, N, O, Ne, Ar and Sare derived for all galactic planetary nebulae (PNe) so far observedwith a relatively high accuracy, in an effort to overcome differences inthese quantities obtained over the years by different authors usingdifferent procedures. These include: ways to correct for interstellarextinction, the atomic data used to interpret the observed line fluxes,the model nebula adopted to represent real objects and the ionizationcorrections for unseen ions. A unique good quality' classical-typeprocedure, i.e. making use of collisionally excited forbidden lines toderive ionic abundances of heavy ions, has been applied to allindividual sets of observed line fluxes in each specific position withineach PN. Only observational data obtained with linear detectors, andsatisfying some quality' criteria, have been considered. Suchobservations go from the mid-1970s up to the end of 2001. Theobservational errors associated with individual line fluxes have beenpropagated through the whole procedure to obtain an estimate of theaccuracy of final abundances independent of an author's prejudices'.Comparison of the final abundances with those obtained in relevantmulti-object studies on the one hand allowed us to assess the accuracyof the new abundances, and on the other hand proved the usefulness ofthe present work, the basic purpose of which was to take full advantageof the vast amount of observations done so far of galactic PNe, handlingthem in a proper homogeneous way. The number of resulting PNe that havedata of an adequate quality to pass the present selection amounts to131. We believe that the new derived abundances constitute a highlyhomogeneous chemical data set on galactic PNe, with realisticuncertainties, and form a good observational basis for comparison withthe growing number of predictions from stellar evolution theory. Owingto the known discrepancies between the ionic abundances of heavyelements derived from the strong collisonally excited forbidden linesand those derived from the weak, temperature-insensitive recombinationlines, it is recognized that only abundance ratios between heavyelements can be considered as satisfactorily accurate. A comparison withtheoretical predictions allowed us to assess the state of the art inthis topic in any case, providing some findings and suggestions forfurther theoretical and observational work to advance our understandingof the evolution of low- and intermediate-mass stars. Sulfur, Chlorine, and Argon Abundances in Planetary Nebulae. IV. Synthesis and the Sulfur AnomalyWe have compiled a large sample of O, Ne, S, Cl, and Ar abundances thathave been determined for 85 Galactic planetary nebulae in a consistentand homogeneous manner using spectra extending from 3600 to 9600Å. Sulfur abundances have been computed using the near-IR lines of[S III] λλ9069, 9532 along with [S III] temperatures. Wefind average values, expressed logarithmically with a standarddeviation, of log(S/O)=-1.91+/-0.24, log(Cl/O)=-3.52+/-0.16, andlog(Ar/O)=-2.29+/-0.18, numbers consistent with previous studies of bothplanetary nebulae and H II regions. We also find a strong correlationbetween [O III] and [S III] temperatures among planetary nebulae. Inanalyzing abundances of Ne, S, Cl, and Ar with respect to O, we find atight correlation for Ne-O, and loose correlations for Cl-O and Ar-O.All three trends appear to be colinear with observed correlations for HII regions. S and O also show a correlation, but there is a definiteoffset from the behavior exhibited by H II regions and stars. We suggestthat this S anomaly is most easily explained by the existence ofS+3, whose abundance must be inferred indirectly when onlyoptical spectra are available, in amounts in excess of what is predictedby model-derived ionization correction factors in PNe. Finally for thedisk PNe, abundances of O, Ne, S, Cl, and Ar all show gradients whenplotted against Galactocentric distance. The slopes are statisticallyindistinguishable from one another, a result which is consistent withthe notion that the cosmic abundances of these elements evolve inlockstep. The relation between Zanstra temperature and morphology in planetary nebulaeWe have created a master list of Zanstra temperatures for 373 galacticplanetary nebulae based upon a compilation of 1575 values taken from thepublished literature. These are used to evaluate mean trends intemperature for differing nebular morphologies. Among the most prominentresults of this analysis is the tendency forη=TZ(HeII)/TZ(HeI) to increase with nebularradius, a trend which is taken to arise from the evolution of shelloptical depths. We find that as many as 87 per cent of nebulae may beoptically thin to H ionizing radiation where radii exceed ~0.16 pc. Wealso note that the distributions of values η and TZ(HeII)are quite different for circular, elliptical and bipolar nebulae. Acomparison of observed temperatures with theoretical H-burning trackssuggests that elliptical and circular sources arise from progenitorswith mean mass ≅ 1 Msolar(although the elliptical progenitors are probably more massive).Higher-temperature elliptical sources are likely to derive fromprogenitors with mass ≅2 Msolar, however, implying thatthese nebulae (at least) are associated with a broad swathe ofprogenitor masses. Such a conclusion is also supported by trends in meangalactic latitude. It is found that higher-temperature ellipticalsources have much lower mean latitudes than those with smallerTZ(HeII), a trend which is explicable where there is anincrease in with increasing TZ(HeII).This latitude-temperature variation also applies for most other sources.Bipolar nebulae appear to have mean progenitor masses ≅2.5Msolar, whilst jets, Brets and other highly collimatedoutflows are associated with progenitors at the other end of the massrange (~ 1 Msolar). Indeed it ispossible, given their large mean latitudes and low peak temperatures,that the latter nebulae are associated with the lowest-mass progenitorsof all.The present results appear fully consistent with earlier analyses basedupon nebular scale heights, shell abundances and the relativeproportions of differing morphologies, and offer further evidence for alink between progenitor mass and morphology. Near-infrared spectroscopy of (proto)-planetary nebulae: molecular hydrogen excitation as an evolutionary tracerWe present an in-depth analysis of molecular excitation in 11H2-bright planetary and protoplanetary nebulae (PN and PPN).From newly acquired K-band observations, we extract a number of spectraat positions across each source. H2 line intensities areplotted on column density ratio' diagrams so that we may examine theexcitation in and across each region. To achieve this, we combine theshock models of Smith, Khanzadyan & Davis with the photodissociationregion (PDR) models of Black & van Dishoeck to yield ashock-plus-fluorescence fit to each data set.Although the combined shock + fluorescence model is needed to explainthe low- and high-energy H2 lines in most of the sourcesobserved (fluorescence accounts for much of the emission from thehigher-energy H2 lines), the relative importance of shocksover fluorescence does seem to change with evolutionary status. We findthat shock excitation may well be the dominant excitation mechanism inthe least evolved PPN (CRL 2688 - in both the bipolar lobes and in theequatorial plane) and in the most evolved PN considered (NGC 7048).Fluorescence, on the other hand, becomes more important at intermediateevolutionary stages (i.e. in young' PN), particularly in the inner coreregions and along the inner edges of the expanding post-asymptotic giantbranch (AGB) envelope. Since H2 line emission seems to beproduced in almost all stages of post-AGB evolution, H2excitation may prove to be a useful probe of the evolutionary status ofPPN and PN alike. Moreover, shocks may play an important role in themolecular gas excitation in (P)PN, in addition to the low- and/orhigh-density fluorescence usually attributed to the excitation in thesesources. Galactic Planetary Nebulae and their central stars. I. An accurate and homogeneous set of coordinatesWe have used the 2nd generation of the Guide Star Catalogue (GSC-II) asa reference astrometric catalogue to compile the positions of 1086Galactic Planetary Nebulae (PNe) listed in the Strasbourg ESO Catalogue(SEC), its supplement and the version 2000 of the Catalogue of PlanetaryNebulae. This constitutes about 75% of all known PNe. For these PNe, theones with a known central star (CS) or with a small diameter, we havederived coordinates with an absolute accuracy of ~0\farcs35 in eachcoordinate, which is the intrinsic astrometric precision of the GSC-II.For another 226, mostly extended, objects without a GSC-II counterpartwe give coordinates based on the second epoch Digital Sky Survey(DSS-II). While these coordinates may have systematic offsets relativeto the GSC-II of up to 5 arcsecs, our new coordinates usually representa significant improvement over the previous catalogue values for theselarge objects. This is the first truly homogeneous compilation of PNepositions over the whole sky and the most accurate one available so far.The complete Table \ref{tab2} is only available in electronic form atthe CDS via anonymous ftp to cdsarc.u-strasbg.fr (130.79.128.5) or viahttp://cdsweb.u-strasbg.fr/cgi-bin/qcat?J/A+A/408/1029} The relation between elemental abundances and morphology in planetary nebulaeAn investigation of the variation of elemental abundances with planetarynebula morphology is of considerable interest, since it has a bearingupon how such sources are formed, and from which progenitors they areejected. Recent advances in morphological classification now enable usto assess such trends for a statistically significant number of sources.We find, as a result, that the distribution N[log(X/H)] of sources withrespect to elemental abundance (X/H) varies between the differingmorphologies. Circular sources tend to peak towards low abundancevalues, whilst bipolar nebulae (BPNe) peak towards somewhat highervalues. This applies for most elemental species, although it is perhapsleast apparent for oxygen. In contrast, elliptical sources appear todisplay much broader functions N[log(X/H)], which trespass upon thedomains of both circular and elliptical planetary nebulae (PNe).We take these trends to imply that circular sources derive fromlower-mass progenitors, bipolar sources from higher-mass stars, and thatelliptical nebulae derive from all masses of progenitor, high and low.Whilst such trends are also evident in values of mean abundance, they are much less clear. Only in the cases of He/H, N/H,Ne/H and perhaps Ar/H is there evidence for significant abundancedifferences.Certain BPNe appear to possess low abundance ratios He/H and Ar/H, andthis confirms that a few such outflows may arise from lower-massprogenitors. Similarly, we note that ratios are quite modestin elliptical planetary nebulae, and not much different from those forcircular and bipolar PNe; a result that conflicts with the expectationsof at least one model of shell formation. Gas temperature and excitation classes in planetary nebulaeEmpirical methods to estimate the elemental abundances in planetarynebulae usually use the temperatures derived from the [O III] and [N II]emission-line ratios, respectively, for the high- and low-ionizationzones. However, for a large number of objects these values may not beavailable. In order to overcome this difficulty and allow a betterdetermination of abundances, we discuss the relationship between thesetwo temperatures. Although a correlation is not easily seen when asample of different PNe types is used, the situation is improved whenthey are gathered into excitation classes. From [OII]/[OIII] andHeII/HeI line ratios, we define four excitation classes. Then, usingstandard photoionization models which fit most of the data, a linearrelation between the two temperatures is obtained for each of the fourexcitation classes. The method is applied to several objects for whichonly one temperature can be obtained from the observed emission linesand is tested by recalculation of the radial abundance gradient of theGalaxy using a larger number of PNe. We verified that our previousgradient results, obtained with a smaller sample of planetary nebulae,are not changed, indicating that the temperature relation obtained fromthe photoionization models are a good approximation, and thecorresponding statistical error decreases as expected. Tables 3-5, 7 and9 are only available in electronic form at http://www.edpsciences.org Classification of planetary nebulae by their departure from axisymmetryWe propose a scheme to classify planetary nebulae (PNe) according totheir departure from axisymmetric structure. We consider only departurealong and near the equatorial plane, i.e. between the two sidesperpendicular to the symmetry axis of the nebula. We consider six typesof departure from axisymmetry: (1) PNe where the central star is not atthe centre of the nebula; (2) PNe having one side brighter than theother; (3) PNe having unequal size or shape of the two sides; (4) PNewhere the symmetry axis is bent, e.g. the two lobes in a bipolar PN arebent toward the same side; (5) PNe where the main departure fromaxisymmetry is in the outer regions, e.g. an outer arc; and (6) PNe thatshow no departure from axisymmetry, i.e. any departure, if it exists, ison scales smaller than the scale of blobs, filaments and otherirregularities in the nebula. PNe that possess more than one type ofdeparture are classified by the most prominent type. We discuss theconnection between departure types and the physical mechanisms that maycause them, mainly resulting from the influence of a stellar binarycompanion. We find that ~50 per cent of all PNe in the analysed samplepossess large-scale departure from axisymmetry. This number is largerthan that expected from the influence of binary companions, namely~25-30 per cent. We argue that this discrepancy comes from many PNewhere the departure from axisymmetry, mainly unequal size, shape orintensity, results from the presence of long-lived and large (hot orcool) spots on the surface of their asymptotic giant branch progenitors.Such spots locally enhance the mass-loss rate, leading to a departurefrom axisymmetry, mainly near the equator, in the descendent PN. The 3-D ionization structure of the planetary nebula NGC 6565A detailed study of the planetary nebula NGC 6565 has been carried outon long-slit echellograms (lambda /Delta lambda =60 000, spectral range= lambda lambda 3900-7750 Å) at six, equally spaced positionangles. The expansion velocity field, the c(Hβ ) distribution andthe radial profile of the physical conditions (electron temperature anddensity) are obtained. The distance, radius, mass and filling factor ofthe nebula and the temperature and luminosity of the central star arederived. The radial ionization structure is analyzed using both theclassical method and the photo-ionization code CLOUDY. Moreover, wepresent the spatial structure in a series of images from differentdirections, allowing the reader to see'' the nebula in 3-D. NGC 6565results to be a young (2000-2500 years), patchy, optically thicktriaxial ellipsoid (a=10.1 arcsec, a/b=1.4, a/c=1.7) projected almostpole-on. The matter close to major axis was swept-up by someaccelerating agent (fast wind? ionization? magnetic fields?), formingtwo faint and asymmetric polar cups. A large cocoon of almost neutralgas completely embeds the ionized nebula. NGC 6565 is in a recombinationphase, because of the luminosity drop of the massive powering star,which is reaching the white dwarf domain (log T* =~ 5.08 K;log L*/Lsun =~ 2.0). The stellar decline startedabout 1000 years ago, but the main nebula remained optically thin forother 600 years before the recombination phase occurred. In the nearfuture the ionization front will re-grow, since the dilution factor dueto the expansion will prevail on the slower and slower stellar decline.NGC 6565 is at a distance of 2.0 (+/-0.5) kpc and can be divided intothree radial zones: the fully ionized'' one, extending up to0.029-0.035 pc at the equator (0.050 pc at the poles), thetransition'' one, up to 0.048-0.054 pc (0.080 pc), the halo'',detectable up to 0.110 pc. The ionized mass ( =~ 0.03 Msun)is only a fraction of the total mass (>= 0.15 Msun), whichhas been ejected by an equatorial enhanced superwind of 4 (+/-2) x10-5 Msun yr-1 lasted for 4 (+/-2) x103 years. Based on observations made with ESO Telescopes atthe La Silla Observatories, under programme ID 65.I-0524, and onobservations made with the NASA/ESA Hubble Space Telescope, obtainedfrom the data archive at the Space Telescope Institute (observingprogram GO 7501; P.I. Arsen Hajian). STScI is operated by theassociation of Universities for Research in Astronomy, Inc. under theNASA contract NAS 5-26555. We have applied the photoionization codeCLOUDY, developed at the Institute of Astronomy of the CambridgeUniversity. High Dispersion Spectra for Planetary Nebula StudiesThe extremely complicated shapes of planetary nebulae revealed throughthe high resolution radio maps, direct imaging with the Hubble SpaceTelescope and observations with adaptive optics at large telescopes, aregreatly different from their imagined simplicity long ago. To addressthe complexity in physical conditions and geometries of planetarynebulae, one must secure spectra of high spatial resolution and highdispersion. It also may require a long exposure even with a largetelescopic aperture to reach faint features. We briefly review plasmadiagnostics and a diagnostic possibility of iron ions based on ourrecent high dispersion spectroscopic work. Study of electron density in planetary nebulae. A comparison of different density indicatorsWe present a comparison of electron density estimates for planetarynebulae based on different emission-line ratios. We have considered thedensity indicators [O Ii]lambda 3729/lambda 3726, [S Ii]lambda6716/lambda 6731, [Cl Iii]lambda 5517/lambda 5537, [Ar Iv]lambda4711/lambda 4740, C Iii]lambda 1906/lambda 1909 and [N I]lambda5202/lambda 5199. The observational data were extracted from theliterature. We have found systematic deviations from the densityhomogeneous models, in the sense that: Ne(ion {N}i) <~Ne(ion {O}{ii}) < Ne(ion {S}{ii}, ion {C}{iii},ion {Cl}{iii} or ion {Ar}{iv}) and Ne(ion {S}{ii}) ~Ne(ion {C}{iii}) ~ Ne(ion {Cl}{iii}) ~Ne(ion {Ar}{iv}). We argue that the lower [O Ii] densityestimates are likely due to errors in the atomic parameters used. Shapes and Shaping of Planetary NebulaeWe review the state of observational and theoretical studies of theshaping of planetary nebulae (PNe) and protoplanetary nebulae (pPNe). Inthe past decade, high-resolution studies of PNe have revealed abewildering array of morphologies with elaborate symmetries. Recentimaging studies of pPNe exhibit an even richer array of shapes. Thevariety of shapes, sometimes multiaxial symmetries, carefully arrangedsystems of low-ionization knots and jets, and the often Hubble-flowkinematics of PNe and pPNe indicate that there remains much tounderstand about the last stages of stellar evolution. In many cases,the basic symmetries and shapes of these objects develop on extremelyshort timescales, seemingly at the end of AGB evolution when the mode ofmass loss abruptly and radically changes. No single explanation fits allof the observations. The shaping process may be related to externaltorques of a close or merging binary companion or the emergence ofmagnetic fields embedded in dense outflowing stellar winds. We suspectthat a number of shaping processes may operate with different strengthsand at different stages of the evolution of any individual object.
Submit a new article
• - No Links Found -
Submit a new link
### Member of following groups:
#### Observation and Astrometry data
Constellation: Sagitta Right ascension: 20h12m42.81s Declination: +19°59'22.7" Apparent magnitude: 12
Catalogs and designations:
Proper Names (Edit) NGC 2000.0 NGC 6886 → Request more catalogs and designations from VizieR
|
{}
|
## Refereed articles in journals
NOTE: The versions of the papers that can be downloaded from this page are often the official journal versions with restricted access. Often, preliminary versions in the form of preprints are available. In any case, we advise you to prefer the published articles as they contain the most recent versions. Electronic reprints of the official versions are available on request.
[J263] Compact Thermo-Mechanical Models for the Fast Simulation of Machine Tools with Nonlinear Component Behavior Julia Vettermann, Alexander Steinert, Christian Brecher, Peter Benner, and Jens Saak at-Automatisierungstechnik, accepted 7 July 2022. [J262] Model Order Reduction for Delayed PEEC Models With Guaranteed Accuracy and Observed Stability Lihong Feng, Luigi Lombardi, Peter Benner, Daniele Romano, and Giulio Antonini IEEE Transactions on Circuits and Systems I: Regular Papers, 14 pages, published online 14 July 2022. DOI: 10.1109/TCSI.2022.3189389 [J261] Discovery of Nonlinear Dynamical Systems using a Runge-Kutta Inspired Dictionary-based Sparse Regression Approach Pawan Goyal and Peter Benner Proceedings of the Royal Society A, Vol. 478, Art. 20210883 (24 pages), 2022. DOI: 10.1098/rspa.2021.0883 Preliminary version appeared as arXiv Preprint arXiv: 2105.04869, [cs.LG], May 2021. [J260] A Low-Rank Solution Method for Riccati Equations with Indefinite Quadratic Terms Peter Benner, Jan Heiland, and Steffen W. R. Werner Numerical Algorithms, 21 pages, published online 31 May 2022. DOI: 10.1007/s11075-022-01331-w Preliminary version appeared as arXiv Preprint arXiv: 2111.06516, [math.NA], 12 November 2021. [J259] Palindromic Linearization and Numerical Solution of Nonsymmetric Algebraic T-Riccati Equations Peter Benner, Bruno Iannazzo, Beatrice Meini, and Davide Palitta BIT Numerical Mathematics, published online 20 May 2022. DOI: 10.1007/s10543-022-00926-y Preliminary version appeared as arXiv Preprint arXiv: 2110.03254, [math.NA], 7 October 2021. [J258] Reduced Basis Method for the Nonlinear Poisson-Boltzmann Equation regularized by the Range-Separated Canonical Tensor Format Cleophas Kweyu, Lihong Feng, Matthias Stein, and Peter Benner International Journal of Nonlinear Sciences and Numerical Simulation, published online 18 May 2022. DOI: 10.1515/ijnsns-2021-0103 Preliminary version appeared as arXiv Preprint arXiv:2103.00245, [math.NA], February 2021. [J257] Convergence Analysis of Vector Extended LOBPCG for Computing Extreme Eigenvalues Peter Benner and Xin Liang Numerical Linear Algebra with Applications, e2445 (24 pages), published online 9 May 2022. DOI: 10.1002/nla.2445 Preliminary version appeared as arXiv Preprint arXiv:2004.14002v1, [math.NA], April 2020. [J256] Convolutional Neural Networks for Very Low-dimensional LPV Approximations of Incompressible Navier-Stokes Equations Jan Heiland, Peter Benner, and Rezvan Bahmani Frontiers in Applied Mathematics and Statistics, section Statistical and Computational Physics, published online 29 April 2022. DOI: 10.3389/fams.2022.879140 [J255] A Greedy Data Collection Scheme For Linear Dynamical Systems Karim Cherifi, Pawan Goyal, and Peter Benner Data-Centric Engineering, Vol. 3, e16 (14 pages), 2022. DOI: 10.1017/dce.2022.16 Preliminary version appeared as arXiv Preprint arXiv: 2107.12950, [eess.SY], 27 July 2021. [J254] Stable and Efficient Computation of Generalized Polar Decompositions Peter Benner, Yuji Nakatsukasa, and Carolin Penke SIAM Journal on Matrix Analysis and Applications, accepted 7 March 2022. Preliminary version appeared as arXiv Preprint arXiv:2104.06659, [math.NA], April 2021. [J253] Robust Output-Feedback Stabilization for Incompressible Flows using Low-Dimensional H∞-Controllers Peter Benner, Jan Heiland, and Steffen W. R. Werner Computational Optimization and Application, published online, 2022. DOI: 10.1007/s10589-022-00359-x Preliminary version appeared as arXiv Preprint arXiv:2103.01608, [math.OC], March 2021. [J252] Fast A Posteriori State Error Estimation for Reliable Frequency Sweeping in Microwave Circuits via the Reduced-Basis Method Valentin de la Rubia, Sridhar Chellappa, Lihong Feng, and Peter Benner IEEE Transactions on Microwave Theory and Techniques, accepted 17 January 2022. Preliminary version appeared as arXiv Preprint arXiv: 2110.05925, [math.NA], 12 October 2021. [J251] Parallel Algorithms for Tensor Train Arithmetic Hussam Al Daas, Grey Ballard, and Peter Benner SIAM Journal on Scientific Computing, Vol. 44, Issue 1, pp. C25-C35, 2022. DOI: 10.1137/20M1387158 Preliminary version appeared as arXiv Preprint arXiv:2011.06532, [math.NA], November 2020. [J250] A Training Set Subsampling Strategy for the Reduced Basis Method Sridhar Chellappa, Lihong Feng, and Peter Benner Journal of Scientific Computing, Vol. 89, Art. 63 (pp. 34), 2021. DOI: 10.1007/s10915-021-01665-y Preliminary version appeared as arXiv Preprint arXiv:2103.06185, [math.NA], March 2021. [J249] A Non-Intrusive Method to Inferring Linear Port-Hamiltonian Realizations using Time-Domain Data Karim Cherifi, Pawan Goyal, and Peter Benner Electronic Transactions on Numerical Analysis, (Special Volume on Scientific Machine Learning), Vol. 56, pp. 102-116, 2022. DOI: 10.1553/etna_vol56s102 Preliminary version appeared as arXiv Preprint arXiv:2005.09371, [eess.SY], May 2020. [J248] Data-Driven Model Order Reduction for Problems with Parameter-Dependent Jump-Discontinuities Neeraj Sarna and Peter Benner Computer Methods in Applied Mechanics and Engineering, Vol. 387, Art. 114168, 2021. DOI: 10.1016/j.cma.2021.114168 Preliminary version appeared as arXiv Preprint arXiv:2105.00547, [math.NA], May 2021. [J247] Index-Aware Model-Order Reduction for a Special Class of Nonlinear Differential-Algebraic Equations Nicodemus Banagaaya, Giuseppe Ali, Sara Grundel, and Peter Benner Journal of Dynamics and Differential Equations, published online, 2021. DOI: 10.1007/s10884-021-10063-9 Preliminary version appeared as arXiv Preprint arXiv:2002.09751v1, [math.NA], February 2020. [J246] Model Order Reduction Methods for Coupled Machine Tool Models Julia Vettermann, Stefan Sauerzapf, Andreas Naumann, Michael Beitelschmidt, Roland Herzog, Peter Benner, and Jens Saak MM Science Journal, pp. 4652-4659, 2021. DOI: 10.17973/MMSJ.2021_7_2021072 [J245] Model Order Reduction for Gas and Energy Networks Christian Himpe, Sara Grundel, and Peter Benner Journal of Mathematics in Industry, Vol. 11, Art. 13 (pp. 46), 2021. DOI: 10.1186/s13362-021-00109-4 Preliminary version appeared as arXiv Preprint arXiv:2011.12099, [math.OC], November 2020. [J244] NFDI for Catalysis-Related Sciences: NFDI4Cat Sara Espinoza, David Linke, Christoph Wulf, Sonja Schimmler, Stephan Andreas Schunk, Peter Benner et al. Bausteine Forschungsdatenmanagement, pp. 57-71, 2021. DOI: 10.17192/bfdm.2021.2.8333 [J243] Gramians, Energy Functionals and Balanced Truncation for Linear Dynamical Systems with Quadratic Outputs Peter Benner, Pawan Goyal, and Igor Pontes Duff IEEE Transactions on Automatic Control, Vol 67, Issue 2, pp. 886-893, 2022. DOI: 10.1109/TAC.2021.3086319 Preliminary version appeared as arXiv Preprint arXiv:1909.04597v1, [math.OC], September 2019. [J242] A Stochastic Maximum Principle for Control Problems Constrained by the Stochastic Navier-Stokes Equations Peter Benner and Christoph Trautwein Applied Mathematics and Optimization, Vol. 84, pp. 1001-1054, 2021. DOI: 10.1007/s00245-021-09792-6 Preliminary version appeared as arXiv Preprint arXiv:1810.12119[math.OC], October 2018. [J241] Efficient and Accurate Algorithms for Solving the Bethe-Salpeter Eigenvalue Problem for Crystalline Systems Peter Benner and Carolin Penke Journal of Computational and Applied Mathematics, Vol. 400, Art. 113650, 2022. DOI: 10.1016/j.cam.2021.113650 Preliminary version appeared as arXiv Preprint arXiv:2008.08825, [math.NA], August 2020. [J240] Galerkin Trial Spaces and Davison-Maki Methods for the Numerical Solution of Differential Riccati Equations Maximilian Behr, Peter Benner, and Jan Heiland Applied Mathematics and Computation, Vol. 410, Art. 126401 (28 pp.), 2021. DOI: 10.1016/j.amc.2021.126401 Preliminary version appeared as arXiv Preprint arXiv:1910.13362, [math.NA], October 2019. [J239] Learning Reduced-order Dynamics for Parametrized Shallow Water Equations from Data Süleyman Yildiz, Pawan Goyal, Peter Benner, and Bülent Karasözen International Journal for Numerical Methods in Fluids, Vol. 93, Issue 8, pp. 2803-2821, 2021. DOI: 10.1002/fld.4998 Preliminary version appeared as arXiv Preprint arXiv:2007.14079, [math.NA, July 2020. [J238] Structure-Preserving Interpolation for Model Reduction of Parametric Bilinear Systems Peter Benner, Serkan Gugercin, and Steffen W. R. Werner Automatica, Vol. 132, Art. 109799 (9 pp.), 2021. DOI: 10.1016/j.automatica.2021.109799 Preliminary version appeared as arXiv Preprint arXiv:2007.11269, [math.NA], July 2020. [J237] A Rational Even-IRA Algorithm for the Solution of T-even Polynomial Eigenvalue Problems Peter Benner, Heike Fassbender, and Philip Saltenberger SIAM Journal on Matrix Analysis and Applications, Vol. 42, Issue 3, pp. 1172-1198, 2021. DOI: 10.1137/20M1364485 Preliminary version appeared as arXiv Preprint arXiv:2009.01762, [math.NA], September 2020. [J236] Operator Inference and Physics-Informed Learning of Low-Dimensional Models for Incompressible Flows Peter Benner, Pawan Goyal, Jan Heiland, and Igor Pontes Duff Electronic Transactions on Numerical Analysis, (Special Volume on Scientific Machine Learning), Vol. 56, pp. 28-51, 2022. DOI: 10.1553/etna_vol56s28 Preliminary version appeared as arXiv Preprint arXiv:2010.06701, [math.DS], October 2020. [J235] A Linear Implicit Euler Method for the Finite Element Discretization of a Controlled Stochastic Heat Equation Peter Benner, Tony Stillfjord, and Christoph Trautwein IMA Journal of Numerical Analysis, published online, 2021. DOI: 10.1093/imanum/drab033 Preliminary version appeared as arXiv Preprint arXiv:2006.05370, [math.NA], June 2020. [J234] Structure-Preserving Interpolation of Bilinear Control Systems Peter Benner, Serkan Gugercin, and Steffen W. R. Werner Advances in Computational Mathematics, Vol. 47, Issue 3, Art. 43 (38 pp.), 2021. DOI: 10.1007/s10444-021-09863-w Preliminary version appeared as arXiv Preprint arXiv:2005.00795v1, [math.NA], May 2020. [J233] Relative Perturbation Theory for Quadratic Hermitian Eigenvalue Problem Peter Benner, Xin Liang, Suzana Miodragović, and Ninoslav Truhar Linear Algebra and its Applications, Vol. 618, pp. 97-128, 2021. DOI: 10.1016/j.laa.2021.01.023 Preliminary version appeared as arXiv Preprint arXiv:1602.03420v2 [math.NA], February 2016. [J232] Factorized Solution of Generalized Stable Sylvester Equations Using Many-Core GPU Accelerators Peter Benner; Ernesto Dufrechou; Pablo Ezzatti; Rodrigo Gallardo; Enrique Quintana-Ortí Journal of Supercomputing, Vol. 77, pp. 10152-10164, 2021. DOI: 10.1007/s11227-021-03658-y [J231] Tomographic X-ray Scattering based on Invariant Reconstruction - Analysis of the 3D Nanostructure of Bovine Bone Paolino De Falco, Richard Weinkamer, Wolfgang Wagermaier, Chenghao Li, Tim Snow, Nicholas J. Terrill, Himadri S. Gupta, Pawan Goyal, Martin Stoll, Peter Benner and Peter Fratzl Journal of Applied Crystallography, Vol. 54, pp. 486-497, 2021. DOI: 10.1107/S1600576721000881 [J230] On Error Estimation for Reduced-Order Modeling of Linear Non-Parametric and Parametric Systems Lihong Feng and Peter Benner ESAIM: Mathematical Modelling and Numerical Analysis (M2AN), Vol. 55, Issue 2, pp. 561-594, 2021. DOI: 10.1051/m2an/2021001 Preliminary version appeared as arXiv Preprint arXiv:2003.14319, [math.NA], March 2020. [J229] Regularization of Poisson-Boltzmann Type Equations with Singular Source Terms using the Range-Separated Tensor Format Peter Benner, Boris Khoromskij, Venera Khoromskaia, Cleophas Kweyu, and Matthias Stein SIAM Journal on Scientific Computing, Vol. 43, Issue 1, pp. A415-A445, 2021. DOI: 10.1137/19M1281435 Preliminary version appeared as arXiv Preprint arXiv:1901.09864, [math.NA], January 2019. [J228] Implicit Higher-Order Moment Matching Technique for Model Reduction of Quadratic-bilinear Systems Mian Muhammad Arsalan Asif, Mian Ilyas Ahmad, Peter Benner, Lihong Feng, and Tatjana Stykel Journal of the Franklin Institute, Vol. 358, Issue 3, pp. 2015-2038, 2021. DOI: 10.1016/j.jfranklin.2020.11.012 Preliminary version appeared as arXiv Preprint arXiv:1911.05400, [eess.SY], November 2019. [J227] Machine Learning for Material Characterization with an Application for Predicting Mechanical Properties Anke Stoll and Peter Benner GAMM Mitteilungen, Vol. 44, Issue 1, e202100003 (21 pp.), 2021. DOI: 10.1002/gamm.202100003 Preliminary version appeared as arXiv Preprint arXiv:2010.06010, [cs.LG], October 2020. [J226] Balanced Truncation of Linear Time-Invariant Systems over Finite-frequency Ranges Peter Benner, Xin Du, Guanghong Yang, and Dan Ye Advances in Computational Mathematics, Vol. 46, Art. 82 (34 pp.), 2020. DOI: 10.1007/s10444-020-09823-w Preliminary version appeared as arXiv Preprint arXiv:1602.04402v1 [cs.SY], February 2016. [J225] Interpolation-Based Model Order Reduction for Polynomial Systems Peter Benner and Pawan Goyal SIAM Journal on Scientific Computing, Vol. 43, Issue 1, A84-A108, 2021. DOI: 10.1137/19M1259171 Preliminary version appeared as arXiv Preprint arXiv:1904.11891, [math.NA], April 2019. [J224] On the Solution of the Nonsymmetric T-Riccati Equation Peter Benner and Davide Palitta Electronic Transactions on Numerical Analysis (ETNA), Vol. 54, pp. 68-88, 2021. DOI: 10.1553/etna_vol54s68 Preliminary version appeared as arXiv Preprint arXiv:2003.03693v1, [math.NA], March 2020. [J223] Model Order Reduction for Delay Systems by Iterative Interpolation Dominik Alfke, Lihong Feng, Luigi Lombardi, Giulio Antonini, and Peter Benner International Journal for Numerical Methods in Engineering, Vol. 122, Issue 3, pp. 684-706, 2021. DOI: 10.1002/nme.6554 [J222] Operator Inference for Non-Intrusive Model Reduction of Systems with Non-Polynomial Nonlinear Terms Peter Benner, Pawan Goyal, Boris Kramer, Benjamin Peherstorfer, and Karen Willcox Computer Methods in Applied Mechanics and Engineering, Vol. 372, 113433 (17 pp.), 2020. DOI: 10.1016/j.cma.2020.113433 Preliminary version appeared as arXiv Preprint arXiv:2002.09726, [math.NA], February 2020. [J221] Fast Solution of the Linearized Poisson-Boltzmann Equation with nonaffine Parametrized Boundary Conditions Using the Reduced Basis Method Cleophas Kweyu, Lihong Feng, Matthias Stein, and Peter Benner Computing and Visualization in Science, Vol. 23, Art. 15 (19 pp.), 2020. DOI: 10.1007/s00791-020-00336-z Preliminary version appeared as arXiv Preprint arXiv:1705.08349, May 2017. [J220] Efficient Numerical Methods for Gas Network Modeling and Simulation Yue Qiu, Sara Grundel, Martin Stoll, and Peter Benner Networks and Heterogeneous Media, Vol. 15, Issue 4, pp. 653-679, 2020. DOI: 10.3934/nhm.2020018 Preliminary version appeard as arXiv Preprint arXiv:1807.07142 [math.NA], July 2018. [J219] Identification of Port-Hamiltonian Systems from Frequency Response Data Peter Benner, Pawan Goyal, and Paul Van Dooren Systems & Control Letters, Vol. 143, 104741 (9 pp.), 2020. DOI: 10.1016/j.sysconle.2020.104741 Preliminary version appeared as arXiv Preprint arXiv:1911.00080v1, [eess.SY], October 2019. [J218] Frequency- and Time-Limited Balanced Truncation for Large-Scale Second-Order Systems Peter Benner and Steffen W. R. Werner Linear Algebra and Its Applications, Special Issue in Honor of Paul Van Dooren, Vol. 623, pp. 68-103, 2021. DOI: 10.1016/j.laa.2020.06.024 Preliminary version appeared as arXiv Preprint arXiv:2001.06185, [math.OC], January 2020. [J217] Low-Rank Linear Fluid-Structure Interaction Discretizations Roman Weinhandl, Peter Benner, and Thomas Richter Zeitschrift für Angewandte Mathematik und Mechanik (ZAMM), Vol. 100, Issue 11, e201900205 (28 pp.), 2020. [Editor's Choice] DOI: 10.1002/zamm.201900205 Preliminary version appeared as arXiv Preprint arXiv:1905.11000, [math.NA], May 2019. [J216] Adaptive Basis Construction and Improved Error Estimation for Parametric Nonlinear Dynamical Systems Peter Benner, Sridhar Chellappa, and Lihong Feng International Journal for Numerical Methods in Engineering, Vol. 121, Issue 23, pp. 5320-5349, 2020. DOI: 10.1002/nme.6462 Preliminary version appeared as arXiv Preprint arXiv:1911.05235, [math.NA], November 2019. [J215] High Performance Solution of Skew-symmetric Eigenvalue Problems with Applications in Solving the Bethe-Salpeter Eigenvalue Problem Carolin Penke, Andreas Marek, Christian Vorwerk, Claudia Draxl, and Peter Benner Parallel Computing, Vol. 96, Art. 102639 (8 pp.), 2020. Preliminary version appeared as arXiv Preprint arXiv:1912.04062v1, [math.NA], December 2019. [J214] Low-Rank Solution of an Optimal Control Problem Constrained by Random Navier-Stokes Equations Peter Benner, Sergey Dolgov, Akwum Onwunta, and Martin Stoll International Journal for Numerical Methods in Fluids, Vol. 92, Issue 11, pp. 1653-1678, 2020. DOI: 10.1002/fld.4843 Preliminary version appeared as arXiv Preprint arXiv:1703.06097, [math.NA], March 2017. [J213] A Subspace Framework for H∞-Norm Minimization Nicat Aliyev, Peter Benner, Emre Mengi, and Matthias Voigt SIAM Journal on Matrix Analysis and Applications, Vol. 41, Issue 2, pp. 928–956, 2020. DOI: 10.1137/19M125892X Preliminary version appeared as arXiv Preprint arXiv:1905.04208, [math.NA], May 2019. [J212] Model Order Reduction for Bilinear Control Systems with Inhomogeneous Initial Conditions Xingang Cao, Peter Benner, Igor Pontes Duff, and Wil Schilders International Journal of Control, Vol. 94, Issue 10, pp. 2886-2895, 2021. DOI: 10.1080/00207179.2020.1740945 [J211] A Numerical Comparison of Different Solvers for Large-Scale, Continuous-Time Algebraic Riccati Equations and LQR Problems Peter Benner, Zvonimir Bujanović, Patrick Kürschner, and Jens Saak SIAM Journal on Scientific Computing, Vol 42, Issue 2, pp. A957-A996, 2020. DOI: 10.1137/18M1220960 Preliminary version appeared as arXiv Preprint arXiv:1811.00850[math.NA], November 2018. [J210] Hankel-Norm Approximation of Large-Scale Descriptor Systems Peter Benner and Steffen W. R. Werner Advances in Computational Mathematics, Vol. 46, Issue 3, Article 40 (31 pp.), 2020. DOI: 10.1007/s10444-020-09750-w Preliminary version appeared as arXiv Preprint arXiv:1612.06205v2, [math.OC], December 2018. [J209] Efficient Solution of Large-Scale Algebraic Riccati Equations Associated with Index-2 DAEs via the Inexact Low-Rank Newton-ADI Method Peter Benner, Matthias Heinkenschloss, Jens Saak, and Heiko K. Weichelt Applied Numerical Mathematics, Vol. 152, pp. 338-354, 2020. DOI: 10.1016/j.apnum.2019.11.016 Preliminary version appeared as arXiv Preprint arXiv:1804.01410[math.NA], April 2018. [J208] Solution Formulas for Differential Sylvester and Lyapunov Equations Maximilian Behr, Peter Benner, and Jan Heiland Calcolo, Vol. 56, Article 51 (33 pp.), 2019. DOI: 10.1007/s10092-019-0348-x Preliminary version appeared as arXiv Preprint arXiv:1811.08327[math.NA], November 2018. [J207] Greedy Low-Rank Algorithm for Spatial Connectome Regression Patrick Kürschner, Sergey Dolgov, Kameron Decker Harris, and Peter Benner The Journal of Mathematical Neuroscience, Vol. 9, Issue 1, Article 9 (22 pp.), 2019. DOI: 10.1186/s13408-019-0077-0 Preliminary version appeared as arXiv Preprint arXiv:1808.05510[math.NA], August 2018. [J206] A New Error Estimator for Reduced-Order Modeling of Linear Parametric Systems Lihong Feng and Peter Benner IEEE Transactions on Microwave Theory and Techniques, Vol. 67, Issue 12, pp. 4848-4859, 2019. DOI: 10.1109/TMTT.2019.2948858 [J205] Cross-Gramian-Based Dominant Subspaces Peter Benner and Christian Himpe Advances in Computational Mathematics, Vol. 45, Issue 5, pp. 2533-2553, 2019. DOI: 10.1007/s10444-019-09724-7 Preliminary version appeared as arXiv Preprint arXiv:1809.08066[math.OC], September 2018. [J204] Extended and Improved Criss-Cross Algorithms for Computing the Spectral Value Set Abscissa and Radius Peter Benner and Tim Mitchell SIAM Journal on Matrix Analysis and Applications, Vol. 40, Issue 4, pp. 1325-1352, 2019. DOI: 10.1137/19M1246213 Preliminary version appeared as arXiv Preprint arXiv:1712.10067 [math.OC], December 2017. [J203] Model Reduction of Controlled Fokker-Planck and Liouville-von Neumann Equations Peter Benner, Tobias Breiten, Carsten Hartmann, and Burkhard Schmidt Journal of Computational Dynamics, Vol. 7, Issue 1, pp. 1-33, 2020. DOI: 10.3934/jcd.2020001 Preliminary version appeared as arXiv Preprint arXiv:1706.09882 , July 2017. [J202] New Gramians for Linear Switched Systems: Reachability, Observability, and Model Reduction Igor Pontes Duff, Sara Grundel, and Peter Benner IEEE Transactions on Automatic Control, Vol. 65, Issue 6, pp. 2526-2535, 2020. DOI: 10.1109/TAC.2019.2934020 Preliminary version appeared as arXiv Preprint arXiv:1806.00406, June 2018. [J201] Reduced-Order Modelling of Parametric Systems via Interpolation of Heterogeneous Surrogates Yao Yue, Lihong Feng, and Peter Benner Advanced Modeling and Simulation in Engineering Sciences, Vol. 6, Art. 10 (33 pp.), 2019. DOI: 10.1186/s40323-019-0134-y [J200] Optimal Distributed and Tangential Boundary Control for the Unsteady Stochastic Stokes Equations Christoph Trautwein and Peter Benner ESAIM: Control, Optimisation and Calculus of Variations, Vol. 26, Art. 62 (32 pp.), 2020. DOI: 10.1051/cocv/2019042 Preliminary version appeared as arXiv Preprint arXiv:1809.00911v1 [math.OC], September 2018. [J199] Balanced Truncation for a Special Class of Bilinear Descriptor Systems Igor Pontes Duff Pereira, Pawan Goyal, and Peter Benner IEEE Control Systems Letters, Vol. 3, Issue 3, pp. 535-540, 2019. DOI: 10.1109/LCSYS.2019.2911904 [J198] A Bilinear H2 Model Order Reduction Approach to Linear Parameter-Varying Systems Peter Benner, Xingang Cao, and Wil Schilders Advances in Computational Mathematics, Vol. 45, Issue 5-6. pp. 2241–2271, 2019. DOI: 10.1007/s10444-019-09695-9 [J197] Computing the Density of States for Optical Spectra of Molecules by Low-Rank and QTT Tensor Approximation Peter Benner, Venera Khoromskaia, Boris N. Khoromskij, and Chao Yang Journal of Computational Physics, Vol. 382, pp. 221-239, 2019. DOI: 10.1016/j.jcp.2019.01.011 Preliminary version appeared as arXiv Preprint arXiv:1801.03852 [math.NA], January 2018. [J196] Periodic Switching Strategies for an Isoperimetric Control Problem with Application to Nonlinear Chemical Reactions Peter Benner, Andreas Seidel-Morgenstern, and Alexander Zuyev Applied Mathematical Modelling, Vol. 69, pp. 287-300, 2019. DOI: 10.1016/j.apm.2018.12.005 Preliminary version appeared as arXiv Preprint arXiv:1811.07421 [math.OC], November 2018. [J195] Interpolatory Model Reduction for Quadratic-Bilinear Systems using Error Estimators Mian Ilyas Ahmad, Peter Benner, and Lihong Feng Engineering Computations, Vol. 36, Issue 1, pp. 25-44, 2019. DOI: 10.1108/EC-04-2018-0162 [J194] Optimal Control Problems Constrained by the Stochastic Navier-Stokes Equations with Multiplicative Lévy Noise Peter Benner and Christoph Trautwein Mathematische Nachrichten, Vol. 292, Issue 7, pp. 1444-1461, 2019. DOI: 10.1002/mana.201700185 [J193] A New Two-Sided Projection Technique for Model Reduction of Quadratic-Bilinear Descriptor Systems Mian Ilyas Ahmad, Peter Benner, and Lihong Feng International Journal of Computer Mathematics, Vol. 96, Issue 10, pp. 1899-1909, 2019. DOI: 10.1080/00207160.2018.1542134 [J192] Adaptive Discontinuous Galerkin Approximation of Optimal Control Problems Governed by Transient Convection-Diffusion Equations Hamdullah Yücel, Martin Stoll, and Peter Benner Electronic Transactions on Numerical Analysis, Vol. 48, pp. 407-434, 2018. DOI: 10.1553/etna_vol48s407 Preliminary version appeared as MPI Magdeburg Preprint MPIMD/15-11, July 2015. [J191] Faster and more Accurate Computation of the H∞-Norm via Optimization Peter Benner and Tim Mitchell SIAM Journal on Scientific Computing, Vol. 40, No. 5, pp. A3609–A3635, 2018. DOI: 10.1137/17M1137966 Preliminary version appeared as arXiv Preprint arXiv:1707.02497, July 2017. [J190] Low-Rank Eigenvector Compression of Posterior Covariance Matrices for Linear Gaussian Inverse Problems Peter Benner, Yue Qiu, and Martin Stoll SIAM/ASA Journal on Uncertainty Quantification, Vol. 6, Issue 2, pp. 965-989, 2018. DOI: 10.1137/17M1121342 Preliminary version appeared as arXiv Preprint arXiv:1703.05638, March 2017. [J189] A Low-Rank Inexact Newton-Krylov Method for Stochastic Eigenvalue Problems Peter Benner, Akwum Onwunta, and Martin Stoll Computational Methods in Applied Mathematics(CMAM), Vol. 19, Issue 1, pp. 5-22, 2019. DOI: 10.1515/cmam-2018-0030 Preliminary version appeared as arXiv Preprint arXiv:1710.09470, October 2017. [J188] Space-time Galerkin POD with Application in Optimal Control of Semi-Linear Parabolic Partial Differential Equations Manuel Baumann, Peter Benner and Jan Heiland SIAM Journal on Scientific Computing, Vol. 40, Issue 3, pp. A1611-A1641, 2018. DOI: 10.1137/17M1135281 Preliminary version appeared as arXiv Preprint arXiv:1611.04050, November 2016. [J187] Model Reduction of Linear Multi-Agent Systems by Clustering and with H2- and H∞-Error Bounds Hidde-Jan Jongsma, Petar Mlinarić, Sara Grundel, Peter Benner, and Harry L. Trentelman Mathematics of Control, Signals, and Systems, Vol. 30, No. 6 (38 pp.), 2018. DOI: 10.1007/s00498-018-0212-6 Preliminary version appeared as arXiv Preprint arXiv:1610.02432, October 2016. [J186] Comparison of Model Order Reduction Methods for Optimal Sensor Placement for Thermo-Elastic Models Peter Benner, Roland Herzog, Norman Lang, Ilka Riedel, and Jens Saak Engineering Optimization, Vol. 51, No. 3, pp. 465-483, 2019. DOI: 10.1080/0305215X.2018.1469133 [J185] H2-Quasi-Optimal Model Order Reduction for Quadratic-Bilinear Control Systems Peter Benner, Pawan Goyal, and Serkan Gugercin SIAM Journal on Matrix Analysis and Applications, Vol. 39, Issue 2, pp. 983-1032, 2018. DOI: 10.1137/16M1098280 Preliminary version appeared as arXiv Preprint arXiv:1610.03279v1, October 2016. [J184] A GPU-aware Mixed-precision Solver for Low-rank Algebraic Riccati Equations Peter Benner, Ernesto Dufrechou, Pablo Ezzatti, Alfredo Remón, and Jens Saak Concurrency and Computation: Practice and Experience, Vol. 31, Issue 6, e4462 (12 pp.), 2019. DOI: 10.1002/cpe.4462 [J183] On Reduced Input-Output Dynamic Mode Decomposition Peter Benner, Christian Himpe, and Tim Mitchell Advances in Computational Mathematics, Vol. 44, Issue 6, pp. 1751-1768, 2018. DOI: 10.1007/s10444-018-9592-x Preliminary version appeared as arXiv Preprint arXiv:1712.08447 [cs.SY], December 2017. [J182] Some Remarks on the Complex J-Symmetric Eigenproblem Peter Benner, Heike Faßbender, and Chao Yang Linear Algebra and Its Applications, Vol. 544, pp. 407-442, 2018. DOI: 10.1016/j.laa.2018.01.014 Preliminary version appeared as MPI Magdeburg Preprint MPIMD/15-12, July 2015. [J181] Model Reduction by Iterative Error System Approximation Athanasios C. Antoulas, Peter Benner, and Lihong Feng Mathematical and Computer Modelling of Dynamical Systems, Vol. 24, Issue 2, pp. 103-118, 2018. DOI: 10.1080/13873954.2018.1427116 [J180] Range-Separated Tensor Format for Many-Particle Modeling Peter Benner, Venera Khoromskaia, and Boris N. Khoromskij SIAM Journal on Scientific Computing, Vol. 40, Issue 2, pp. A1034–A1062, 2017. DOI: 10.1137/16M1098930 Preliminary version appeared as arXiv Preprint arXiv:1606.09218, June 2016. [J179] L∞-Norm Computation for Large-Scale Descriptor Systems Using Structured Iterative Eigensolvers Peter Benner, Ryan Lowe, and Matthias Voigt Numerical Algebra, Control and Optimization (NACO), Vol. 8, Issue 1, pp. 119-133, 2018. DOI: 10.3934/naco.2018007 [J178] Large-Scale Computation of L∞-Norms by a Greedy Subspace Method Nicat Aliyev, Peter Benner, Emre Mengi, Paul Schwerdtner, and Matthias Voigt SIAM Journal on Matrix Analysis and Applications, Vol. 38, No. 4, pp. 1496–1516, 2017. DOI: 10.1137/16M1086200 Preliminary version appeared as arXiv Preprint arXiv:1705.10086, May 2017. [J177] A Linear Quadratic Control Problem for the Stochastic Heat Equation Driven by Q-Wiener Processes Peter Benner and Christoph Trautwein Journal of Mathematical Analysis and Applications, Vol. 457, No. 1, pp. 776-802, 2018. DOI: 10.1016/j.jmaa.2017.08.052 [J176] Exponential Stability and Stabilization of Extended Linearizations via Continuous Updates of Riccati-Based Feedback Peter Benner and Jan Heiland International Journal of Robust and Nonlinear Control, Vol. 28, No. 4, pp. 1218-1232, 2018. DOI: 10.1002/rnc.3949 Preliminary version appeared as arXiv Preprint arXiv:1607.08441, July 2016. [J175] Singular Perturbation Approximation for Linear Systems with Lévy Noise Peter Benner and Martin Redmann Stochastics and Dynamics, Vol. 18, No.4, 1850033 (23 pp.), 2018. DOI: 10.1142/S0219493718500338 Preliminary version appeared as MPI Magdeburg Preprint MPIMD/15-22, December 2015. [J174] RADI: A Low-Rank ADI-type Algorithm for Large Scale Algebraic Riccati Equations Peter Benner, Zvonimir Bujanović, Patrick Kürschner, and Jens Saak Numerische Mathematik, Vol. 138, No. 2, pp. 301-330, 2018. DOI: 10.1007/s00211-017-0907-5 Preliminary version appeared as "A Low-Rank Quadratic ADI Algorithm for Algebraic Riccati Equations", arXiv Preprint arXiv:1510.00040 [math.NA], September 2015. [J173] Optimal Control of a Stefan Problem Fully Coupled with Incompressible Navier-Stokes Equations and Mesh Movement Björn Baran, Peter Benner, Jan Heiland, and Jens Saak Analele Stiintifice ale Universitatii Ovidius Constanta: Seria Matematica, Vol. 26, No. 2, pp. 11-40, 2018. DOI: 10.2478/auom-2018-0016 [J172] Computing the Stochastic H∞-Norm by a Newton Iteration Tobias Damm, Peter Benner, and Jan Hauth IEEE Control Systems Letters, Vol. 1, No. 1, pp. 92-97, 2017. DOI: 10.1109/LCSYS.2017.2707409 Preliminary version appeared as arXiv Preprint arXiv:1703.04440, March 2017. [J171] An H2-Type Error Bound for Balancing-Related Model Order Reduction of Linear Systems with Lévy Noise Peter Benner and Martin Redmann Systems & Control Letters, Vol. 105, pp. 1-5, 2017. DOI: 10.1016/j.sysconle.2017.04.004 Preliminary version appeared as MPI Magdeburg Preprint MPIMD/15-20, December 2015. [J170] An Index-aware Parametric Model Order Reduction Method for Parametrized Quadratic Differential-Algebraic Equations Nicodemus Banagaaya, Peter Benner, Lihong Feng, Peter Meuris, and Wim Schoenmaker Applied Mathematics and Computation, Vol. 319, pp. 409-424, 2018. DOI: 10.1016/j.amc.2017.04.024 [J169] Moment-Matching Based Model Reduction for Navier-Stokes Type Quadratic-Bilinear Descriptor Systems Mian Ilyas Ahmad, Peter Benner, Pawan Goyal, and Jan Heiland Zeitschrift für Angewandte Mathematik und Mechanik (ZAMM), Vol. 97, No. 10, pp. 1252–1267, 2017. DOI: 10.1002/zamm.201500262 Preliminary version appeared as MPI Magdeburg Preprint MPIMD/15-18, October 2015. [J168] Adaptive POD-DEIM Basis Construction and its Application to a Nonlinear Population Balance System Lihong Feng, Michael Mangold, and Peter Benner AIChE Journal, Vol. 63, No. 9, pp. 3832–3844, 2017. DOI: 10.1002/aic.15749 [J167] Some a Posteriori Error Bounds for Reduced-Order Modelling of (Non-)Parametrized Linear Systems Lihong Feng, Athanasios C. Antoulas, and Peter Benner ESAIM: Mathematical Modelling and Numerical Analysis (M2AN), Vol. 51, No. 6, pp. 2127-2158, 2017. DOI: 10.1051/m2an/2017014 Preliminary version appeared as MPI Magdeburg Preprint MPIMD/15-17, October 2015. [J166] POD-DEIM for Efficient Reduction of a Dynamic 2D Catalytic Reactor Model Peter Benner, Jens Bremer, Lihong Feng, Pawan Goyal, and Kai Sundmacher Computers & Chemical Engineering, Vol. 106, pp. 777-784, 2017. DOI: 10.1016/j.compchemeng.2017.02.032 [J165] Adaptive Symmetric Interior Penalty Galerkin Method for Boundary Control Problems Peter Benner and Hamdullah Yücel SIAM Journal on Numerical Analysis, Vol. 55, No. 2, pp. 1101–1133, 2017. DOI: 10.1137/15M1034507 Preliminary version appeared as MPI Magdeburg Preprint MPIMD/15-13, August 2015. [J164] Structure Preserving Iterative Methods for Periodic Projected Lyapunov Equations and their Application in Model Reduction of Periodic Descriptor Systems Peter Benner and Mohammad-Sahadet Hossain Numerical Algorithms, Vol. 76, Issue 4, pp. 881-904, 2017. DOI: 10.1007/s11075-017-0288-y Preliminary version appeared as MPI Magdeburg Preprint MPIMD/15-08, July 2015. [J163] Fast Iterative Solution of the Bethe-Salpeter Eigenvalue Problem Using Low-Rank and QTT Tensor Approximation Peter Benner, Sergey Dolgov, Venera Khoromskaia, and Boris N. Khoromskij Journal of Computational Physics, Vol. 334, pp. 221–239, 2017. DOI: 10.1016/j.jcp.2016.12.047. Preliminary version appeared as arXiv Preprint arXiv:1602.02646v1 [math.NA], February 2016. [J162] An Isoperimetric Optimal Control Problem for a Non-Isothermal Chemical Reactor with Periodic Inputs Alexander Zuyev, Andreas Seidel-Morgernstern, and Peter Benner Chemical Engineering Science, Vol. 161, pp. 206–214, 2017. DOI: 10.1016/j.ces.2016.12.025. [J161] Extending the Gauss-Huard Method for the Solution of Lyapunov Matrix Equations and Matrix Inversion Peter Benner, Pablo Ezzatti, Enrique S. Quintana-Ortí, and Alfredo Remón Concurrency and Computation: Practice and Experience, Vol. 29, No. 9, pp. e4076 (13pp.), 2017. DOI: 10.1002/cpe.4076. [J160] Krylov Subspace-based Model Reduction for a Class of Bilinear Descriptor Systems Pawan Goyal, Mian Ilyas Ahmad, and Peter Benner Journal of Computational and Applied Mathematics, Vol. 315, pp. 303–318, 2017. DOI: 10.1016/j.cam.2016.11.009. Preliminary version appeared as MPI Magdeburg Preprint MPIMD/15-07, May 2015. [J159] Implicit Volterra Series Interpolation for Model Reduction of Bilinear Systems Mian Ilyas Ahmad, Ulrike Baur, and Peter Benner Journal of Computational and Applied Mathematics, Vol. 316, pp. 15-28, 2017. DOI: 10.1016/j.cam.2016.09.048. Preliminary version appeared as MPI Magdeburg Preprint MPIMD/15-21, December 2015. [J158] Accelerating Optimization and Uncertainty Quantification of Nonlinear SMB Chromatography Using Reduced-Order Models Yongjin Zhang, Lihong Feng, Andreas Seidel-Morgenstern, and Peter Benner Computers & Chemical Engineering, Vol. 96, pp. 237-247, 2017. DOI: 10.1016/j.compchemeng.2016.09.017. [J157] Parametric Modeling and Model Order Reduction for (Electro-)Thermal Analysis of Nanoelectronic Structures Lihong Feng, Yao Yue, Nicodemus Banagaaya, Peter Meuris, Wim Schoenmaker, and Peter Benner Journal of Mathematics in Industry, Vol. 6, Art. 10 (16 pp.), 2016. DOI: 10.1186/s13362-016-0030-8. [J156] Multipoint Interpolation of Volterra Series and H2-Model Reduction for a Family of Bilinear Descriptor Systems Peter Benner and Pawan Goyal Systems & Control Letters, Vol. 97, pp. 1-11, 2016. DOI: 10.1016/j.sysconle.2016.08.008. Preliminary version appeared as MPI Magdeburg Preprint MPIMD/15-16, September 2015. [J155] Structure Preserving Model Order Reduction of Large Sparse Second-Order Index-1 Systems and Application to a Mechatronics Model Peter Benner, Jens Saak, and M. Monir Uddin Mathematical and Computer Modelling of Dynamical Systems, Vol. 22, Issue 6, pp. 509-523, 2016. DOI: 10.1080/13873954.2016.1218347. Preliminary version appeared as MPI Magdeburg Preprint MPIMD/14-23, December 2014. [J154] Nanoelectronic COupled Problems Solutions - nanoCOPS: Modelling, Multirate, Model Order Reduction, Uncertainty Quantification, Fast Fault Simulation E. J. W. ter Maten, P. A. Putek, M. Günther, R. Pulch, C. Tischendorf, Ch. Strohm, W. Schoenmaker, P. Meuris, B. De Smedt, P. Benner, L. Feng, N. Banagaaya, Y. Yue et al. Journal of Mathematics in Industry, Vol. 7, Art. 2 (19 pp.), 2016. DOI: 10.1186/s13362-016-0025-5. Updated and extended version of OP17. [J153] Krylov Subspace Methods for Model Reduction of Quadratic-Bilinear Systems Mian Ilyas Ahmad, Peter Benner, and Imad M. Jaimoukha IET Control Theory & Applications, Vol. 10, Issue 16, pp. 2010–2018, 2016. DOI: 10.1049/iet-cta.2016.0415. [J152] An Inexact Low-Rank Newton-ADI Method for Large-Scale Algebraic Riccati Equations Peter Benner, Matthias Heinkenschloss, Jens Saak, and Heiko K. Weichelt Applied Numerical Mathematics, Vol. 108, pp. 125-142, June 2016. DOI: 10.1016/j.apnum.2016.05.006. Preliminary version appeared as MPI Magdeburg Preprint MPIMD/15-06, May 2015. [J151] Dual Pairs of Generalized Lyapunov Inequalities and Balanced Truncation of Stochastic Linear Systems Peter Benner, Tobias Damm, and Yolanda Rocio Rodriguez Cruz IEEE Transactions on Automatic Control, Vol. 62, No. 2, pp. 782-791, 2017. DOI: 10.1109/TAC.2016.2572881. [J150] Numerical Solution of the Infinite-Dimensional LQR Problem and the Associated Riccati Differential Equations Peter Benner and Hermann Mena Journal of Numerical Mathematics, Vol. 26, Issue 1, pp. 1-20, 2018. DOI: 10.1515/jnma-2016-1039. Preliminary version appeared as MPI Magdeburg Preprint MPIMD/12-13, August 2012. [J149] Computation of Thermo-Elastic Deformations on Machine Tools - A study of Numerical Methods Andreas Naumann, Norman Lang, Marian Partzsch, Michael Beitelschmidt, Peter Benner, Axel Voigt, and Jörg Wensch Production Engineering Research & Development, Vol. 10, Issue 3, pp. 253-263, 2016. DOI: 10.1007/s11740-016-0674-7. [J148] Trigonometric Spline and Spectral Bounds for the Solution of Linear Time-Periodic Systems Peter Benner, Jonas Denißen, and Ludwig Kohaupt Journal of Applied Mathematics and Computing, Vol. 54, Issue 1, pp. 127–157, 2017. DOI: 10.1007/s12190-016-1001-3. Preliminary version appeared as MPI Magdeburg Preprint MPIMD/16-01, January 2016. [J147] Low-Rank Solvers for Unsteady Stokes-Brinkman Optimal Control Problem with Random Data Peter Benner, Sergey Dolgov, Akwum Onwunta, and Martin Stoll Computer Methods in Applied Mechanics and Engineering, Vol. 304, pp. 26-54, 2016. DOI: 10.1016/j.cma.2016.02.004. Preliminary version appeared as MPI Magdeburg Preprint MPIMD/15-10, July 2015. [J146] Balancing Energy and Performance in Dense Linear System Solvers for Hybrid ARM+GPU Juan Pablo Silva, Ernesto Dufrechou, Pablo Ezzatti, Enrique S. Quintana-Ortí, Alfredo Remón, and Peter Benner CLEI Electronic Journal, Vol. 19, Issue 1, Paper 2 (13 pp.), 2016. DOI: 10.19153.19.1.2. [J145] Block-diagonal Preconditioning for Optimal Control Problems Constrained by PDEs with Uncertain Inputs Peter Benner, Akwum Onwunta, and Martin Stoll SIAM Journal on Matrix Analysis and Applications, Vol. 37, No. 2, pp. 491–518, 2016. DOI: 10.1137/15M1018502. Preliminary version appeared as MPI Magdeburg Preprint MPIMD/15-05, April 2015. [J144] A Reduced Basis Approach for Calculation of the Bethe-Salpeter Excitation Energies using Low-Rank Tensor Factorizations Peter Benner, Venera Khoromskaia, and Boris N. Khoromskij Molecular Physics, Vol. 114, Issue 7-8, pp. 1148-1161, 2016. DOI: 10.1080/00268976.2016.1149241. [J143] Frequency-Limited Balanced Truncation with Low-Rank Approximations Peter Benner, Patrick Kürschner, and Jens Saak SIAM Journal on Scientific Computing, Vol. 38, Issue 1, pp. A471–A499, 2016. DOI: 10.1137/15M1030911. Preliminary version appeared as MPI Magdeburg Preprint MPIMD/15-09, July 2015. [J142] Time-dependent Dirichlet Conditions in Finite Element Discretizations Peter Benner and Jan Heiland ScienceOpen Research, available online 3 October 2015. DOI: 10.14293/S2199-1006.1.SOR-MATH.AV2JW3.v1. Preliminary version appeared as MPI Magdeburg Preprint MPIMD/15-03, September 2012. [J141] On the Solution of Large-Scale Algebraic Riccati Equations by Using Low-Dimensional Invariant Subspaces Peter Benner and Zvonimir Bujanović Linear Algebra and Its Applications, Vol. 488, pp. 430-459, 2016. DOI: 10.1016/j.laa.2015.09.027. Preliminary version appeared as MPI Magdeburg Preprint MPIMD/14-15, August 2014. [J140] A Fully Adaptive Scheme for Model Order Reduction Based on Moment-Matching Lihong Feng, Jan G. Korvink, and Peter Benner IEEE Transactions on Components, Packaging, and Manufacturing Technology, Vol. 5, No. 12, pp. 1872-1884, 2015. DOI: 10.1109/TCPMT.2015.2491341. Preliminary version appeared as MPI Magdeburg Preprint MPIMD/12-14, September 2012. [J139] An Efficient Output Error Estimation for Model Order Reduction of Parametrized Evolution Equations Yongjin Zhang, Lihong Feng, and Peter Benner SIAM Journal on Scientific Computing, Vol. 37, No. 6, pp. B910–B936, 2015. DOI: 10.1137/140998603. [J138] Unleashing GPU Acceleration for Symmetric Band Linear Algebra Kernels and Model Reduction Peter Benner, Ernestro Dufrechou, Pablo Ezzatti, Enrique S. Quintana-Ortí, and Alfredo Remón Cluster Computing, Vol. 18, Issue 4, pp. 1351-1362, 2015. DOI: 10.1007/s10586-015-0489-x. [J137] A Discontinous Galerkin Method for Optimal Control Problems Governed by a System of Convection-Diffusion PDEs with Nonlinear Reaction Terms Hamdullah Yücel, Martin Stoll, and Peter Benner Computers and Mathematics with Applications, Vol. 70, Issue 10, pp. 2414-2431, 2015. DOI: 10.1016/j.camwa.2015.09.006. MPI Magdeburg Preprint MPIMD/13-10, July 2013. [J136] Estimating the Inf-Sup Constant in Reduced Basis Methods for Time-Harmonic Maxwell's Equations Martin W. Hess, Sara Grundel, and Peter Benner IEEE Transactions on Microwave Theory and Techniques, Vol. 63, Issue 11, pp. 3549-3557, 2015. DOI: 10.1109/TMTT.2015.2473157. Preliminary version appeared as MPI Magdeburg Preprint MPIMD/14-19, November 2014. [J135] Algorithm 961 - FORTRAN 77 Subroutines for the Solution of Skew-Hamiltonian/Hamiltonian Eigenproblems Peter Benner, Vasile Sima, and Matthias Voigt ACM Transactions on Mathematical Software (TOMS), Vol. 42, Issue 3, no. 24 (26 pp.), 2016. DOI: 10.1145/2818313. [J134] Balancing Based Model Reduction for Structured Index-2 Unstable Descriptor Systems with Application to Flow Control Peter Benner, Jens Saak, and M. Monir Uddin Numerical Algebra, Control and Optimization (NACO), Vol. 6, Issue 1, pp. 1-20, 2016. DOI: 10.3934/naco.2016.6.1. Preliminary version appeared as MPI Magdeburg Preprint MPIMD/14-20, November 2014. [J133] Modeling Glyphosate Aerial Spray Drift at the Ecuador-Colombia Border Peter Benner, Hermann Mena, and René Schneider Applied Mathematical Modelling, Vol. 40, No. 1, 373–387. 2016. DOI: 10.1016/j.apm.2015.04.057. Preliminary version appeared as MPI Magdeburg Preprint MPIMD/14-08, April 2014. [J132] Recycling BiCGSTAB with an Application to Parametric Model Order Reduction Kapil Ahuja, Eric de Sturler, Lihong Feng, and Peter Benner SIAM Journal on Scientific Computing, Vol. 37, Issue 5, pp. S429-S446, 2015. DOI: 10.1137/140972433. Preliminary version appeared as MPI Magdeburg Preprint MPIMD/13-21, November 2013. [J131] Model Reduction for Stochastic Systems Peter Benner and Martin Redmann Stochastic Partial Differential Equations: Analysis and Computations, Vol. 3, Issue 3, pp. 291-338, 2015. DOI: 10.1007/s40072-015-0050-1. Preliminary version appeared as MPI Magdeburg Preprint MPIMD/14-03, February 2014. [J130] Accelerating PDE Constrained Optimization by the Reduced Basis Method: Application to Batch Chromatography Yongjin Zhang, Lihong Feng, Suzhou Li, and Peter Benner International Journal for Numerical Methods in Engineering, Vol. 104, Issue 11, pp. 983-1007, 2015. DOI: 10.1002/nme.4950. Preliminary version appeared as MPI Magdeburg Preprint MPIMD/14-09, May 2014. [J129] A Survey of Projection-Based Model Reduction Methods for Parametric Dynamical Systems Peter Benner, Serkan Gugercin, and Karen Willcox SIAM Review, Vol. 57, Issue 4, pp. 483–531, 2015. DOI: 10.1137/130932715. Preliminary version appeared as MPI Magdeburg Preprint MPIMD/13-14, August 2013. [J128] Low-Rank Newton-ADI Methods for Large Nonsymmetric Algebraic Riccati Equations Peter Benner, Patrick Kürschner, and Jens Saak Journal of the Franklin Institute, Vol. 353, Issue 5, pp. 1147-1167, 2016. DOI: 10.1016/j.jfranklin.2015.04.016. Preliminary version appeared as MPI Magdeburg Preprint MPIMD/14-21, November 2014. [J127] Semi-Active Damping Optimization of Vibrational Systems using the Parametric Dominant Pole Algorithm Peter Benner, Patrick Kürschner, Zoran Tomljanović, and Ninoslav Truhar Zeitschrift für Angewandte Mathematik und Mechanik (ZAMM), Vol. 96, Issue 5, pp. 604-619, 2016. DOI: 10.1002/zamm.201400158. MPI Magdeburg Preprint MPIMD/14-12, July 2014. [J126] Towards the Identification of Heat Induction in Chip Removing Processes via an Optimal Control Approach Norman Lang, Jens Saak, Peter Benner, Steffen Ihlenfeldt, Steffen Nestmann, and Klaus Schädlich Production Engineering: Development & Research, Vol. 9, Issue 3, pp. 343-349, 2015. DOI: 10.1007/s11740-015-0608-9. [J125] Low Rank Solution of Unsteady Diffusion Equations with Stochastic Coefficients Peter Benner, Akwum Onwunta, and Martin Stoll SIAM/ASA Journal on Uncertainty Quantification, Vol. 3, No. 1, pp. 622–649, 2015. DOI: 10.1137/130937251. Preliminary version appeared as MPI Magdeburg Preprint MPIMD/13-13, August 2013. [J124] Model Order Reduction for Coupled Problems: Survey Peter Benner and Lihong Feng Applied and Computational Mathematics: An International Journal, Vol. 14, No. 1, pp. 3-22, 2015. Preliminary version appeared as MPI Magdeburg Preprint MPIMD/15-02, February 2015. [J123] Riccati-Based Boundary Feedback Stabilization of Incompressible Navier-Stokes Flow Eberhard Bänsch, Peter Benner, Jens Saak, and Heiko K. Weichelt SIAM Journal on Scientific Computing, Vol. 37, No. 2, pp. A832–A858, 2015. DOI: 10.1137/140980016. [J122] Parametric Model Order Reduction with a Small H2-Error Using Radial Basis Functions Peter Benner, Sara Grundel, Nils Hornung Advances in Computational Mathematics, Vol. 41, Issue 5, pp. 1231-1253, 2015. DOI: 10.1007/s10444-015-9410-7. Preliminary version appeared as MPI Magdeburg Preprint MPIMD/14-01, January 2014. [J121] Nonlinear Model Reduction of a Continuous Fluidized Bed Crystallizer Michael Mangold, Dmytro Khlopov, Stefan Palis, Lihong Feng, Peter Benner, Daniel Binev, and Andreas Seidel-Morgenstern Journal of Computational and Applied Mathematics, Vol. 289, pp. 253-266, 2015. DOI: 10.1016/j.cam.2015.01.028. [J120] Two-Sided Projection Methods for Nonlinear Model Order Reduction Peter Benner and Tobias Breiten SIAM Journal on Scientific Computing, Vol. 37, Issue 2, pp. B239–B260, 2015. DOI: 10.1137/14097255X. Preliminary version appeared as MPI Magdeburg Preprint MPIMD/12-12, June 2012. [J119] Positive Operators and Stable Truncation Peter Benner, Tobias Damm, Martin Redmann, and Yolanda Rocio Rodriguez Cruz Linear Algebra and its Applications, Vol. 498, pp. 74-87, 2016. DOI: 10.1016/j.laa.2014.12.005. Preliminary version appeared as MPI Magdeburg Preprint MPIMD/14-11, July 2014. [J118] Fast and Reliable Noise Estimation for Hyperspectral Subspace Identification Peter Benner, Vedran Novaković, Antonio Plaza, Enrique S. Quintana-Ortí, and Alfredo Remón IEEE Geoscience and Remote Sensing Letters, Vol. 12, Issue 6, pp. 1199-1203, June 2015. DOI: 10.1109/LGRS.2014.2388133. [J117] Uncertainty Quantification for Maxwell's Equations Using Stochastic Collocation and Model Order Reduction Peter Benner and Judith Schneider International Journal for Uncertainty Quantification, Vol. 5, No. 3, pp. 195–208, 2015. DOI: 10.1615/Int.J.UncertaintyQuantification.2015010170. Preliminary version appeared as MPI Magdeburg Preprint MPIMD/13-19, October 2013. [J116] Computing the Eigenvalues of Symmetric H2-Matrices by Slicing the Spectrum Peter Benner, Steffen Börm, Thomas Mach, and Knut Reimer Computing and Visualization in Science, Vol. 16, No. 6, pp. 271-282, published online 4 March 2015. DOI: 10.1007/s00791-015-0238-y. Preliminary version appeared as MPI Magdeburg Preprint MPIMD/14-06, March 2014. [J115] Extending Lyapack for the Solution of Band Lyapunov Equations on Hybrid CPU-GPU Platforms Peter Benner, Ernesto Dufrechou, Pablo Ezzatti, Enrique S. Quintana-Ortí, and Alfredo Remón The Journal of Supercomputing, Vol. 71, No. 2, pp. 740-750, 2015. DOI: 10.1007/s11227-014-1322-7. [J114] Self-Generating and Efficient Shift Parameters in ADI Methods for Large Lyapunov and Sylvester Equations Peter Benner, Patrick Kürschner, and Jens Saak Electronic Transactions on Numerical Analysis, Vol. 43, pp. 142-162, 2014. DOI: 10.17617/2.2071065. Preliminary version appeared as MPI Magdeburg Preprint MPIMD/13-18, October 2013. [J113] Parallel Model Order Reduction of Sparse Electromagnetic/Circuit Models Giovanni De Luca, Giulio Antonini, and Peter Benner ACES Journal, Vol. 30, No. 1, S1-21, January 2015. [J112] Adaptive Discontinuous Galerkin Methods for State Constrained Optimal Control Problems Governed by Convection Diffusion Equations Hamdullah Yücel and Peter Benner Computational Optimization and Applications, Vol. 62, pp. 291-321, 2015. DOI: 10.1007/s10589-014-9691-7. Preliminary version appeared as MPI Magdeburg Preprint MPIMD/13-16, September 2013. [J111] Model Order Reduction for a Family of Linear Systems with Application in Parametric and Uncertain Systems Peter Benner and Sara Grundel Applied Mathematics Letters, Vol. 39, pp. 1-6, 2015. DOI: 10.1016/j.aml.2014.08.002. Preliminary version appeared as MPI Magdeburg Preprint MPIMD/14-05, March 2014. [J110] Rational Interpolation Methods for Symmetric Sylvester Equations Peter Benner and Tobias Breiten Electronic Transactions on Numerical Analysis, Vol. 42, pp. 147-164, 2014. DOI: 10.17617/2.2060858. Preliminary version appeared as MPI Magdeburg Preprint MPIMD/13-23, December 2013. [J109] Model Reduction for Linear Simulated Moving Bed Chromatography Systems Using Krylov-Subspace Methods Suzhou Li, Yao Yue, Lihong Feng, Andreas Seidel-Morgenstern, and Peter Benner AIChE Journal, Vol. 60, No. 11, pp. 3773–3783, 2014. DOI: 10.1002/aic.14561. [J108] Model Order Reduction for Linear and Nonlinear Systems: a System-Theoretic Perspective Peter Benner, Ulrike Baur, and Lihong Feng Archives of Computational Methods in Engineering, Vol. 21, No. 4, pp. 331–358, 2014. DOI: 10.1007/s11831-014-9111-2. Preliminary version appeared as MPI Magdeburg Preprint MPIMD/14-07, March 2014. [J107] Model Order Reduction for Systems with Moving Loads Norman Lang, Jens Saak, and Peter Benner at-Automatisierungstechnik, Vol. 62, No. 7, pp. 512-522, 2014. DOI: 10.1515/auto-2014-1095. [J106] Parametric Model Order Reduction of Thermal Models Using the Bilinear Interpolatory Rational Krylov Algorithm Peter Benner and Angelika Bruns Mathematical and Computer Modelling of Dynamical Systems, Vol. 21, Issue 2, pp. 103-129, 2015. DOI: 10.1080/13873954.2014.924534. The hyperlink to the paper (title link) allows 50 free downloads of the full text. Afterwards, it directs to the paper abstract. [J105] A Reduced Basis Method for Microwave Semiconductor Devices with Geometric Variations Martin W. Hess and Peter Benner COMPEL: The International Journal for Computation and Mathematics in Electrical and Electronic Engineering, Vol. 33, No. 4, pp. 1071-1081, 2014. DOI: 10.1108/COMPEL-12-2012-0377. [J104] A Structured Pseudospectral Method for H∞-Norm Computation of Large-Scale Descriptor Systems Peter Benner and Matthias Voigt Mathematics of Control, Signals, and Systems, Vol. 26, No. 2, pp. 303-338, 2014. DOI: 10.1007/s00498-013-0121-7. Preliminary version appeared as MPI Magdeburg Preprint MPIMD/12-10, May 2012. [J103] Using Surrogate Models for Efficient Optimization of Simulated Moving Bed Chromatography Suzhou Li, Lihong Feng, Peter Benner, and Andreas Seidel-Morgenstern Computers & Chemical Engineering, Vol. 67, pp. 121-132, 2014. DOI: 10.1016/j.compchemeng.2014.03.024. [J102] Numerical Solution of Projected Algebraic Riccati Equations Peter Benner and Tatjana Stykel SIAM Journal on Numerical Analysis, Vol. 52, No. 2, pp. 581-600, 2014. DOI: 10.1137/130923993 Preliminary version appeared as MPI Magdeburg Preprint MPIMD/13-06, May 2013. [J101] Trading off Performance for Energy in Linear Algebra Operations with Applications in Control Theory Peter Benner, Pablo Ezzatti, Enrique S. Quintana-Ortí, and Alfredo Remón CLEI Electronic Journal, Vol. 17, No. 1, Paper 3 (12 pp.), 2014. DOI: 10.19153/cleiej.17.1.3 [J100] Computing Real Low-Rank Solutions of Sylvester Equations by the Factored ADI Method Peter Benner and Patrick Kürschner Computers and Mathematics with Applications, Vol. 67, Issue 9, pp. 1656–1672, 2014. DOI: 10.1016/j.camwa.2014.03.004 Preliminary version appeared as MPI Magdeburg Preprint MPIMD/13-05, May 2013. [J99] On Optimality of Approximate Low-Rank Solutions of Large-Scale Matrix Equations Peter Benner and Tobias Breiten Systems & Control Letters Vol. 67, pp. 55-64, 2014. DOI: 10.1016/j.sysconle.2014.02.005 Preliminary version appeared as MPI Magdeburg Preprint MPIMD/11-10, December 2011. [J98] Low-Rank Iterative Methods of Periodic Projected Lyapunov Equations and their Application in Model Reduction of Periodic Descriptor Systems Peter Benner, Mohammad-Sahadet Hossain, and Tatjana Stykel Numerical Algorithms, Vol. 67, No. 3, pp. 669–690, 2014. DOI: 10.1007/s11075-013-9816-6 Preliminary version appeared as Chemnitz Scientific Computing Preprints 11-01, TU Chemnitz, February 2011. [J97] Solving Matrix Equations on Multi-Core and Many-Core Architectures Peter Benner, Pablo Ezzatti, Hermann Mena, Enrique S. Quintana-Ortí, and Alfredo Remón Algorithms, Vol. 6, No. 4, pp. 857-870, 2013. DOI: 10.3390/a6040857 Preliminary version appeared as MPI Magdeburg Preprint MPIMD/11-07, October 2011. [J96] On the Squared Smith Method for Large-Scale Stein Equations Peter Benner, Grece El Khoury, and Miloud Sadkane Numerical Linear Algebra with Applications, Vol. 21, Issue 5, pp. 645-665, 2014. DOI: 10.1002/nla.1918 Preliminary version appeared as MPI Magdeburg Preprint MPIMD/12-15, September 2012. [J95] Fast solution of Cahn-Hilliard Variational Inequalities using Implicit Time Discretization and Finite Elements Jessica Bosch, Martin Stoll, and Peter Benner Journal of Computational Physics, Vol. 262, pp. 38-57, 2014. DOI: 10.1016/j.jcp.2013.12.053 Preliminary version appeared as MPI Magdeburg Preprint MPIMD/13-01, January 2013. [J94] A Factored Variant of the Newton Iteration for the Solution of Algebraic Riccati Equations via the Matrix Sign Function Peter Benner, Pablo Ezzatti, Enrique S. Quintana-Ortí, and Alfredo Remón Numerical Algorithms, Vol. 66, No. 2, pp. 363-377, 2014. DOI: 10.1007/s11075-013-9739-2 [J93] Numerical Solution of Large and Sparse Continuous Time Algebraic Matrix Riccati and Lyapunov Equations: A State of the Art Survey Peter Benner and Jens Saak GAMM Mitteilungen, Vol. 36, No. 1, pp. 32-52, 2013. DOI: 10.1002/gamm.201310003 [J92] Discontinuous Galerkin Finite Element Methods with Shock-Capturing for Nonlinear Convection Dominated Models Hamdullah Yücel, Martin Stoll, and Peter Benner Computers & Chemical Engineering, Vol. 58, pp. 278-287, 2013. DOI: 10.1016/j.compchemeng.2013.07.011 Preliminary version appeared as MPI Magdeburg Preprint MPIMD/13-03, January 2013. [J91] Rosenbrock Methods for Solving Differential Riccati Equations Peter Benner and Hermann Mena IEEE Transactions on Automatic Control, Vol. 58, No. 11, pp. 2950-2956, 2013. DOI: 10.1109/TAC.2013.2258495 Preliminary version appeared as MPI Magdeburg Preprint MPIMD/11-06, October 2011. © IEEE. [J90] A Non-Conforming Composite Quadrilateral Finite Element Pair for Feedback Stabilization of the Stokes Equations Peter Benner, Jens Saak, Friedhelm Schieweck, Piotr Skrzypacz, and Heiko K. Weichelt Journal of Numerical Mathematics, Vol. 22, No. 3, pp. 191-220, 2014. DOI: 10.1515/jnma-2014-0009 Preliminary version appeared as MPI Magdeburg Preprint MPIMD/12-19, October 2012. [J89] Fast Evaluation of Time-Harmonic Maxwell's Equations Using the Reduced Basis Method Peter Benner and Martin W. Hess IEEE Transactions on Microwave Theory and Techniques, Vol. 61, No. 6, pp. 2265-2274, 2013. DOI: 10.1109/TMTT.2013.2258167 Preliminary version appeared as MPI Magdeburg Preprint MPIMD/12-17, October 2012. [J88] An Improved Numerical Method for Balanced Truncation for Symmetric Second-Order Systems Peter Benner, Patrick Kürschner, and Jens Saak Mathematical and Computer Modelling of Dynamical Systems, Vol. 19, No. 6, pp. 593-615, 2013. DOI: 10.1080/13873954.2013.794363. Preliminary version appeared as MPI Magdeburg Preprint MPIMD/12-20, November 2012. [J87] Efficient Solution of Large-Scale Saddle Point Systems Arising in Riccati-Based Boundary Feedback Stabilization of Incompressible Stokes Flow Peter Benner, Jens Saak, Martin Stoll and Heiko K. Weichelt SIAM Journal on Scientific Computing, Vol. 35, No. 5, pp. S150-S170, 2013. DOI: 10.1137/120881312. Preliminary version appeared as Preprint SPP1253-130 of the DFG Priority Programme 1253 Optimization with Partial Differential Equations, June 2012. [J86] The LR Cholesky Algorithm for Symmetric Hierarchical Matrices Peter Benner and Thomas Mach Linear Algebra and its Applications, Vol. 439, No. 4, pp. 1150-1166, 2013. DOI: 10.1016/j.laa.2013.03.001. Preliminary version appeared as MPI Magdeburg Preprint MPIMD/12-05, February 2012. [J85] Spectral Characterization and Enforcement of Negative Imaginariness for Descriptor Systems Peter Benner and Matthias Voigt Linear Algebra and its Applications, Vol. 439, No. 4, pp. 1104-1129, 2013. DOI: 10.1016/j.laa.2012.12.044. Preliminary version appeared as MPI Magdeburg Preprint MPIMD/12-09, April 2012. [J84] Low Rank Methods for a Class of Generalized Lyapunov Equations and Related Issues Peter Benner and Tobias Breiten Numerische Mathematik, Vol. 124, No. 3, pp. 441-470, 2013. DOI: 10.1007/s00211-013-0521-0. Preliminary version appeared as MPI Magdeburg Preprint MPIMD/12-03, February 2012. [J83] Subspace Recycling Accelerates the Parametric Macro‐modeling of MEMS Lihong Feng, Peter Benner, and Jan G. Korvink International Journal for Numerical Methods in Engineering, Vol. 94, No. 1, pp. 84-110, 2013. DOI: 10.1002/nme.4449. Preliminary version appeared as Fast Model Reduction of Parametric Systems using Subspace Recycling MPI Magdeburg Preprint MPIMD/12-08, March 2012. [J82] Model Reduction of an Elastic Crankshaft for Elastic Multibody Simulations Christine Nowakowski, Patrick Kürschner, Peter Eberhard, and Peter Benner Zeitschrift für Angewandte Mathematik und Mechanik (ZAMM), Vol. 93, No. 4, pp. 198-216, 2013. DOI: 10.1002/zamm.201200054. Preliminary version appeared as MPI Magdeburg Preprint MPIMD/12-07, March 2012. [J81] Computation of a Compact State Space Model for an Adaptive Spindle Head Configuration with Piezo Actuators using Balanced Truncation M. Monir Uddin, Jens Saak, Burkhard Kranz, and Peter Benner Production Engineering, Vol. 6, No. 6, pp. 577-586, 2012. DOI: 10.1007/s11740-012-0410-x. [J80] Matrix Inversion on CPU-GPU Platforms with Applications in Control Theory Peter Benner, Pablo Ezzatti, Enrique S. Quintana-Ortí, and Alfredo Remón Concurrency and Computation: Practice and Experience, Vol. 25, No. 8, pp. 1170-1182, 2013. DOI: 10.1002/cpe.2933. Preliminary version appeared as MPI Magdeburg Preprint MPIMD/12-02, February 2012. [J79] Interpolation-Based H2-Model Reduction of Bilinear Control Systems Peter Benner and Tobias Breiten SIAM Journal on Matrix Analysis and Applications, Vol. 33, No. 3, pp. 859-885, 2012. DOI: 10.1137/110836742. Preliminary version appeared as MPI Magdeburg Preprint MPIMD/11-02, June 2011. [J78] HPC en simulación y control a gran escala [in Spanish] Peter Benner, Pablo Ezzatti, Hermann Mena, Enrique S. Quintana-Ortí, and Alfredo Remón Revista ELEMENTOS, Vol. 3, No. 3, pp. 9-35, 2013. [J77] Efficient Handling of Complex Shift Parameters in the Low-Rank Cholesky Factor ADI method Peter Benner, Patrick Kürschner, and Jens Saak Numerical Algorithms, Vol. 62, No. 2, pp. 225-251, 2013. DOI: 10.1007/s11075-012-9569-7. Preliminary version appeared as MPI Magdeburg Preprint MPIMD/11-08, October 2011. [J76] A Mathematical Biography of Danny C. Sorensen Peter Benner, Mark Embree, Richard B. Lehoucq, and C.T. Kelley Linear Algebra and its Applications, Vol. 436, Issue 8, pp. 2717-2724, 2012. DOI: 10.1016/j.laa.2012.01.031. [J75] The Preconditioned Inverse Iteration for Hierarchical Matrices Peter Benner and Thomas Mach Numerical Linear Algebra with Applications, Vol. 20, No. 1, pp. 150-166, 2013. DOI: 10.1002/nla.1830. Preliminary version appeared as MPI Magdeburg Preprint MPIMD/11-01, February 2011. [J74] Optimal Damping of Selected Eigenfrequencies Using Dimension Reduction Peter Benner, Zoran Tomljanović, and Ninoslav Truhar Numerical Linear Algebra with Applications, Vol. 20, No. 1, pp. 1-17, 2013. DOI: 10.1002/nla.833. Preliminary version appeared as Preprint 1/2011, Department of Mathematics, J. J. Strossmayer University of Osijek, March 2011. [J73] Computing All or Some Eigenvalues of Symmetric ℋℓ-Matrices Peter Benner and Thomas Mach SIAM Journal on Scientific Computing, Vol. 34, Issue 1, pp. A485-A496, 2012. DOI: 10.1137/100815323. Preliminary version appeared as MPI Magdeburg Preprint MPIMD/10-01, November 2010. [J72] L∞-Norm Computation for Continuous-Time Descriptor Systems Using Structured Matrix Pencils Peter Benner, Vasile Sima, and Matthias Voigt IEEE Transactions on Automatic Control, Vol. 57, No. 1, pp. 233-238, 2012. DOI: 10.1109/TAC.2011.2161833. [J71] Interpolatory Projection Methods for Parameterized Model Reduction Ulrike Baur, Christopher Beattie, Peter Benner, and Serkan Gugercin SIAM Journal on Scientific Computing, Vol. 33, Issue 5, pp. 2489-2518, 2011. DOI: 10.1137/090776925. Preliminary version appeared as Chemnitz Scientific Computing Preprints 09-08, TU Chemnitz, November 2009. [J70] Generalized Tangential Interpolation for Model Reduction of Discrete-Time MIMO Bilinear Systems Peter Benner, Tobias Breiten, and Tobias Damm International Journal of Control, Vol. 84, No. 8 pp. 1398-1407, 2011. DOI: 10.1080/00207179.2011.601761. Preliminary version appeared as MPI Magdeburg Preprint MPIMD/10-02, December 2010. [J69] Robust Formulas for Optimal H∞ Controllers Peter Benner, Ralph Byers, Philip Losse, Volker Mehrmann, and Hongguo Xu Automatica, Vol. 47, No. 12, pp. 2639-2646, 2011. DOI: 10.1016/j.automatica.2011.09.013. Preliminary version appeared as Preprint 20-2010, Preprint series of the Institute of Mathematics, Technische Universität Berlin, September 2010. [J68] Lyapunov Equations, Energy Functionals, and Model Order Reduction of Bilinear and Stochastic Systems Peter Benner and Tobias Damm SIAM Journal on Control and Optimization, Vol. 49, No. 2, pp. 686-711, 2011. DOI: 10.1137/09075041X. [J67] A Mixed-Precision Algorithm for the Solution of Lyapunov Equations on Hybrid CPU-GPU Platforms Peter Benner, Pablo Ezzatti, Daniel Kressner, Enrique S. Quintana-Ortí, and Alfredo Remón Parallel Computing, Vol. 37, Issue 8, pp. 439–450, 2011. DOI: 10.1016/j.parco.2010.12.002. Preliminary version appeared as Research Report No. 2009-40, Seminar für Angewandte Mathematik, ETH Zürich, December 2009. [J66] Efficient Balancing-Based MOR for Large-Scale Second-Order Systems Peter Benner and Jens Saak Mathematical and Computer Modelling of Dynamical Systems, Vol. 17, No. 2, pp. 123-143, 2011. Author Posting. (c) Taylor & Francis, 2011. This is the author's version of the work. It is posted here by permission of Taylor & Francis for personal use, not for redistribution. The definitive version was published in January 2011. DOI: 10.1080/13873954.2010.540822. [J65] Dimension Reduction for Damping Optimization in Linear Vibrating Systems Peter Benner, Zoran Tomljanović, and Ninoslav Truhar Zeitschrift für Angewandte Mathematik und Mechanik (ZAMM), Vol. 91, No. 3, pp. 179–191, 2011. Editor's choice of this ZAMM issue! DOI: 10.1002/zamm.201000077. Preliminary version appeared as Preprint 2/2010, Department of Mathematics, J. J. Strossmayer University of Osijek, April 2010. [J64] On the Numerical Solution of Large-Scale Sparse Discrete-Time Riccati Equations Peter Benner and Heike Faßbender Advances in Computational Mathematics, Vol. 35, No. 2-4, pp. 119-147, 2011. DOI: 10.1007/s10444-011-9174-7. Preliminary version appeared as Chemnitz Scientific Computing Preprints 09-11, TU Chemnitz, December 2009. [J63] Parameter Preserving Model Order Reduction for MEMS Applications Ulrike Baur, Peter Benner, Andreas Greiner, Jan G. Korvink, Jan Lienemann, and Christian Moosmann Mathematical and Computer Modelling of Dynamical Systems, Vol. 17, No. 4, pp. 297-317, 2011. DOI: 10.1080/13873954.2011.547658. [J62] A Hamiltonian Krylov-Schur-type method based on the symplectic Lanczos process Peter Benner, Heike Faßbender, and Martin Stoll Linear Algebra and its Applications, Vol. 435, Issue 3, pp. 578-600, 2011. DOI: 10.1016/j.laa.2010.04.048. Preliminary version appeared as OCCAM Report Number 09/32, Oxford Centre for Collaborative Applied Mathematics, July 2009. [J61] On the QR Decomposition of H-Matrices. Peter Benner and Thomas Mach Computing, Vol. 88, No. 3-4, pp. 111-129, 2010. DOI: 10.1007/s00607-010-0087-y. See also Chemnitz Scientific Computing Preprints09-04, TU Chemnitz, July 2009. [J60] Die SLICOT-Toolboxen für Matlab (The SLICOT-Toolboxes for Matlab) [German] Peter Benner, Daniel Kressner, Vasile Sima, and Andras Varga at-Automatisierungstechnik, Vol. 58, Issue 1, pp. 15-25, 2010. DOI: 10.1524/auto.2010.0814. For an English version, see The SLICOT Toolboxes - a Survey, SLICOT Working Note 2009-1, August 2009. [J59] Modale versus moderne Ordnungsreduktionsverfahren: Effiziente Simulation von Werkzeugmaschinen [German] Thomas Bonin, Michael Zäh, Andreas Soppa, Heike Faßbender, Jens Saak, and Peter Benner MECHATRONIK, Vol. 11-12/2009, pp. 46-47, 2009. [J58] On the ADI Method for Sylvester Equations Peter Benner, Reng-Cang Li, and Ninoslav Truhar Journal of Computational and Applied Mathematics, Vol. 233, No. 4, pp. 1035-1045, 2009. DOI: 10.1016/j.cam.2009.08.108. [J57] Model Reduction for Parametric Systems Using Balanced Truncation and Interpolation [German] Ulrike Baur and Peter Benner at-Automatisierungstechnik Vol. 57, No. 8, pp. 411-419, 2009. DOI: 10.1524/auto.2009.0787. [J56] A Rational SHIRA Method for the Hamiltonian Eigenvalue Problem Peter Benner and Cedric Effenberger Taiwanese Journal of Mathematics, Vol. 14, No. 3A, pp. 805-823, 2010. See also Chemnitz Scientific Computing Preprints 08-08, TU Chemnitz, December 2008. [J55] Solving Large-Scale Quadratic Eigenvalue Problems with Hamiltonian Eigenstructure using a Structure-Preserving Krylov Subspace Method Peter Benner, Heike Faßbender, and Martin Stoll Electronic Transactions on Numerical Analysis, Vol. 29, pp. 212-229, 2008. See also Numerical Analysis Group Research Report NA-07/03, Oxford University, February 2007. [J54] Cross-Gramian Based Model Reduction for Data-Sparse Systems Ulrike Baur and Peter Benner Electronic Transactions on Numerical Analysis, Vol. 31, pp. 256-270, 2008. [J53] Numerical Solution of Large Lyapunov Equations, Riccati Equations, and Linear-Quadratic Control Problems Peter Benner, Jing-Rebecca Li, and Thilo Penzl Numerical Linear Algebra with Applications, Vol. 15, No. 9, pp. 755-777, 2008. DOI: 10.1002/nla.622 [J52] Gramian-Based Model Reduction for Data-Sparse Systems Ulrike Baur and Peter Benner SIAM Journal on Scientific Computing, Vol. 31, No. 1, pp. 776-798, 2008. DOI: 10.1137/070711578 See also Chemnitz Scientific Computing Preprints 07-01, TU Chemnitz, February 2007. [J51] Solving Linear-Quadratic Optimal Control Problems on Parallel Computers Peter Benner, Enrique S. Quintana-Ortí, and Gregorio Quintana-Ortí Optimization Methods and Software, Vol. 23, No. 6, pp. 879-909, 2008. DOI: 10.1080/10556780802058721 See also: Preprint SFB393/05-19, Sonderforschungsbereich 393 - Numerische Simulation auf massiv parallelen Rechnern, TU Chemnitz, March 2006. [J50] On the Parameter Selection Problem in the Newton-ADI Iteration for Large-Scale Riccati Equations Peter Benner, Hermann Mena, and Jens Saak Electronic Transactions on Numerical Analysis, Vol. 29, pp. 136-149, 2008. [J49] Efficient Algorithms for Generalized Algebraic Bernoulli Equations based on the Matrix Sign Function Sergio Barrachina, Peter Benner, and Enrique S. Quintana-Ortí Numerical Algorithms, Vol. 46, No. 4, pp. 351-368, 2007. DOI: 10.1007/s11075-007-9143-x Partially extracted from [TR15]. [J48] A Robust Numerical Method for the $\gamma$-Iteration in $H_\infty$ Control Peter Benner, Ralph Byers, Volker Mehrmann, and Hongguo Xu Linear Algebra and its Applications, Vol. 425, No. 2-3, pp. 548-570, 2007. DOI: 10.1016/j.laa.2007.03.026 Revised version of [TR13]. [J47] On the Solution of the Rational Matrix Equation X = Q + LX^{-1}L^T Peter Benner and Heike Faßbender EURASIP Journal on Advances in Signal Processing, special issue on "Numerical Linear Algebra in Signal Processing Application", Vol. 2007, Article ID 21850, 10 pages, 2007. DOI: 10.1155/2007/21850 See also: Chemnitz Scientific Computing Preprints 06-02, TU Chemnitz, September 2006. [J46] Factorized Solution of Lyapunov Equations Based on Hierarchical Matrix Arithmetic Ulrike Baur and Peter Benner Computing, Vol. 78, No. 3, pp. 211-234, 2006. DOI: 10.1007/s00607-006-0178-y [J45] Stabilizing Large-Scale Generalized Systems on Parallel Computers Using Multithreading and Message-Passing Peter Benner, Maribel Castillo, Rafael Mayo, Enrique S. Quintana-Ortí, and Gregorio Quintana-Ortí Concurrency and Computation: Practice and Experience, Vol. 19, No. 4, pp. 531-542, 2007. DOI: 10.1002/cpe.1148 [J44] Partial Realization of Descriptor Systems Peter Benner and Viatcheslav I. Sokolov Systems & Control Letters, Vol. 55, Issue 11, pp. 929-938, 2006. DOI: 10.1016/j.sysconle.2006.06.009 [J43] Evaluation of the Linear Matrix Equation Solvers in SLICOT Martin Slowik, Peter Benner, and Vasile Sima Journal of Numerical Analysis, Industrial and Applied Mathematics, Vol. 2, No. 1-2, pp. 11-34, 2007. Extendend version appeared as SLICOT Working Note 2004-1, August 2004. [J42] An Arithmetic for Matrix Pencils: Theory and New Algorithms Peter Benner and Ralph Byers Numerische Mathematik, Vol. 103, No. 4, pp. 539-573, 2006. DOI: 10.1007/s00211-006-0001-x [J41] Numerische Methoden zur passivitätserhaltenden Modellreduktion (Numerical Methods for Passivity Preserving Model Reduction) (in German) Peter Benner and Heike Faßbender at-Automatisierungstechnik, Vol. 54, No. 4, pp. 153-160, 2006. DOI: 10.1524/auto.2006.54.4.153 [J40] Numerical Linear Algebra for Model Reduction in Control and Simulation Peter Benner GAMM Mitteilungen, Vol. 29, No. 2, pp. 275-296, 2006. DOI: 10.1002/gamm.201490034 See also Preprint 18/2005, Fakultät für Mathematik, TU Chemnitz. [J39] Parallel Solution of Large-Scale Algebraic Bernoulli Equations with the Matrix Sign Function Method Sergio Barrachina, Peter Benner, and Enrique S. Quintana-Ortí International Journal of Computational Science and Engineering, Vol. 4, No. 2, pp. 88-93, 2009. DOI: 10.1504/IJCSE.2009.027000 Extended version of [CA45]. [J38] Algorithm 854: Fortran 77 Subroutines for Computing the Eigenvalues of Hamiltonian Matrices Peter Benner and Daniel Kressner ACM Transactions on Mathematical Software, Vol. 32, No. 2, pp. 352-373, 2006. DOI: 10.1145/1141885.1141895 For the corresponding software visit the HAPACK homepage. [J37] Solving Stable Sylvester Equations via Rational Iterative Schemes Peter Benner, Enrique S. Quintana-Ortí, and Gregorio Quintana-Ortí Journal of Scientific Computing, Vol. 28, No. 1, pp. 51-83, 2006. DOI: 10.1007/s10915-005-9007-2 Note: in Algorithm 3, Step 9 must read: U := U-CB; in Example 1, the entries of the solution matrix X need to be negated. [J36] Balancing Sparse Hamiltonian Eigenproblems Peter Benner and Daniel Kressner Linear Algebra and its Applications, Vol. 415, Issue 1, pp. 3-19, 2006. Original available from ScienceDirect, DOI: 10.1016/j.laa.2004.09.023 [J35] Partial Stabilization of Large-Scale Discrete-Time Linear Control Systems Peter Benner, Maribel Castillo, and Enrique S. Quintana-Ortí International Journal of Computational Science and Engineering, Vol. 1, No. 1, pp. 15-21, 2005. DOI: 10.1504/IJCSE.2005.008907 See also [CA19] [J34] Solving Large-Scale Control Problems Peter Benner IEEE Control Systems Magazine, Vol. 24, No. 1, pp. 44-59, 2004. DOI: 10.1109/MCS.2004.1272745 [J33] Parallel Algorithms for Model Reduction of Discrete-Time Systems Peter Benner, Enrique S. Quintana-Ortí, and Gregorio Quintana-Ortí International Journal of System Sciences, Vol. 34, No. 5, pp. 319-333, 2003. DOI: 10.1080/0020772031000158564 The retrievable version appeared as Berichte aus der Technomathematik, Report 00-18, December 2000. [J32] State-Space Truncation Methods for Parallel Model Reduction of Large-Scale Systems Peter Benner, Enrique S. Quintana-Ortí, and Gregorio Quintana-Ortí Parallel Computing, Vol. 29 (special issue on "Parallel and Distributed Scientific and Engineering Computing"), Issue 11-12, pp. 1701-1722, 2003. DOI: 10.1016/j.parco.2003.05.013 [J31] Structure Preservation: A Challenge in Computational Control Peter Benner, Daniel Kressner, and Volker Mehrmann Future Generation Computer Systems, Vol. 19, Issue 7, pp. 1243-1252, 2003. DOI: 10.1016/S0167-739X(03)00049-9 [J30] The Kleinman Iteration for Nonstabilizable Systems Peter Benner, Vicente Hernández, and Antonio Pastor Mathematics of Control, Signals, and Systems , Vol. 16, Issue 1,pp. 76-93, 2003. DOI: 10.1007/s00498-003-0130-z The retrievable version appeared as Berichte aus der Technomathematik, Report 99-11, October 1999. [J29] Numerical Solution of Discrete Stable Linear Matrix Equations on Multicomputers Peter Benner, Enrique S. Quintana-Ortí, and Gregorio Quintana-Ortí Parallel Algorithms and Applications (since 2005: International Journal of Parallel, Emergent and Distributed Systems) Vol. 17, No. 2, pp. 127-146, 2002. DOI: 10.1080/10637190208941436 Free preprint available at Berichte aus der Technomathematik, Report 99-13, November 1999. [J28] Numerical Computation of Deflating Subspaces of Skew-Hamiltonian/Hamiltonian Pencils Peter Benner, Ralph Byers, Volker Mehrmann, and Hongguo Xu SIAM Journal on Matrix Analysis and Applications, Vol. 24, No. 1, pp. 165-190, 2002. DOI: 10.1137/S0895479800367439 For an extended version, including detailed algorithms, see [TR10] . [J27] Parallel Algorithms for LQ Optimal Control of Discrete-Time Periodic Linear Systems Peter Benner, Ralph Byers, Rafael Mayo, Enrique S. Quintana-Ortí, and Vicente Hernández Journal of Parallel and Distributed Computing, Vol. 62, Issue 2, pp. 306-325, 2002. DOI: 10.1006/jpdc.2001.1790 The retrievable version appeared as Berichte aus der Technomathematik, Report 01-02, February 2001. [J26] Efficient Numerical Algorithms for Balanced Stochastic Truncation Peter Benner, Enrique S. Quintana-Ortí, and Gregorio Quintana-Ortí International Journal of Applied Mathematics and Computer Science, Vol. 11, No. 5, pp. 1123-1150, 2001. The retrievable version appeared as Berichte aus der Technomathematik, Report 01-03, March 2001. [J25] Perturbation Analysis for the Eigenvalue Problem of a Formal Product of Matrices Peter Benner, Volker Mehrmann, and Hongguo Xu BIT, Vol. 42, No. 1, pp. 1-43, 2002. DOI: 10.1023/A:1021966001542 The retrievable version appeared as Berichte aus der Technomathematik, Report 00-01, January 2000. [J24] A Hybrid Method for the Numerical Solution of Discrete-Time Algebraic Riccati Equations Peter Benner and Heike Faßbender Contemporary Mathematics, Vol. 280, pp. 255-269, 2001. The retrievable version appeared as Berichte aus der Technomathematik, Report 99-12, October 1999. [J23] Cholesky-like Factorizations of Skew-Symmetric Matrices Peter Benner, Ralph Byers, Heike Faßbender, Volker Mehrmann, and David S. Watkins Electronic Transactions on Numerical Analysis, Vol. 11, pp. 85-93, 2000. [J22] Symplectic Balancing of Hamiltonian Matrices Peter Benner SIAM Journal on Scientific Computing, Vol. 22, No. 5, pp. 1885-1904, 2001. DOI: 10.1137/S1064827500367993 [J21] Evaluating Products of Matrix Pencils and Collapsing Matrix Products Peter Benner and Ralph Byers Numerical Linear Algebra with Applications, Vol. 8, No. 6-7, pp. 357-380, 2001. DOI: 10.1002/nla.251 [J20] Balanced Truncation Model Reduction of Large-Scale Dense Systems on Parallel Computers Peter Benner, Enrique S. Quintana-Ortí, and Gregorio Quintana-Ortí Mathematical and Computer Modelling of Dynamical Systems, Vol. 6, No.4, pp. 383-405, 2000. DOI: 10.1076/mcmd.6.4.383.3658 The retrievable version appeared as Berichte aus der Technomathematik, Report 99-07, September 1999. [J19] An Implicitly Restarted Symplectic Lanczos Method for the Symplectic Eigenvalue Problem Peter Benner and Heike Faßbender SIAM Journal on Matrix Analysis and Applications, Vol. 22, No. 3, pp. 682-713, 2001. DOI: 10.1137/S0895479898343115 Retrieve preprint: Berichte aus der Technomathematik, Report 98-01, July 1998. [J18] Algorithm 800: Fortran 77 Subroutines for Computing the Eigenvalues of Hamiltonian Matrices I: The Square-Reduced Method Peter Benner, Ralph Byers, and Eric Barth ACM Transactions on Mathematical Software, Vol. 26, No. 1, pp. 49-77, 2000. DOI: 10.1145/347837.347852 For the corresponding software, click here (shell script, compressed using gzip). See also: [TR5] and Peter Benner's software. [J17] Solving Algebraic Riccati Equations on Parallel Computers Using Newton's Method with Exact Line Search Peter Benner, Ralph Byers, Enrique S. Quintana-Ortí, and Gregorio Quintana-Ortí Parallel Computing, Vol. 26, No. 10, pp. 1345-1368, 2000. DOI: 10.1016/S0167-8191(00)00012-0 The retrievable version appeared as Berichte aus der Technomathematik, Report 98-05, August 1998. [J16] A Note on the Numerical Solution of Complex Hamiltonian and Skew-Hamiltonian Eigenvalue Problems Peter Benner, Volker Mehrmann, and Hongguo Xu Electronic Transactions on Numerical Analysis, Vol. 8, pp. 115-126, 1999. [J15] Solving Stable Generalized Lyapunov Equations with the Matrix Sign Function Peter Benner and Enrique S. Quintana-Ortí Numerical Algorithms, Vol. 20, No. 1, pp. 75-100, 1999. DOI: 10.1023/A:1019191431273 Preliminary version appeared as Preprint SFB393/97-23, Sonderforschungsbereich 393 - Numerische Simulation auf massiv parallelen Rechnern, TU Chemnitz, October 1997. [J14] Parallel Partial Stabilizing Algorithms for Large Linear Control Systems Peter Benner, Maribel Castillo, Enrique S. Quintana-Ortí, and Vicente Hernández The Journal of Supercomputing, Vol. 15, Issue 2, pp. 193-206, 2000. DOI: 10.1023/A:1008108004247 The retrievable version appeared as Berichte aus der Technomathematik, Report 98-03, July 1998. [J13] Parallel Distributed Solvers for Large Stable Generalized Lyapunov Equations Peter Benner, José M. Claver, and Enrique S. Quintana-Ortí DOI: 10.1142/S0129626499000165 Parallel Processing Letters, Vol. 9, No. 1, pp. 147-158, 1999. [J12] SR and SZ Algorithms for the Symplectic (Butterfly) Eigenproblem Peter Benner, Heike Faßbender, and David S. Watkins Linear Algebra and its Applications, Vol. 287, Issue 1-3, pp. 41-76, 1999. DOI: 10.1016/S0024-3795(98)10090-3 [J11] SLICOT - A Subroutine Library in Systems and Control Theory Peter Benner, Volker Mehrmann, Vasile Sima, Sabine Van Huffel, and Andras Varga Applied and Computational Control, Signals, and Circuits, Vol. 1, pp. 505-546, 1999. The retrievable version appeared as NICONET Report 97-03, Working Group on Software, June 1997. [J10] Computational Methods for Linear-Quadratic Optimization Peter Benner Rendiconti del Circolo Matematico di Palermo, Supplemento, Serie II, No. 58, pp. 21-56, 1999. Extendend version appeared as Berichte aus der Technomathematik, Report 98-04, August 1998. [J9] The Symplectic Eigenvalue Problem, the Butterfly Form, the SR Algorithm, and the Lanczos Method Peter Benner and Heike Faßbender Linear Algebra and its Applications, Vol. 275/276, pp. 19-47, 1998. DOI: 10.1016/S0024-3795(97)10049-0 [J8] A Numerically Stable, Structure Preserving Method for Computing the Eigenvalues of Real Hamiltonian or Symplectic Pencils Peter Benner, Volker Mehrmann, and Hongguo Xu Numerische Mathematik, Vol. 78, No. 3, pp. 329-358, 1998. DOI: 10.1007/s002110050315 [J7] An Exact Line Search Method for Solving Generalized Continuous-Time Algebraic Riccati Equations Peter Benner and Ralph Byers IEEE Transactions on Automatic Control, Vol. 43, No. 1, pp. 101-107, 1998. See also: [TR3] . [J6] Two Connections between the SR and HR Eigenvalue Algorithms Peter Benner, Heike Faßbender, and David S. Watkins Linear Algebra and its Applications, Vol. 272, Issue 1-3, pp. 17-32, 1998. DOI: 10.1016/S0024-3795(97)00279-6 [J5] A New Method for Computing the Stable Invariant Subspace of a Real Hamiltonian Matrix Peter Benner, Volker Mehrmann and Hongguo Xu Journal of Computational and Applied Mathematics, Vol. 86, Issue 1, pp. 17-43, 1997. DOI: 10.1016/S0377-0427(97)00146-5 [J4] NICONET, a Network for Numerically Reliable Software in CACSD Ad J.W. van den Boom, Sabine Van Huffel, and Peter Benner Journal A - Benelux Quarterly Journal on Automatic Control, Vol. 38, No. 3, pp. 20-21, 1997. [J3] Benchmarks for the Numerical Solution of Algebraic Riccati Equations Peter Benner, Alan J. Laub, and Volker Mehrmann IEEE Control Systems Magazine, Vol. 17, No. 5, pp. 18-28, 1997. DOI: 10.1109/37.621466 See also: [TR1], [TR2] [J2] An Implicitly Restarted Symplectic Lanczos Method for the Hamiltonian Eigenvalue Problem Peter Benner and Heike Faßbender Linear Algebra and its Applications, Vol. 263, pp. 75-111, 1997. DOI: 10.1016/S0024-3795(96)00524-1 See also: [TR4] [J1] A Multishift Algorithm for the Numerical Solution of Algebraic Riccati Equations Gregory Ammar, Peter Benner, and Volker Mehrmann Electronic Transactions on Numerical Analysis, Vol. 1, pp. 33-48, 1993.
|
{}
|
# A Submillimeter View of Circumstellar Dust Disks in $ρ$ Ophiuchus - Astrophysics
A Submillimeter View of Circumstellar Dust Disks in $ρ$ Ophiuchus - Astrophysics - Descarga este documento en PDF. Documentación en PDF para descargar gratis. Disponible también para leer online.
Abstract: We present new multiwavelength submillimeter continuum measurements of thecircumstellar dust around 48 young stars in the $ho$ Ophiuchus dark clouds.Supplemented with previous 1.3 mm observations of an additional 99 objects fromthe literature, the statistical distributions of disk masses and submillimetercolors are calculated and compared to those in the Taurus-Auriga region. Thesebasic submillimeter properties of young stellar objects in both environmentsare shown to be essentially identical. As with their Taurus counterparts, the$ho$ Oph circumstellar dust properties are shown to evolve along an empiricalevolution sequence based on the infrared spectral energy distribution. Thecombined $ho$ Oph and Taurus Class II samples 173 sources are used to setbenchmark values for basic outer disk characteristics: M disk ~ 0.005 solarmasses, M disk-M star ~ 1%, and $\alpha$ ~ 2 where $F { u} \propto u^{\alpha}$ between 350 microns and 1.3 mm. The precision of these numbersare addressed in the context of substantial solid particle growth in theearliest stages of the planet formation process. There is some circumstantialevidence that disk masses inferred from submillimeter emission may beunder-estimated by up to an order of magnitude.
Autor: Sean M. Andrews, Jonathan P. Williams
Fuente: https://arxiv.org/
|
{}
|
# Displacement Operator definition
1. Aug 13, 2012
### McLaren Rulez
Hi,
Could someone explain how the following two definitions of the displacement operator are equal? The first is the standard one
1) $e^{\alpha a^{\dagger}-\alpha*a}$
2) $e^{\frac{-|\alpha^{2}|}{2}}e^{\alpha a^{\dagger}}$
What is the reasoning behind not using the second equation to define the operator that acts on a vacuum state to produce a coherent state? Or if it is fine to use definition 2), then can someone help me show that the two definitions are equivalent?
Thank you
2. Aug 13, 2012
### strangerep
They're not equal as operators, but if you act on the vacuum with them, e.g.,
$$e^{\alpha a^{\dagger}-\bar\alpha a} \, |0\rangle$$
you should be able to use one of the BCH theorems (Baker-Campbell-Hausdorff) -- or is it Zassenhaus? -- to convert the exponential into a product of exponentials, where the rightmost one involves only the annihilation operator, and hence leaves the vacuum unchanged.
http://en.wikipedia.org/wiki/Baker–Campbell–Hausdorff_formula
The advantage of version (1) is that its formally unitary. Both are useful in different circumstances. Try Mandel & Wolf for lots more on this.
3. Aug 13, 2012
### McLaren Rulez
Using the BCH equations, the article arrived at $e^{\frac{|\alpha^{2}|}{2}}e^{\alpha a^{\dagger}}e^{\alpha^{*} a}$ for the displacement operator. Now, I expand the last exponential term in powers of a.
But if the rightmost one only involves the annihilation operator, then we get zero. Isn't it true that $a\left|0\right\rangle = 0$, not $\left|0\right\rangle$?
4. Aug 13, 2012
### Bill_K
It's an exponential whose exponent involves only a. And e0 = 1.
5. Aug 14, 2012
### McLaren Rulez
Oh of course! I forgot the identity term in the Taylor expansion. Thank you very much Bill_K and strangerep
|
{}
|
## anonymous 4 years ago Evaluate w/o using a calulator: sec (11 pi/6) My answer was (2 sqrt3/3) Can someone check if it's right?
1. anonymous
sec (11π / 6) is the same as 1 / cos(11π / 6)
2. anonymous
sec(11pi/6) = 1/cos(11pi/6) cos(11pi/6) we can see from the graph that cos(2pi - x) = cos(x) cos(11pi/6) = cos(2pi - pi/6) = cos(pi/6) = sqrt(3)/2
3. anonymous
so sec(11pi/6) = 1/(sqrt(3)/2) = 2/sqrt(3) = 2sqrt(3)/3
4. anonymous
yes you are correct
5. anonymous
$1 / \cos(11π / 6) = 1 / \sqrt{3}/2$ $= 2\sqrt{3}/3$
6. anonymous
Thanks smart people lol i thought i had it wrong. :)
|
{}
|
# Taking the derivative
1. Aug 2, 2005
### Cyrus
In stewart, page 806 he says:
"In the special case in which the equation of a surface S is of the form z=f(x,y) (that is, S is the graph of a function f of two variables), we can write the equation as
F(x,y,z) = f(x,y) - z = 0
and regard S as a level surface (with k=0) of F. Then
Fx(x0,y0,z0) = fx(x0,y0)
Fy(x0,y0,z0) = fy(x0,y0)
Fz(x0,y0,z0)= -1 "
end quote
I understand his moving z to the other side.
But when the take the derivative W.R.T x, what about the z? z is not a variable, z is a functin of x and y, so why dont u have some dz/dx term in there, Fx(x0,y0,z0)=fx(x0,y0) -dz/dx . How did z no longer become a dependent variable on x and y?
2. Aug 2, 2005
### matt grime
becuase we are taking a level curve (when F=0), or if you like, F, a fucntion of three variables, and f are completetly different things, F=0 implicitly defines z as a function of x and y (f).
Last edited: Aug 2, 2005
3. Aug 2, 2005
### Cyrus
Right, F=0 is a level surface of three variables, which means that for any value of x,y or z=f(x,y) the function spits out the value zero.
" F=0 implicitly defines z as a function of x and y (f)"
I dont quite know what you mean in this part. z is implicitely defined as a function of x, and y. I dont know what you mean by y(f) though. How does making F(x,y,z) change the fact that z is no longer depended on x any y? I could not figure that out from your explination, sorry.
4. Aug 2, 2005
### matt grime
What's not to figure out? You appear to have a common theme in your threads of simply leaping to a conclusion and not thinking about it. We have two different ways of thinking about the same thing, that's all. I can either have z as a fucntion of x and y explicitly (your dependent situation) or I can take a *different* function of three (independent) variables and use that to implicitly define z as a function of x and y.
The f in brackets was to indicate how F=0 implicitly defines z as a function of y, ie the function is z=f(x,y).
You are used to doing this for two variables, surely.
Consider the equation
xy+y^2+x=1
this defines a locus in the xy plane. Would you be happy to find me dy/dx from this?
Seeing as you are assigning too much meaning to symbols, cnosider this, let f be a fuction of x and y (look no mention of z) and now define a new function F of three variables F(x,y,z) = f(x,y)-z
now, the level surface F=0 is the set of points x,y,z satisfying f(x,y)=z. I can plot this as a surface.
See, no mention of dependent or indepenent variables, which is all a red herring anyway.
The nature of the z changes exactly because we declare them to be differnt in the different expressions. Again, just a notational convention that you need to learn. There is no smoke and mirrors going on.
perhaps it would help if you used completely different letters: F(u,v,w)=f(u,v)-w
Last edited: Aug 2, 2005
5. Aug 2, 2005
### Cyrus
Hmmmm, I like your explination, becuase through your way, z is has no relation to f(x,y). By setting f(x,y)-z=0, we can still have any value of x or y, but it forces us to limit our scope to value of the (variable) z, so that the equation is satisfied. So in that sense, z is truely a variable now, it can be whatever it wants to be, but the equality forces a restriction on it. Is that an ok way to think about it?
Ok, here is a better question to ask. When we have z=f(x,y), then z is a function of x and y. So z is not an independent variable. If I take the derivative w.r.t x or y, id get dz/dx or dz/dy.
But when I declare F(x,y,z), does that change z from being a dependent variable into a dependent variable, but with a restirction imposed on it, since it must satisfy f(x,y)-z=0?
Last edited: Aug 2, 2005
6. Aug 2, 2005
### matt grime
Gah! Whether or not something is or is not dependent would depend upon what you're doign. IN THIS CASE x,y and z are INDEPENDENT variables of a function F, absolutely indpendent, totally independent.
From this we can create a function z=f(x,y) by NOW taking a level curve and from this we can think of z as a fucntion of the independent variables x and y (z is not a variable now, if you like, dependent or otherwise, z is a function of x and y). this level curve also may define y as a fucntion of x and z or even x as a function of y and z.
"when i declare F(x,y,z)". What does that mean?
7. Aug 2, 2005
### Cyrus
"when i declare F(x,y,z)". What does that mean?
That F is a function of three imput variables, x,y,z which have no dependance on one another.
8. Aug 2, 2005
### Hurkyl
Staff Emeritus
It might help you to realize this:
defining F(x, y, z) = f(x, y) - z
means exactly the same thing as
defining $F(\spadesuit, q, \xi) = f(\spadesuit, q) - \xi$.
9. Aug 4, 2005
### Cyrus
Hey matt. I thought about what you said. Could you help me out with this please.
Lets say we have, a function defined as F(x,y,z) = f(x,y) -z =0
In this situation, x,y,and z are independent variables. So I could pick any value for x and y, but then I am limited in what I can choose for z in order for the equation to equal zero. Would It be incorrect for me to say that I could pick a value for x and z, but then be limited in my choice of y, while still keeping the equation the same, f(x,y)-z=0 , or would I have to rewrite it as f(x,z)-y=0?
10. Aug 4, 2005
### matt grime
No, F(x,y,z)=0 is not a function of 3 independent variables at all. it is an equality. F(x,y,z) is a function of 3 independent vairables.
11. Aug 4, 2005
### Cyrus
I see, could I write it as F(x,y,f(x,y))=0 ?
12. Aug 4, 2005
### matt grime
That would depend what you meant when you wrote it. If you meant a function of three inputs, then no, in fact that makes even less sense now. The whole point of this is that we define a fucntion
F(x,y,z)=f(x,y)-z
that is the fucntion. OK?
then we take the level curve of that fucntion given by requiring F=0.
Last edited: Aug 4, 2005
|
{}
|
# Accelerated rest frame
1. Feb 14, 2014
### analyst5
I was reading some posts on one of my previous threads (great discussion there btw) and I red something that I really didn't understand.
So the basic premise was that if we have 2 clocks that undergo acceleration on the same level, to that they are mutually at rest, that they will see their time rates differently. Can somebody explain this? I've always thought that an observer that is accelarating with an object will see the time pass on that object by the same rate as in his own frame. What are the rules here? How will the time flow relative to each other's frame (both frames are mutually at rest and accelerating at the sam rate)?
2. Feb 14, 2014
### stevendaryl
Staff Emeritus
Here's an intuitive argument that explains time dilation on board an accelerating rocket.
Imagine that you have a rocket that is initially at rest in some inertial reference frame. There is a clock at the rear of the rocket, and another clock at the front of the rocket. Suppose that every $T$ seconds, the rear clock sends a light signal toward the front clock. If the rocket is at rest, then the first signal will arrive at time $t=\frac{L}{c}$, where $L$ is the length of the rocket, and $c$ is the speed of light. Thereafter, signals from the rear clock will arrive every $T$ seconds.
Now, suppose the rocket is accelerating at rate $a$. Then, using nonrelativistic physics (which we can do, because the rocket will take a while before it gets going fast enough for relativity to be important), we can estimate how long the signals will take to travel from the rear of the rocket to the front. The first signal will take approximately time $\frac{L}{c}$ as before. The second signal will take a little longer, because by then, the front of the rocket will be moving away from the rear, and it will take a little longer for the light signal to catch up. If we assume that $a T$ is much smaller than $c$, and that $a L$ is much smaller than $c^2$, then we find that the time between arrival of signals coming from the rear clock is approximately:
$T' = T(1+\frac{a L}{c^2})$
If the people aboard the rocket interpret their situation as if they were at rest in a fictitious gravitational field, what they would conclude is that the rear clock is running slower than the front clock, by a factor of $1+\frac{a L}{c^2}$.
3. Feb 14, 2014
### WannabeNewton
4. Feb 14, 2014
### 94JZA80
i think i completely understand your explanation, but i also think one needs to be very careful with the wording when discussing motion, relativistic or not. to say that the front of the rocket will have moved away from the rear implies that the rocket is in fact stretching in length. if anything, the rear of the rocket will move toward the front of the rocket (at the instant the engine is fired) before the front of the rocket starts to move too (b/c there is no such thing as a perfectly rigid body, and the impulse of the thrust generated by the engine takes time to reach the front of the rocket). now, if we ignore the fact that nothing in this universe is perfectly rigid, and that impulse travels at the speed of light (and not instantaneously), then the relationship between the front of the rocket and rear of the rocket is no different than the scenario the OP suggested in the first place...that is, regardless of whether the rocket is at rest, undergoing uniform motion, or undergoing accelerated motion, the front of the rocket and the rear of the rocket are simply mutually at rest with each other the entire time.
thus, it would probably be more appropriate to say that the second signal will take a bit longer to arrive at the front of the rocket because at T > 0, the front of the rocket will have moved some distance from its original position. hence, the front of the rocket will have also moved some distance from the position at which the rear of the rocket resided at T = 0 (specifically, the front of the rocket will have moved directly AWAY from the rear of the rocket's previous position, creating a distance longer than L). thus the signal takes longer to traverse the length of the rocket not because the rocket stretched its length, but because the distance from the rear of the rocket at T = 0 to the front of the rocket at T > 0 is greater than L (the distance from the rear of the rocket at T = 0 to the front of the rocket at T = 0).
5. Feb 14, 2014
### WannabeNewton
You're thinking of Born rigid acceleration of the rocket. While it's true that in Born rigid acceleration the front and rear ends of the rocket will accelerate at different proper rates in order to remain relatively at rest and furthermore clocks at rest at the respective locations will tick at different rates relative to one another, even in the Newtonian approximation wherein all parts of the rocket have the same proper acceleration whilst maintaining rigidity one will still be able to derive time dilation of resting clocks at different locations in the rocket so what you're proposing as the reason isn't quite correct.
What Steven said is in fact perfectly fine.
6. Feb 14, 2014
### stevendaryl
Staff Emeritus
No, from the point of view of the initial rest frame, the position of the front of the rocket is given, for small values of $t$, by
$x = L + \frac{a t^2}{2}$
The rear of the rocket will have a position given by:
$x= \frac{a t^2}{2}$
A light signal sent from the rear after $T$ seconds will have a position given by:
$x = \frac{a T^2}{2} + c (t - T)$
So, the light signal will catch up with the front when
$\frac{a T^2}{2} + c (t - T) = L + \frac{a t^2}{2}$
We can rearrange this to get:
$(\frac{a (T+t)}{2} + c) (t-T) = L$
which has the approximate solution
$t = T (1+\frac{aL}{c^2})$
That is certainly true, but pretty much irrelevant. If you assume that rear clock is sending out signals at a rate of one every $T$ seconds, and assume that $c T$ is much greater than $L$, then the front of the rocket will be moving by the time the second signal gets to it.
The notion of two accelerating objects "being mutually at rest" is a frame-dependent notion. They may be at rest according to one frame, but not according to another.
You're the one who said that it had something to do with the rocket stretching, I didn't.
7. Feb 15, 2014
### analyst5
I think you didn't understand my question. I was refering to the situation where we have a collection of observers that are at rest but accelerating at the same rate. In inertial frames, if we are at rest with some object, time will flow with the same rate in in as in our system. What is the difference for accelerated objects mutually at rest? That's what I don't understand.
8. Feb 15, 2014
### Bill_K
In the initial frame, all the rockets have the same velocity at the same time t. But events that appear simultaneous in one frame are not simultaneous in another. An observer on one of the rockets will have his own definition of simultaneous, and in his frame at t' = const the other rockets will be at different stages of their acceleration. Those ahead of him will have traveled longer and gained more velocity, while those behind him will have just begun, or not even started yet.
9. Feb 15, 2014
### stevendaryl
Staff Emeritus
There is no such thing as accelerating observers accelerating identically. It's one of the quirks of special relativity that makes this impossible. If, according to the initially "launch" frame, two observers are accelerating identically, then the observers themselves will see the forward observer move steadily away. They would not see themselves as "at rest" relative to one another.
Alternatively, if the two observers try to remain "at rest" relative to one another, it would require that the forward observer accelerate less strongly than the rear observer. So the forward observer would (from the point of view of the initial rest frame) be traveling slower than the rear observer, and would experience less time dilation. His clock would run faster than the one of the rear observer.
10. Feb 15, 2014
### analyst5
How do we measure proper time elapsed during acceleration, since a clock at rest with an accelerated object would tick differently?
11. Feb 15, 2014
### WannabeNewton
It looks like you completely missed Steven and Bill's important and related points. Have you worked with Rindler space-time before? Do you know how to compute invariants of time-like congruences? Do you know the relationship between clock synchronization and lack of Born rigidity of a time-like congruence (e.g. compressibility of a fluid)?
If you do then it will be really easy to explicitly see why two observers accelerating at the same rate along the line joining them cannot be at rest relative to one another in either of their rest frames and therefore will notice their clocks ticking at different rates of proper time due to kinematical time dilation. On the other hand if you have two observers accelerating at different rates along the line joining them in exactly the manner needed to maintain Born rigidity then the same formalism of Rindler space-time and the kinematical decomposition of time-like congruences will make it easy to see that the observers remain at rest relative to one another in their rest frames but will notice their clocks ticking at different rates of proper time due to "gravitational" time dilation.
12. Feb 17, 2014
### johnny_bohnny
So can an object accelerate in a manner that all of its parts accelerate simultaneously viewed from its own rest frame? This is really confusing. What would happen if an inertial, launch frame, saw the object accelerating part by part?
13. Feb 17, 2014
### WannabeNewton
Well for starters this assumes the extended object has a rest frame to begin with. If the constituents of the object are moving radially relative to one another, such as in an arbitrary fluid, then it doesn't make much sense to talk about the rest frame of the object itself, right? In fact in general the best we can do is talk about the individual rest frames of the constituents of the objects. In a fluid these individual rest frames would correspond to those of the individual fluid elements. When can we talk about the global rest frame of an extended object itself? Precisely when the object is rigid.
But the definition of rigidity in relativity is quite different from the definition of rigidity in Newtonian mechanics. The precise definition requires some mathematical machinery but intuitively rigidity in relativity, often termed Born rigidity, means that given any constituent of the extended object, the spatial distances in the rest frame of this constituent to all neighboring constituents of the object remain constant i.e. there are no relative radial velocities between the neighboring constituents of the object. Note that this definition still allows a rigidly rotating object to have an extended rest frame.
Now coming back to your question, imagine we have a rod at rest in an inertial frame and say we want to accelerate the rod along its length. For the reasons explained above, if we even want to talk about the rest frame of the entire rod itself during the acceleration phase, we better make sure the acceleration is done in a Born rigid manner i.e. the rod must remain rigid during the acceleration phase. Then we can indeed talk about the global rest frame of the rod i.e. an extended frame in which all points of the rod are at rest. In order to do this, each point of the rod must receive a different proper acceleration-in fact we must impart a proper acceleration to each point of the rod inversely proportional to the fixed spatial location of the point in the rest frame of the rod. Now obviously in this case if all points of the rod are accelerated simultaneously in the inertial frame in which the rod is initially at rest then all points of the rod will accelerate simultaneously in the rest frame of the rod by construction*
On the other hand, say we don't accelerate the rod Born rigidly. Say we impart to all points of the rod the same proper acceleration and do so simultaneously in the inertial frame in which the rod is initially at rest. Then in fact it no longer makes sense to talk about the extended rest frame of the rod itself because in the individual rest frames of the points of the rod, the neighboring points will have radial velocities. Furthermore, now if we go to the rest frame of any given point of the rod, then relative to this frame the other points of the rod will have actually accelerated at different times i.e. even though all points were accelerated simultaneously in the initial inertial frame, because we gave all points the same proper acceleration we find that in the rest frame of any given point of the rod the other points will not have accelerated simultaneously. This is in fact the content of the famous Bell spaceship paradox: http://en.wikipedia.org/wiki/Bell_spaceship_paradox
*We have implicitly assumed here that we can actually talk about global Einstein simultaneity in the extended rest frame of the rod. Rigidity is actually not enough to guarantee this. The reason we can make this assumption is due to the irrotationality of the rod motion.
14. Feb 18, 2014
### johnny_bohnny
I think I get it Wannabe Newton, thanks for your answer and the work you put into it.
So we may consider an object to be our frame of reference when it is Born rigid accelerated so that defines its own rest frame? I didn't knew this was possible. Interesting.
15. Feb 19, 2014
### pervect
Staff Emeritus
"Rest frame" is a confusing topic. But I think that whatever significance you ascribe to a "rest frame", I believe can be understood by the less ambiguous idea of a coordinate system.
The rules you are asking for are then are just a mathematical expression for what a clock measures (proper time) given the details of the coordinate system. This is just
for timelike separations:
($\Delta$ proper time) ^2 = $\sum_{ij} g^{ij} \Delta x_i \Delta x_j$
If the RHS is negataive we have a spacelike separation
($\Delta$ proper distance) ^2 = -$\sum_{ij} g^{ij} \Delta x_i \Delta x_j$
Here $g^{ij}$ are the metric coefficients associated with your particular choice of coordinates.
(note: there's a sign convention here that is sometimes variable, it won't matter in this post but it might if you compare to other posts or textbooks).
We've specified the change in a clock reading between two nearby points, (or the change in distance if the separation is a space-like one). To get the change in a clock reading between two distant points you need to plot a particular path or curve between the two distant points and add up (integrate) the change in clock reading between nearby points on the connecting curve. You use the same integration process to find the distance between two points given a connecting curve.
I'm sure you have been told this before, but it doesn't seem to be getting through to you. I'm not sure if just repeating this is going to help, but I'm not sure what about the explanation you may not be getting, so I'll try.
The choice of coordinates to use is arbitrary, in your example in one case you choose inertial coordinates and in the other case you choose "accelerated coordinates".
The numbers associated to events change when you change coordinate systems. The readings of actual clocks do NOT change when you change coordinate system - whatever coordinate system you choose, the clocks report the same amount of time. Note that actual clocks follow some specific path through space-time, you need that path to integrate along.
The notion of clocks "speeding up" or "slowing down", unlike the actual readings of the clocks, DOES depend on the coordinate system you choose. The observer-independent (coordinate independent) notion of what is going on is the actual readings of actual clocks. The observer dependent notion of what is going on is what coordinates you assign to specific clock readings.
If you want more about how the metric coefficients are determined given your choice of frame or coordinates, we can perhaps give a few examples. The simplest thing to note is that inertial observers always have a metric of diag(-1,1,1,1) - (I hope that notation makes sense?).
Before discussing further the details of the metric coefficients, I want to see if the preliminaries make sense.
I rather get the impression you are viewing things backwards - giving observer independence and/or mental priority to the notion of "frames" - which I think is likely the fundamental issue that's confusing you. What's really observer independent are the actual readings of clocks. The "frame" is a mental construct you use to compactly describe all of these observer-independent things, and different observers have different mental constructs.
16. Feb 19, 2014
### johnny_bohnny
So how do we measure the proper time of an accelerating object when all points are accelerated at different rates?
17. Feb 19, 2014
### PAllen
If you imagine a radioactive mass of some size, accelerated for a long time, then the forward (direction of the acceleration) parts of the object will have decayed more than the rear. Different parts of the object experience different proper time elapsed. This is no different from the fact that with current super accurate clocks, we can measure that more time passes for your head than your feet (assuming you are mostly upright).
18. Feb 19, 2014
### johnny_bohnny
So in accelerated bodies there is no possibility that different points on the worltube experience same time dilation, like they do in inertially moving bodies?
19. Feb 19, 2014
### WannabeNewton
Sure there is but again it depends entirely on the acceleration profile i.e. on the 4-acceleration field of the world-tube.
As a simple example take a thin ring in free space rotating about its symmetry axis with a constant angular velocity relative to an inertial observer at the center. All points of the ring experience the same time dilation relative to this observer.
20. Feb 19, 2014
### johnny_bohnny
What about the acceleration that occurs after or before coming into inertial motion? Like in the twins paradox scenario (where the space twin starts his trip). Or for example when somebody starts his car?
21. Feb 19, 2014
### WannabeNewton
You can do it but the motion might not be Born rigid. You can accelerate a rod along its length in such a way that all points of the rod receive the same proper acceleration and hence have the same velocity and time dilation factor relative to the inertial frame the rod was initially at rest in but in the rest frame of any one point of the rod, the proper length of the rod will be increasing, meaning neighboring points will have radial velocities relative to this rest frame, until the stresses break the rod.
Fun fact: you cannot, even in principle, put a thin disk into uniform rotation in a Born rigid manner-the tangential acceleration applied will necessarily violate Born rigidity.
22. Feb 19, 2014
### PAllen
However, for realistic acceleration profiles (and turns), you would expect different parts of you and your car to undergo slightly different aging in the course of a trip.
23. Feb 19, 2014
### johnny_bohnny
This is basically the scenario you mentioned when you said that different points will disagree in simultaneity, the 'forward' ones will already be in different stages of their acceleration, and the 'rear' ones will be perhaps even not starting their acceleration. If I'm understanding you correctly?
24. Feb 19, 2014
### WannabeNewton
Well which points start accelerating when depends on which point's rest frame you choose to boost to but yes the idea is the acceleration of all the points won't be simultaneous in the rest frame of any given point of the rod even if it was simultaneous in the initial inertial frame. It's the exact same set of calculations as in the Bell spaceship paradox.
|
{}
|
Program to find length of the largest subset where one element in every pair is divisible by other in Python
PythonServer Side ProgrammingProgramming
Suppose we have a list of unique numbers called nums, so we have to find the largest subset such that every pair of elements in the subset like (i, j) satisfies either i % j = 0 or j % i = 0. So we have to find the size of this subset.
So, if the input is like nums = [3, 6, 12, 24, 26, 39], then the output will be 4, as the largest valid subset is [3, 6, 12, 24].
To solve this, we will follow these steps −
• dp := a list of size nums and fill with 1
• sort the list nums
• n := size of nums
• if n <= 1, then
• return n
• ans := 0
• for i in range 1 to n, do
• for j in range 0 to i, do
• if nums[i] is divisible by nums[j], then
• dp[i] := maximum of dp[i] and dp[j] + 1
• ans := maximum of ans and dp[i]
• return ans
Example (Python)
Let us see the following implementation to get better understanding −
Live Demo
class Solution:
def solve(self, nums):
dp = [1] * len(nums)
nums.sort()
n = len(nums)
if n <= 1:
return n
ans = 0
for i in range(1, n):
for j in range(0, i):
if nums[i] % nums[j] == 0:
dp[i] = max(dp[i], dp[j] + 1)
ans = max(ans, dp[i])
return ans
ob = Solution()
nums = [3, 6, 12, 24, 26, 39]
print(ob.solve(nums))
Input
[3, 6, 12, 24, 26, 39]
Output
4
Published on 12-Dec-2020 09:13:56
Advertisements
|
{}
|
Home > documentation, numerical analysis, Sage, symbolic computation > Lagrange interpolating polynomials
## Lagrange interpolating polynomials
I’ve uploaded a draft of an expository paper presenting some basics of the Lagrange interpolating polynomials, with examples using Sage. The paper also covers Neville’s method for recursively generating Lagrange interpolating polynomials. The PDF version of the paper is available for your reading pleasure. The $\LaTeX$ source tarball is also available under the GNU FDL in case you want to modify or extend the paper to suit your purposes.
|
{}
|
# Programmer at Large: How old is this?
This is the latest chapter in my web serial, Programmer at Large. The first chapter is here and you can read the whole archives here. This chapter is also mirrored at Archive of Our Own.
I awoke to the usual morning startup screen – diagnostics for my sleep, news, patches to my software, etc. Nothing very exciting. I was being gently chided for not having had a proper dinner yesterday – two protein bars after the gym really don’t count – but the system had replaced the worst of it while I slept and informed me I should just have a large breakfast to compensate for the rest. Other than that, all systems looked good.
I acknowledged the instruction for breakfast and spent another hundred or so seconds reviewing the data, but it was all pretty boring.
The correct thing to do at that point would of course have been to shower, get dressed, and go get some food as instructed, but something was nagging at me from yesterday’s work… I checked, and it was fine for me to put off breakfast for another 5 ksec or so (though I got the irrational sense that the system was judging me for it), so I decided to just go straight to work.
“Ide, how many temporary patches do we have in the system?”
“Four million, five hundred thousand, one hundred and seven.”
I squeaked slightly.
“WHAT?”
“Four million, five hundred thousand, one hundred and seven.”
OK, fine. I was expecting a large number. I wasn’t expecting it to be that large – I’d have thought maybe ten thousand at the upper limit – but it didn’t actually make a difference. Either way it was too large to handle manually, so it was just a question of getting the software right.
Still, more than four million? I know there’s nothing so permanent as a temporary fix, but that’s just ridiculous.
“Ide, what sort of time frame does that span?”
“The oldest is approximately 0.26 teraseconds old.”
“Wow, OK.”
That didn’t make any sense. The current patch system isn’t even that old. The trade fleet is barely that old.
“Ide, how are you defining temporary patch?”
“I have a complex heuristic subprogram that indexes logical patches from either the fleet system or imported software strata and looks at metadata and comments to flag them as temporary.”
“Oh, right. If you just look at trade fleet authored patches in the modern system which have been explicitly flagged as temporary how many are there?”
“One million, sixty two thousand and eight.”
“Ugh. And when is the oldest of those?”
“Approximately 0.15 teraseconds old.”
This was not going to be a productive line of inquiry, but curiousity got the better of me once again.
“OK, show me the oldest.”
ID: 3b2ca7b197f9c65e883ef177178e20e6bb14b...
Author: Demeter [bim-bof 3 of the Entropy's Plaything née Wild Witless Bird]
Revert-On: Closure of f265957e0a2...
Add a flag that deliberately ties the theorem prover's hands by restricting
the set of valid interleavings when running in time travel mode.
Why? Well it turns out that *some* Linux descendants have a very peculiar idea
of how x86 is supposed to work. This idea is backed up by neither the spec,
nor by the actual physical machines that existed back when x86 was still a
real thing rather than an interop standard.
How did that happen? Well this wasted code comes from a descendant of the
"Corewards Bound", who at some point introduced a bug in their
implementation which made things run faster and didn't obviously break any of their
software. When they found this problem a few hundred gigaseconds later they decided
to patch Linux instead of their misguided grounder-inspired broken emulation software.
Nobody backed it out later, it got passed down through three generations of ships,
and finally got handed over to us and now we're stuck with it.
This patch is stupid and should go away once the referenced issue is resolved.
I looked at the patch. It was some really hairy logic right down at the formalism layer for one of our emulators. I had absolutely no idea whatsoever what was going on in it. I didn’t know the language it was written in, I can barely read either x86 or the intermediate representation it was being compiled to, even with Ide’s assist, and besides I don’t know half of the maths required to understand why it was doing what it was doing.
The referenced issue was about patching Linux to not depend on this broken behaviour. It had reached almost a million words of discussion before just trailing off – I think given the timescales involved everybody who cared about it had died of old age. Or maybe we just lost touch with them – neither this code nor the code it patched were from anywhere in our direct lineage.
“Ide, how many services are running with this flag set?”
“Seven”
I breathed a sigh of relief. That was a much better answer than I was afraid of.
“When was the last time any of them had a bug?”
“No bugs have ever been reported in these services.”
“OK. How about in their service categories?”
“Approximately 80 gigaseconds ago.”
“What was it?”
“The Untranslatable Word passed through an area with an unusually high helium mix in the local interstellar medium. This increased the crash rate of the process by 15%, which triggered a maintenance alarm.”
“How was it fixed?”
“The alarm threshold was raised by 50%.”
OK. So I’d found a weird hack in the implementation of some extremely reliable services. My duty was clear: Do nothing, touch nothing, make like a diasporan and leave at extreme velocity. I was more likely to break something by trying to “fix” it than do any good by touching this.
Time to back up and look at the actual problem.
“OK. How many temporary patches are there that were created on the Eschaton Arbitrage which apply to some but not all running processes in a category, have a trigger to revert on a hardware change, but predate our last planetary stop?”
“Nine”
OK. Now we were getting somewhere.
I spent the next few ksecs triaging those nine manually. They all looked pretty harmless, but I bet there were some gremlins that they’d flush out when the relevant teams looked into them. That definitely wasn’t going to be my job though.
After that, I wrote up a wiki entry about this problem and filed an issue making some general suggestions for how we could improve the systems around this to stop it happening again. I wasn’t very optimistic it would go anywhere, but it was at least worth making the effort.
At which point, I decided it really was time for breakfast, and headed to the showers to get ready for the day.
I showered quickly, dressed, and spent a few hundred seconds dealing with my hair. I found the style my schema gave me a bit boring so I spent a little while tweaking the parameters to give something less symmetrical.
Eventually I got my hair decisions resolved, and headed for the common area for my much delayed breakfast.
Next chapter will be in two weeks (April 7th).
Like this? Why not support me on Patreon! It’s good for the soul, and will result in you seeing chapters as they’re written (which is currently not very much earlier than they’re published, but hopefully I’ll get back on track soon).
This entry was posted in Fiction, Programmer at Large on by .
# How and why to learn about data structures
There’s a common sentiment that 99% of programmers don’t need to know how to build basic data structures, and that it’s stupid to expect them to.
There’s certainly an element of truth to that. Most jobs don’t require knowing how to implement any data structures at all, and a lot of this sentiment is just backlash against using them as part of the interview process. I agree with that backlash. Don’t use data structures as part of your interview process unless you expect the job to routinely involve writing your own data structures (or working on ones somebody has already written). Bad interviewer. No cookie.
But setting aside the interview question, there is still a strong underlying sentiment of this not actually being that useful a thing to spend your time on. After all, you wouldn’t ever implement a hash table when there’s a great one in the standard library, right?
This is like arguing that you don’t need to learn to cook because you can just go out to restaurants.
A second, related, point of view is that if you needed to know how this worked you’d just look it up.
That is, you don’t need to learn how to invent your own recipes because you can just look it up in a cook book.
In principle both of these arguments are fine. There are restaurants, there are cook books, not everybody needs to know how to cook and they certainly don’t need to become a gourmet chef.
But nevertheless, most people’s lives will be improved by acquiring at least a basic facility in the kitchen. Restaurants are expensive and may be inconvenient. You run out of ingredients and can’t be bothered to go to the store so you need to improvise or substitute. Or you’re just feeling creative and want to try something new for the hell of it.
The analogy breaks down a bit, because everybody needs to eat but most people don’t really need to implement custom data structures. It’s not 99%, but it might be 90%. Certainly it’s more than 50%.
But “most” isn’t “all”, and there’s a lot of grey areas at the boundary. If you’re not already sure you need to know this, you can probably get on fine without learning how to implement your own data structures, but you might find it surprisingly useful to do so anyway. Even if you don’t, there are some indirect benefits.
I’m not using this analogy just to make a point about usefulness, I also think it’s a valuable way of looking at it: Data structures are recipes. You have a set of techniques and tools and features, and you put them together in an appropriate way to achieve the result you want.
I think a lot of the problem is that data structures are not usually taught this way. I may be wrong about this – I’ve never formally taken a data structures course because my academic background is maths, not computer science, but it sure doesn’t look like people are being taught this way based on the books I’ve read and the people I’ve talked to.
Instead people are taught “Here’s how you implement an AVL tree. It’s very clever” (confession: I have no idea how you implement an AVL tree. If I needed to know I’d look it up, right?). It’s as if you were going to cookery school and they were taking you through a series of pages from the recipe book and teaching you how to follow them.
Which is not all bad! Learning some recipes is a great way to learn to cook. But some of that is because you already know how to eat food, so you’ve got a good idea what you’re trying to achieve. It’s also not sufficient in its own right – you need to learn to adapt, to combine the things you’ve already seen and apply the basic skills you’ve learned to solve new constraints or achieve new results.
Which is how I would like data structures to be taught. Not “Here is how to implement this named data structure” but “Here is the set of operations I would like to support, with these sorts of complexities as valid. Give me some ideas.”
Because this is the real use of learning to implement data structures: Sometimes the problem you’re given doesn’t match the set of data structures you have in the standard library or any of the standard ones. Maybe you need to support some silly combination of operations that you don’t normally do, or you have an unusual workload where some operations are very uncommon and so you don’t mind paying some extra cost there but some operations are very common so need to be ultra fast.
At that point, knowing the basic skills of data structure design becomes invaluable, because you can take what you’ve learned and put it together in a novel way that supports what you want.
And with that, I’ll finish by teaching you a little bit about data structures.
First lets start with a simple problem: Given a list of N items, I want to sample from them without replacement. How would I do that with an O(N) initialisation and O(1) sample?
Well, it’s easy: You create a copy of the list as an array. Now when you want to sample, you pick an index into the array at random.
Now that you have that index that gives you the value to return. Replace the value at that index with the value that’s at the end of the array, and reduce the array length by one.
Here’s some python:
def sample(ls, random): i = random.randint(0, len(ls) - 1) result = ls[i] ls[i] = ls[-1] ls.pop() return result
Now I’ve given you a recipe to build on, lets see you improve upon it!
1. If you assume the list you are given is immutable and you can hang onto it, can you improve the initialisation to O(1)? (you may need to make sampling only O(1) amortised and/or expected time to do this. Feel free to build on other standard data structures rather than inventing them from scratch).
2. How would I extend that data structure to also support a “Remove the smallest element” operation in O(log(n))? (You may wish to read about how binary heaps work). You’ll probably have to go back to O(n) initialisation, but can you avoid that if you assume the input list is already sorted?
3. How would you create a data structure to support weighted sampling with rejection? i.e. you start with a list of pairs of values and weights, and each value is sampled with probability proportionate to its weight. You may need to make sample O(log(n)) to do this (you can do it in expected O(1) time, but I don’t know of a data structure that does so without quite a lot of complexity). You can assume the weights are integers and/or just ignore questions of numerical stability.
4. How would add an operation to give a key selected uniformly at random to a hash table? (If you haven’t read about how pypy dicts work you may wish to read that first)
5. How would you extend a hash table to add an O(log(n)) “remove and return the smallest key” operation with no additional storage but increasing the insert complexity to O(log(n))? Can you do it without adding any extra storage to the hash table?
These aren’t completely arbitrary examples. Some of them are ones I’ve actually needed recently, others are just applications of the tricks I figured out in the course of doing so. I do recommend working through them in order though, because each will give you hints for how to do later ones.
You may never need any of these combinations, but that doesn’t matter. The point is not that these represent some great innovations in data structures. The point is to learn how to make your own data structures so that when you need to you can.
If you want to learn more, I recommend just playing around with this yourself. Try to come up with odd problems to solve that could be solved with a good data structure. It can also be worth learning about existing ones – e.g. reading about how the standard library in your favourite language implements things. What are the tricks and variations that it uses?
If you’d like to take a more formal course that is structured like this, I’m told Tim Roughgarden’s Coursera specialization on algorithms follows this model, and the second course in it will cover the basics of data structures. I’ve never taken it though, so this is a second hand recommendation. (Thanks @pozorvlak for the recommendation).
(And if you want to learn more things like this by reading more about it from me, support me on Patreon and say so! Nine out of ten cats prefer it, and you’ll get access to drafts of upcoming blog posts)
This entry was posted in programming, Python on by .
# An untapped family of board game mechanics
Have you noticed how board games are in dire need of electoral reform?
No, wait, seriously. Hear me out. This makes sense.
In most board games, through a series of events, you acquire points (votes), where each point (vote) goes to at most one player, and then at the end the person with the most points (votes) wins outright, regardless of how narrow their margin is.
It’s literally plurality voting, and it leads to a number of the same problems.
One of those problems is that it amplifies small leads quite substantially. e.g. take Scrabble. I play a lot of Scrabble because it’s more or less our family game. I also win a lot of Scrabble. Sometimes comfortably if I managed to get that second bingo out, but often by fairly small margins – 10-20 points in a 300-400 point game is not a large victory, but with plurality victory that doesn’t matter, it’s still a victory.
It also creates spoiler effects, where you have players who obviously can’t win but still participate in pile-ons to people in the lead. e.g. if you’ve ever played a Steve Jackson game (Munchkin, Illuminati, etc) you’ve probably seen this in action – “They’re about to win! Get them!”.
Certainly not all games match this description: e.g. you get a lot of games where rather than scoring points there is a defined victory condition – war games where you take a rather different view of politics and must kill all your opponents (e.g. Risk), or hidden mission games where you must achieve some specific goal unknown to the others (e.g. Chrononauts). I’ll consider that sort of game out of scope here.
Some games you play until someone has hit some defined score and then they immediately win (e.g. Love Letter). This is a bit like majoritarian vote systems (where you only win if you get more than 50% of the vote), but not really.
So the analogy isn’t perfect, but I think it has enough merit for the games it applies to to be worth exploring how things can be different. If nothing else it might be an untapped source of interesting game mechanics.
Even within the scope of games which look like elections if you squint at them hard enough there’s some variation.
For example, I’m aware of at least one game which uses random ballot: Killer Bunnies and the Quest for the Magic Carrot.
In this game you acquire carrots (votes), then at the end of the game a winning carrot (vote) is drawn at random and the person who owns that carrot (had that vote cast for them) wins (is elected). So if you own 60% of the carrots then you have a 60% chance of winning.
Given that I’m generally really rather keen on random ballot it may come as some surprise that I think this is a terrible game mechanic.
The problem with random ballot in this context is both that Random Ballot isn’t great for presidential style single winner elections, and also that in the context of a game players (politicians) matter more than carrots (voters). Voters have a right to be well represented in the decision of which politician eats them, carrots don’t.
If the game were very short it would probably be OK, but getting to the end of a long game and then having victory be decided by quite such a large amount of luck is rather frustrating. I think there are ways to fix it and make random ballot a fun game mechanism, but I also worry that it’s a bit too close to existing scoring mechanisms and where it produces different answers players will mostly find it frustrating.
So instead I’d like to look at how you could use an entirely different class of electoral system to produce this effect: Condorcet systems.
The idea of a Condorcet system is that instead of focusing on point scoring you focus on pairwise victories: Which player beats which other player. If there is a player who beats every other player in a one on one victory, they are the Condorcet winner and win outright.
Depending on who you ask, electing the Condorcet winner is either a terrible or a great idea (my personal biases are “It depends what you’re electing them for but all else being equal the Condorcet winner is probably a pretty good choice”), but that doesn’t actually matter here, because the question is not whether it’s a great election system but whether it leads to an interesting game design!
And I think it does.
When there is no Condorcet winner interesting things happen where you get rock-paper-scissors like events where the majority prefers A to B, B to C and C to A. These are called Condorcet cycles. In this case you need some sort of alternate rule to decide which player is most like the Condorcet winner (this can also happen if you get candidates who are tied because an equal number of people prefer each to the other, but you can avoid this possibility by just having an odd number of voters).
There are a wide variety of systems for deciding who the “most Condorcet” candidate is when there is no true Condorcet winner. These range from the very simple to the very complicated, but there’s a particularly simple Condorcet system that I think is very suitable for game design.
It works as follows: You pick an incumbent (usually at random). Then every other candidate (player) challenges the incumbent in some order (usually random). If the majority strictly prefers them (they beat them according to some as of yet unspecified mechanism) then the incumbent drops out and the challenger becomes the incumbent, else the challenger drops out. Once all but one candidate has dropped out, the remaining incumbent is the winner.
There are a number of free variables in how you could turn this into a game mechanic:
1. Who gets to be starting incumbent?
3. What order do people challenge in?
4. Who wins ties?
However you pick these free variables though, I think it’s most interesting if you do this in such a way that allows the possibility of Condorcet cycles (if you don’t it’s really just another scoring system). In order to do this you need something that looks like at least three voters.
The easiest way to do this is might be something like the following:
The game consists of resource acquisition in some manner. You have three resources and treat each as a vote. Each resource is claimed by whichever of the two players owns strictly more of it. If either player claims more resources than the other, that player wins. Otherwise, apply some tie breaking procedure.
Starting from that concept, you can elaborate itinto a full game. The following is a toy game that might work well (but probably would need significant refining through play testing) based this. It’s called “The King is Dead!”
The king is on his death bed and has no natural heir. He has named one, but the nobles are competing to change his mind, and regardless of who he chooses they might not last long enough to take the throne.
There are a few key nobles who might win, but their chances of victory are slim without the support of the two key factions of the kingdom: The peasants, and the clergy. The nobles must use their influence at court to curry favour with these two factions, but be careful! If you use all your influence outside the court, there may be no-one left to support you when you fall afoul of court intrigue.
Game setup:
• An heir is picked at random from the players.
• Three cards per player are drawn from a larger deck and are shuffled together to form the issue deck.
• Each player is given twelve influence tokens.
Each issue card has a number of clergy points and a number of peasant points on it. It may also have a crown on it
The game is played in rounds which proceed as follows:
1. A card from the top of the deck is revealed.
2. The card is now auctioned off with an English auction: The heir bids zero on the card and then play proceeds clockwise, with each player either placing a higher bid or dropping out. Once everyone has dropped out except for the last bidder, that bidder gets that card, places it in front of them, and puts their bid in the centre temporarily.
3. If the card had a crown on it, the person who won it now becomes heir.
4. The bid is now redistributed amongst the players: One token at a time, starting from the heir (the new one if it changed places!) and proceeding clockwise until all the tokens have been redistributed
This process completes until the deck is out of cards. The king then dies, and a swift but bloody war of succession occurs.
Starting to the left of the current heir and proceeding clockwise, each player (not counting the player who was heir when the king died) gets the opportunity to challenge the current heir.
A challenge pits the two players against each other as follows:
Each player accrues a victory on each of three scores: Their total support from the peasantry, their total support from the clergy, and their total influence. If one player has a higher score than the other on each of these, they get a victory.
If one of the two has more victories than the other, that player wins. If they are tied, the current heir wins.
Whichever player won is now the new heir. If there are any more challengers left, the next challenger steps up and challenges the heir, otherwise they win and are crowned. Long live the king!
This game definitely needs play-testing and will almost certainly have a number of problems with it, but I think it’s an interesting corner of game design that I haven’t seen anything much like before.
The thing I like about it is also that it creates a much more interesting victory dynamic in most cases: In the case where someone wins really decisively then they still cream everybody else, but the back and forth between the three resources more or less guarantees that nobody will be a Condorcet winner unless everyone else played really badly – the more they spent to get that faction support, the more influence they gave to other players that they could use to claim some other support. This makes things tense right up until the end game.
More generally, I think there’s a rich seam of game design ideas in electoral theory that hasn’t been tapped much. There’s a theorem that says that almost any voting system admits tactical voting. For electoral system design this is a real pain, because systems which encourage tactical voting have a number of problems, but for game design it’s perfect because it means that there’s this giant body of well studied and powerful mechanics that are full of opportunities for tactical play.
(Like my writing? Want to see more of it? Support me on Patreon! Patreon subscribers saw this posts months ago, because all my drafts get posted there when they’re ready)
This entry was posted in Games, voting on by .
# Programmer at Large: Who wrote this?
This is the latest chapter in my web serial, “Programmer at Large”. Previous chapters are best read at the Archive of our Own mirror. This chapter is also mirrored there.
I was only asleep for about a kilosecond before I started getting an alert from the system. It was quite reasonably informing me that if I insisted on sleeping in the hot tub then perhaps I should consider doing it face up?
I certainly hadn’t started face down in the water, so apparently I’d rolled over in my sleep at some point. I drifted for another couple of tens of seconds and then finally decided to acknowledge that OK yes breathing was useful. I rolled back over and sighed dramatically (important to get the order of those right).
I really hadn’t wanted to fall asleep there. Unsupervised sleep is awful even in zero gravity. In gravity you also have to deal with nonsense like which way up you sleep. I mean really, why should that matter?
Between the exercise, the heat, and the bad sleep I now had an annoying nagging headache and an overwhelming urge for food, water and painkillers, more or less in increasing order of priority.
I put in an order to the nearest delivery slot and heaved my way out of the hot tub to go have a shower while I waited for them to arrive.
I’m not too proud to admit that I shrieked when the cold water hit my head. It’s supposed to be very good for you after the hot tub but I was still half asleep and even at the best of times I usually manage to forget that these showers aren’t kept at a sensible temperature.
I lasted about 40 seconds before I decided enough was enough. I did feel better afterwards, but I swear that’s mostly because of how glad I was to have it stop.
Yes, I know that what’s good for you isn’t the same as what you enjoy. I’ve heard it enough times by now.
I dried off and quickly put my hair up into a bun – I really couldn’t be bothered to do it properly at that point – and by the time I was done with that the delivery had arrived, so I padded over to the nearby delivery point and tore into it.
I put the painkillers in my shunt and gulped down most of the electrolytic drink. Once my thirst had been slaked and the painkillers had kicked in, I turned my attention to the protein bars and devoured them in a few bites.
None of it would be particularly tasty in normal circumstances, but post-gym hunger is a harsh master and salt, water, sugar and protein was exactly what I needed at that point.
I removed the empty painkiller capsule and put it and the empty packaging back in the compartment to be taken for recycling. I figured it would take a while for things to kick in, so now was as good a time as any to get around to that meditation I’d been putting off.
I sat down cross legged (I can do a full lotus, but I couldn’t really be bothered. I know it’s cultural, but I also know the science says it doesn’t help) and called up my program. In the end it took me almost two kiloseconds to work through it – I’m not very good at meditation in the first place, and I was still feeling a bit twitchy from my impromptu nap, but eventually I got my mind into the right state and after that it proceeded more smoothly.
By the end of my meditation I was feeling a lot more human. My headache had subsided, along with the hunger and thirst. I went through some finishing stretches to undo the sitting – yet another reason why gravity is awful. Those finished, I fetched a clean uniform from the wall and changed into it.
I called up an image of myself and quickly checked my hair – I still couldn’t be bothered with more than a bun, but there’s no point in looking outright scruffy – and fixed a few stray bits at the back that I’d missed.
I decided I’d really had enough of people for now, so I got the transit chair back to the main ship and tucked myself away in a quiet pod to work on my Evolve strategies for a kilosecond or five.
Eventually, though, I got curious about work, and my schedule told me I was unlocked for it again and was welcome to resume if I wanted to, so I did.
I’d left some of my prototype zombie hunters running while I was away. They weren’t reporting anywhere except privately to me – I was sure they had bugs in them, but looking at some of the answers they gave now and seeing if they were right would help me figure out what those bugs were.
“Ide, how many potential zombies have my new hunters flagged up?”
“147”
“Ugh. All right, show me.”
I spent some time looking through the list and filtering things down. A bunch were false positives as expected – some interfaces that I had treated as read only in my original criteria were actually read/write but used some obscure different convention due to historical reasons – but eventually after some filtering I’d narrowed it down to 31 that were probably legitimate.
After a while of scanning through them at a high level and doing some basic triage I spotted one that looked interesting. I dug into it for a couple of kiloseconds until I was sure I understood what was going on, but it was exactly what it looked like.
Which left me with a dilemma: I was going to have to tell Kimiko about this. I didn’t want to seem too needy though, so it felt a bit soon to get in touch with them.
I dithered for a couple hundred seconds, but eventually concluded that I was being stupid. Even if I’d never met them I’d want to contact them about this, so putting off telling them about it because I did know who they were was just ridiculous.
I checked their status and they were apparently awake and working, so I opened up a line.
Arthur [vic-taf]: Hey, Kimiko?
Kimiko [jad-nic]: Oh, hey Arthur. What’s up?
Arthur [vic-taf]: I found some broken processes when I was looking into that bug for you, which has sent me off on a zombie hunt. It’s been running for the last couple of tens of kiloseconds and it surfaced something you should probably know about.
Kimiko [jad-nic]: That’s great, I’d love to hear about it later, but I’m kinda in the middle of figuring this yeast problem so do you mind if we take a pause on you telling me about it?
Arthur [vic-taf]: Actually this is about the yeast problem. I think. Maybe. Did you know the nutrient feed for the vat it’s in isn’t working properly?
Arthur [vic-taf]: The feedback loop isn’t running properly – the process that’s monitoring the nutrient levels in the vat has the control part of it patched out, so the feed is just defaulting to a standard rate of flow.
Kimiko [jad-nic]: Argh, waste it. That would do it. This yeast uses slightly more feed than the normal batch, so it’s eating through the available feed stock and then doesn’t have enough to replace it. No wonder the little wasters are going sexual.
Arthur [vic-taf]: Oh good. I wasn’t sure it was relevant, but it seemed too much of a coincidence to ignore.
Kimiko [jad-nic]: Yeah this is absolutely relevant and you’ve probably just saved me a couple tens of kiloseconds of work debugging this. Thanks! [200 millivotes kudos attached]. But why on the ground is that happening?
Arthur [vic-taf]: Apparently the control sensor on it broke when we were last interstellar and we didn’t have replacement parts, so it was patched out as a temporary fix.
Kimiko [jad-nic]: Yeah, I remember that, but we replaced the sensors in-system and that should have triggered the reset condition, right?
Arthur [vic-taf]: Well it should have, but we picked up a new design from the locals and it uses a new interface, but the patch was expecting the old interface so the reset didn’t trigger.
Kimiko [jad-nic]: So who did the patch anyway? I should give them a rap on the knuckles.
Arthur [vic-taf]: Oh, uh, heh. Funny story… [Patch reference attached]
Kimiko [jad-nic]: Huh. I do not remember doing that at all.
Arthur [vic-taf]: Sorry.
Kimiko [jad-nic]: No biggie. Anyway, I’m going to go untangle this and see if I can prove this was all that was going on. Thanks again!
Arthur [vic-taf]: No problem! Happy to help. Good luck.
That was satisfying. Even if the plumbing bug turned out to be completely innocuous, this line of work had proven definitely useful – sure they’d have figured out the bug with the vat in the end, but that wasn’t the point. The point was that the zombie detection worked and it worked well enough that if we’d been running it we’d have caught this bug before it actually ruined a batch of yeast.
Which, I decided, made this a good point to down tools. I was still feeling a bit off from my nap earlier, but some proper sleep would fix that.
I shut down my workspace, plugged into the wall, an initiated my sleep program. The lights dimmed, and after a few tens of seconds I was once again fast asleep.
Like my work and want to say thanks? Or just want to see more of it in advance? Support me on Patreon! You’ll get access to early drafts as and when they’re completed (although right now I’m a bit busy with PhD applications, so e.g. this chapter only got written late last night, but you’ll also see drafts of other blog posts).
This entry was posted in Fiction, Programmer at Large on by .
# Looking into doing a PhD
As regular readers of this blog have probably figured out, I’m a researchy sort of person.
A lot of my hobbies – maths, voting theory, weird corners of programming, etc – are research oriented, and most of my work has had some sort of research slant to it.
The last two years I’ve basically been engaged in a research project working on Hypothesis. It’s come quite far in that time, and I feel reasonably comfortable saying that it’s the best open source property based testing library on most metrics you’d care to choose. It has a number of novel features and implementation details that advance the state of the art.
It’s been pretty great working on Hypothesis like this, but it’s also been incredibly frustrating.
The big problem is that I do not have an academic background. I have a masters in mathematics (more technically I have a BA, an MA, and a CASM. Cambridge is weird. It’s entirely equivalent to a masters in mathematics though), but that’s where I stopped. Although it says “DR” in my online handle and the domain of this blog, those are just my initials and not my qualification.
As a result, I have little to no formal training or experience in doing academic research, and a similarly low understanding of who’s who and what’s what within the relevant fields. So I’ve been reading papers and trying to figure out the right people to talk to all on my own, and while it’s gone OK it’s still felt like fumbling around in the dark.
Which leads to the obvious solution that I spoilered in the title: If the problem is that I’m trying to do research outside of an academic context, the solution is to do research in an academic context.
So I’d like to do a PhD that is either about Hypothesis, or about something close enough to Hypothesis that each can benefit from the other.
There’s probably enough novel work in Hypothesis already that I could “just” clean it up, factor it out, and turn it into a PhD thesis as it is, but I’m not really expecting to do that (though I’d like that to be part of it). There are a number of additional directions that I think it would be worth exploring, and I expect most PhD funding will come with a focus subject attached which I would be happy to adapt to (a lot of the most interesting innovations in Hypothesis came because some external factor forced me to think about things in ways I wouldn’t otherwise have!). If you’d like to know more, I’ve written up a fairly long article about Hypothesis and why I think it’s interesting research on the main Hypothesis site.
Which, finally, brings me to the main point of the post: What I want from you.
I’m already looking into and approaching potential universities and interesting researchers there who might be good supervisors or able to recommend people who are. I’ve been in touch with a couple (some of whom might be reading this post. Hi), but I would also massively appreciate suggestions and introductions.
So, if you work in relevant areas or know of people who do and think it would be useful for me to talk to, please drop me an email at [email protected]. Or just leave a comment on this blog post, tweet at me, etc.
This entry was posted in Hypothesis, life, programming, Python on by .
|
{}
|
Do 2 Kilowatt Nuclear Reactor Still Exist
i was told that the smallest nuclear reactor power
station is 2 Kilowatts and they take up about
half a city block?
is that correct?
can they be built?
Have A Nice Day!
PhysOrg.com science news on PhysOrg.com >> Leading 3-D printer firms to merge in $403M deal (Update)>> LA to give every student an iPad;$30M order>> CIA faulted for choosing Amazon over IBM on cloud contract
Recognitions: Homework Help What makes you think there is a lower limit to the power output to a nuclear power plant? If nothing else, you could just use a normal one to drive a really really inefficient generator :) Even sub-critical masses of radioisotopes can produce enough heat to do useful work.
simon your question/answer: 1 i don't know 2 doesn't sound like a good idea. 3 that went over my head. can a 2 Kilowatt Nuclear Power Generator be built the size of a city block? Have A Nice Day!
Recognitions:
Gold Member
Do 2 Kilowatt Nuclear Reactor Still Exist
Quote by average guy simon your question/answer: 1 i don't know 2 doesn't sound like a good idea. 3 that went over my head. can a 2 Kilowatt Nuclear Power Generator be built the size of a city block? Have A Nice Day!
2 KW is extremely small in terms of power generation. They sell gasoline generators at the store that you can put in your trunk that will do that. I would be very surprised if you couldn't build a reactor of that size in a very small foot print. There would be considerations for shielding and a lot of other things, but I don't see any theoretical reason why it would not be possible. Hell, how big are nuclear reactors in submarines? They produce a couple dozen megawatts
Recognitions: Homework Help [after a sleep and a cup of coffee] I suppose I could break this into two parts: 1. is the historical 2kW plant still running? (which one?) 2. what is the minimum power from a supra-critical core? eg. how far can you damp a critical mass and still have a chain reaction? close enough? note: we usually think of a nuclear reactor as like the one in Fukushima and Chernobyl[1] - these exploit a nuclear chain reaction to produce heat to run a steam engine, which is hooked to an electric generator. To get a chain reaction, you need a minimum mass of fuel ... called a "critical mass". However, the fuel gets warm before it goes critical. You can get a few tens of grams of plutonium, for eg, and insulate it, and it can get red-hot quite quickly. You can use this to drive, say, a stirling engine ... or a thermo-couple. These things are normally called nuclear "batteries" rather than reactors but I would argue that this is just spin. "Battery" sounds nicer than "reactor". Now I could take a nuclear battery and just connect it to something that will glow - so it becomes a nuclear powered light. One way of doing this, rather than go through a generator and wires and stuff, is to just surround the fissioning material with glass coated with phosphor (or whatever) so it glows when the fission fragments hit it. This is how beta-lights work... from beta-decay (a nuclear reaction). So if I am going to be really general by defining a nuclear power reactor as any machine which generates useful power from nuclear reactions, I'll have to include beta-lights as nuclear reactors. Which may be pushing the definition a tad :) ------------------------ [1] the building you see on the news are much much bigger than the actual reactor itself. Similarly, the "city block" sized plant would just be the building. iirc the first atomic pile fit comfortably inside a squash court.
Mentor I suppose you could take a thousand megawatt reactor, pull 2 kW off it and dump the other 999,998 kW as waste heat...
Quote by average guy i was told that the smallest nuclear reactor power station is 2 Kilowatts and they take up about half a city block? is that correct? can they be built? Have A Nice Day!
I've not heard of a 2 kWe nuclear power plant. Would it be a simple thermal plant for district heating. A capacity of 2 kWe is an appropriate capacity for 1 home. Half a city block would be wasted on 2 kWe.
I'll have to check my records, but the smallest nuclear electrical plant of which I'm aware was Big Rock Point at about 60 MWe.
Recognitions:
Homework Help
I suppose you could take a thousand megawatt reactor, pull 2 kW off it and dump the other 999,998 kW as waste heat...
Yeah - the really really really inefficient generator idea.
Just cause you are producing the energy does not mean you have to use it ... but where's the fun in that?
It's been too long since I did any nuclear chain reaction stuff ... not actually being interested in nuclear power (being in NZ and all) at the time.
iirc the reaction rate broadly depends on the range of the fast neutrons v how diffuse the fuel is (and the kind of moderator)? Wasn't there a naturally occurring reactor in prehistory that was very slow?
simon i may have the rating number wrong, it's been more than 8 years since i talked to this person. what is the smallest true nuclear power plant that can be built and it's power output? Have A Nice Day!
Recognitions: Homework Help define "nuclear power plant" The Radioisotope thermoelectric generators in spacecraft are of order of 100s of Watts and about man-sized. Voyagers were 470W. Pic from Cassini probe: ... weighs 57kg and outputs 300W
Quote by average guy simon i may have the rating number wrong, it's been more than 8 years since i talked to this person. what is the smallest true nuclear power plant that can be built and it's power output? Have A Nice Day!
The AGN-201M has a power of 5 W. It's nice for practicing reactor startup, or analyzing neutron source strength, differential rod worth, approach to criticality.
It would be impractical for any power generation.
simon what most people would consider a nuclear power plant. the guy said it can be built in less than a city block and it generates electricity and is nuclear powered. astro nuke think bigger! Have A Nice Day!
Well, the research reactor at my school is rated at 5MW(t), and the reactor facility takes up MUCH less than half a block. I know there are SLOWPOKE reactors that run around 20kW(t), so there's no reason that I can see preventing one from making a 2kW(e) nuclear power station in half a block or less. There's just no point...
Recognitions:
Gold Member
Quote by Astronuc I'll have to check my records, but the smallest nuclear electrical plant of which I'm aware was Big Rock Point at about 60 MWe.
I love it.
schroding sure there is! there's that 2! 20 kilowatt was probably what he was talking about! the idea is to have people work at them and get used to them. the neighborhood GE Nuclear Power Plant! sounds like a winner to me! penguin what do you love? Have A Nice Day!
Recognitions:
Homework Help
Quote by average guy simon what most people would consider a nuclear power plant. the guy said it can be built in less than a city block and it generates electricity and is nuclear powered. astro nuke think bigger! Have A Nice Day!
Oh yeah but there were already others chiming in who knew about the other kinds.
eg Astronuc was checking records...
I think these ones are cool.
Admittedly the AGN-201M probably takes the prize.
But mine is much cooler to look at xD
I mean it is black and shiny and has fins and it sits in a precision engineered clamp - not some old bits of 2x4... look, it's the porsche of nuclear technology - you don't see any women checking out the AGN in that pic do you? No! Case closed. ;)
Mentor Blog Entries: 1 To put the 2KW figure into perspective; I'm sitting next to an electric heater that consumes so much. A "city block" sized reactor to fuel that would indicate a gargantuan waste. IIRC a normal nuclear reactor operates at 1 million times this.
Similar discussions for: Do 2 Kilowatt Nuclear Reactor Still Exist Thread Forum Replies Nuclear Engineering 18 Nuclear Engineering 6 Biology, Chemistry & Other Homework 0 High Energy, Nuclear, Particle Physics 2 High Energy, Nuclear, Particle Physics 8
|
{}
|
# SBS Transit Express Bus Service 851e
SBS Transit Express Bus Service 851e is a Limited Stop Express route variant of Service 851, plying between Yishun and Outram Park at selected hours daily. It passes through Khatib, Ang Mo Kio, Little India, Bugis and Chinatown, skipping Lentor, Marymount and Thomson. The bus service only calls at selected bus stops along its parent route. It was introduced on Sunday, 27 May 2018.
The timetable for Express 851e was changed from 16 Jan 2022.
851e
59009 Yishun Int B5 NS13 Yishun Ave 2
59069 Opp Blk 757 Yishun Ave 2
59059 Blk 608 Yishun Ave 2
59049 Opp Khatib Stn NS14 Yishun Ave 2
59039 Yishun Sports Hall Yishun Ave 2
55189 Yio Chu Kang Stn NS15 Ang Mo Kio Ave 6
55209 Opp Blk 646 Ang Mo Kio Ave 6
54059 Bef Ang Mo Kio Lib Ang Mo Kio Ave 6
54049 Courts Ang Mo Kio Ang Mo Kio Ave 6
54039 Bef Al-Muttaqin Mque Ang Mo Kio Ave 6
54019 Blk 307A Ang Mo Kio Ave 6
40019 Little India Stn NE7DT12 Bt Timah Rd
07539 Opp Rochor Stn DT13 Sungei Rd
07589 Ch of Our Lady of Lourdes Ophir Rd
01139 Parkview Sq Nth Bridge Rd
01039 Bugis Cube Nth Bridge Rd
01029 Opp National Library Nth Bridge Rd
04168 Aft City Hall Stn Exit B EW13NS25 Nth Bridge Rd
04249 Opp The Treasury Nth Bridge Rd
05029 Boat Quay Sth Bridge Rd
05131 Opp Hong Lim Cplx NE4DT19 Upp Cross St
05039 New Bridge Ctr New Bridge Rd
05019 Aft Duxton Plain Pk New Bridge Rd
05012 Bef Pearl's Hill Terr Eu Tong Sen St
05022 Aft Chinatown Stn Exit D NE4DT19 Eu Tong Sen St
04222 Clarke Quay Stn Exit E NE5 Eu Tong Sen St
04223 Old Hill St Police Stn Hill St
04151 Cath Of The Good Shepherd EW13NS25 Victoria St
01013 St. Joseph's Church Victoria St
01112 Opp Bugis Stn Exit C EW12DT14 Victoria St
07551 Fu Lu Shou Cplx Rochor Rd
07531 Rochor Stn DT13 Rochor Canal Rd
40011 Little India Stn Exit A NE7DT12 Bt Timah Rd
54011 Blk 207 Ang Mo Kio Ave 6
54031 Opp Al-Muttaqin Mque Ang Mo Kio Ave 6
54041 Opp Courts Ang Mo Kio Ang Mo Kio Ave 6
54051 Bef Ang Mo Kio Ave 5 Ang Mo Kio Ave 6
55201 Aft Blk 648 Ang Mo Kio Ave 6
55181 Opp Yio Chu Kang Stn NS15 Ang Mo Kio Ave 6
59031 Opp Yishun Sports Hall Yishun Ave 2
59041 Bef Khatib Stn NS14 Yishun Ave 2
59051 Blk 790 Yishun Ave 2
58991 Blk 767 Yishun Ave 2
59009 Yishun Int NS13 Yishun Ave 2
Route Overview Route [Dir 1]: Yishun Bus Interchange → New Bridge Rd (Aft Duxton Plain Pk) [Dir 2]: Eu Tong Sen St (Bef Pearl’s Hill Terr) → Yishun Bus Interchange Passes Through Yishun Ave 2, Ang Mo Kio Ave 6, Nth Bridge Rd / Victoria St Route Length Towards New Bridge Rd: 21.1 km Towards Yishun: 20.6 km Travelling Time 60 – 64 mins Operator Information BCM Route Package Seletar Bus Package Current Operator SBS Transit Ltd Current Depot Seletar Bus Depot (SEDEP) Current Fleet 12-metre Single Deck Bus – MAN A22 12-metre Double Deck Bus – MAN A95 Operating Hours Departure Times from Yishun Weekdays Saturdays & Sundays / Public Holidays 06:00 – 18:00 06:30 – 18:30 Departure Times from Eu Tong Sen St Daily 12:30 – 23:00 Operating Frequency From Yishun 30 – 60 mins From Eu Tong Sen St 30 mins Fare Information Fare Charges express distance fares
Bus Service 851e is a Limited-stop Express Bus Service which offers quicker journeys for commuters at Yishun, Khatib and Ang Mo Kio, linking them with the Central areas of Little India, Bugis, Chinatown and Outram Park. Operating during selected hours in the day, it only calls at selected bus stops served by its parent Bus Service 851, and charges Express Fares as a result.
In between Ang Mo Kio and Little India, the bus plies an express sector along the Central Expressway (CTE) in both directions. In addition, it does not call at bus stops along Lentor Avenue. The route operates daily at 30-minute intervals.
Service 851e is the sixth new route to be introduced under the Bus Contracting Model (BCM) after Service 974, as well as the third limited-stop service to operate daily (after 12e and 147e). The use of the lowercase e as the route suffix has its origins from Fast-Forward services introduced by SBS Transit in 2005, which are limited-stop services operating only during the peak hours. However, unlike other limited-stop routes introduced by SBS Transit in the past, this service carries the Express classification.
The scheduled run-time for Service 851e between Yishun and Outram Park is about 15 minutes faster than Service 851.
Bus Service 851e performs layovers at Kampong Bahru Bus Terminal in between City-bound and Yishun-bound trips.
Along with several other express services, operating hours for Express 851e will be revised from 9 February 2020. Trips from Yishun Bus Interchange will operate till 6:30pm, while trips from Eu Tong Sen Street only start operations at 12:30pm.
The service was degraded further from 4 January 2021, with the frequency adjusted to be 30 minutes throughout operating hours daily instead of 20 minutes during weekday peak hours. From 16 January 2022, operating hours were cut short with up to 1 hour frequency for the service.
###### Departure Timings:
• From Yishun Int:
Weekdays – 06:00, 06:30, 07:00, 07:30, 08:00, 08:30, 09:00, 09:30, 10:00, 11:00, 12:00, 13:00, 14:00, 15:00, 16:00, 17:00 & 18:00
Weekends & Public Holidays – 06:30, 07:30, 08:30, 09:00, 09:30, 10:00, 10:30, 11:00, 11:30, 12:00, 12:30, 13:00, 13:30, 14:00, 14:30, 15:00, 15:30, 16:00, 16:30, 17:00, 17:30, 18:00 & 18:30
• From Eu Tong Sen St (Bef Pearl’s Hill Terr):
Daily – 12:30, 13:00, 13:30, 14:00, 14:30, 15:00, 15:30, 16:00, 16:30, 17:00, 17:30, 18:00, 18:30, 19:00, 19:30, 20:00, 20:30, 21:00, 21:30, 22:00, 22:30 & 23:00
###### Operator History
• 2018 – Present: SBS Transit Ltd
The Bus Service Operating License (BSOL) for this route will be renewed in 2023 under the Seletar Bus Package.
Past Routings
###### Changes in Departure Times:
Day Type / Departures From Yishun From Eu Tong Sen St From 27 May 2018 Weekdays 06:00, 06:30, 07:00, 07:20, 07:40, 08:00, 08:20, 08:40, 09:00, 09:30, 10:00, 10:30, 11:00, 11:30, 12:00, 12:30, 13:00, 13:30, 14:00, 14:30, 15:00, 15:30, 16:00, 16:30, 17:00, 17:30, 17:50, 18:10, 18:30, 18:50, 19:10, 19:30, 20:00, 20:30, 21:00, 21:30, 22:00 07:30, 07:50, 08:10, 08:30, 08:50, 09:10, 09:30, 10:00, 10:30, 11:00, 11:30, 12:00, 12:30, 13:00, 13:30, 14:00, 14:30, 15:00, 15:30, 16:00, 16:30, 17:00, 17:20, 17:40, 18:00, 18:20, 18:40, 19:00, 19:30, 20:00, 20:30, 21:00, 21:30, 22:00, 22:30, 23:00, 23:30 Weekends / Public Holidays 06:00, 06:30, 07:00, 07:30, 08:00, 08:30, 09:00, 09:30, 10:00, 10:30, 11:00, 11:30, 12:00, 12:30, 13:00, 13:30, 14:00, 14:30, 15:00, 15:30, 16:00, 16:30, 17:00, 17:30, 18:00, 18:30, 19:00, 19:30, 20:00, 20:30, 21:00, 21:30, 22:00 07:30, 08:00, 08:30, 09:00, 09:30, 10:00, 10:30, 11:00, 11:30, 12:00, 12:30, 13:00, 13:30, 14:00, 14:30, 15:00, 15:30, 16:00, 16:30, 17:00, 17:30, 18:00, 18:30, 19:00, 19:30, 20:00, 20:30, 21:00, 21:30, 22:00, 22:30, 23:00, 23:30 From 9 Feb 2020 Weekdays 06:00, 06:30, 07:00, 07:20, 07:40, 08:00, 08:20, 08:40, 09:00, 09:30, 10:00, 10:30, 11:00, 11:30, 12:00, 12:30, 13:00, 13:30, 14:00, 14:30, 15:00, 15:30, 16:00, 16:30, 17:00, 17:30, 18:00, 18:30 12:30, 13:00, 13:30, 14:00, 14:30, 15:00, 15:30, 16:00, 16:30, 17:00, 17:20, 17:40, 18:00, 18:20, 18:40, 19:00, 19:30, 20:00, 20:30, 21:00, 21:30, 22:00, 22:30, 23:00, 23:30 Weekends / Public Holidays 06:00, 06:30, 07:00, 07:30, 08:00, 08:30, 09:00, 09:30, 10:00, 10:30, 11:00, 11:30, 12:00, 12:30, 13:00, 13:30, 14:00, 14:30, 15:00, 15:30, 16:00, 16:30, 17:00, 17:30, 18:00, 18:30 12:30, 13:00, 13:30, 14:00, 14:30, 15:00, 15:30, 16:00, 16:30, 17:00, 17:30, 18:00, 18:30, 19:00, 19:30, 20:00, 20:30, 21:00, 21:30, 22:00, 22:30, 23:00, 23:30 From 4 Jan 2021 Daily 06:00, 06:30, 07:00, 07:30, 08:00, 08:30, 09:00, 09:30, 10:00, 10:30, 11:00, 11:30, 12:00, 12:30, 13:00, 13:30, 14:00, 14:30, 15:00, 15:30, 16:00, 16:30, 17:00, 17:30, 18:00, 18:30 12:30, 13:00, 13:30, 14:00, 14:30, 15:00, 15:30, 16:00, 16:30, 17:00, 17:30, 18:00, 18:30, 19:00, 19:30, 20:00, 20:30, 21:00, 21:30, 22:00, 22:30, 23:00, 23:30 From 16 Jan 2022 Weekdays 06:00, 06:30, 07:00, 07:30, 08:00, 08:30, 09:00, 09:30, 10:00, 11:00, 12:00, 13:00, 14:00, 15:00, 16:00, 17:00, 18:00 12:30, 13:00, 13:30, 14:00, 14:30, 15:00, 15:30, 16:00, 16:30, 17:00, 17:30, 18:00, 18:30, 19:00, 19:30, 20:00, 20:30, 21:00, 21:30, 22:00, 22:30, 23:00 Weekends / Public Holidays 06:30, 07:30, 08:30, 09:00, 09:30, 10:00, 10:30, 11:00, 11:30, 12:00, 12:30, 13:00, 13:30, 14:00, 14:30, 15:00, 15:30, 16:00, 16:30, 17:00, 17:30, 18:00, 18:30 12:30, 13:00, 13:30, 14:00, 14:30, 15:00, 15:30, 16:00, 16:30, 17:00, 17:30, 18:00, 18:30, 19:00, 19:30, 20:00, 20:30, 21:00, 21:30, 22:00, 22:30, 23:00
Back to Express Services
Back to Bus Services
### 16 thoughts on “SBS Transit Express Bus Service 851e”
• 21 June 2021 at 9:49 AM
Hope Express 851e should be withdrawn due to low demand. Commuters will have to take alternative Service 851 to and from Yishun, Khatib, Ang Mo Kio to the city areas of Little India, Bugis, City Hall and Chinatown.
• 22 June 2021 at 1:16 AM
Are u even taking this service ah ? This bus has a very gd loading and demand is gd. I myself take this bus on a daily basis. And i hope this bus can be made oprate up to kampang baru terminal.
• 28 January 2019 at 1:01 PM
Service 851e buses layover at Kampong Bahru bus terminal.
• 29 May 2018 at 12:27 PM
I hope that 851e will also have man a24
• 15 October 2020 at 3:58 PM
MAN A24 only allowed to operate on Yishun Feeder services. 851e already have majority Double Deckers.
|
{}
|
# pysal.explore.giddy.markov.homogeneity¶
pysal.explore.giddy.markov.homogeneity(transition_matrices, regime_names=[], class_names=[], title='Markov Homogeneity Test')[source]
Test for homogeneity of Markov transition probabilities across regimes.
Parameters: transition_matrices : list of transition matrices for regimes, all matrices must have same size (r, c). r is the number of rows in the transition matrix and c is the number of columns in the transition matrix. regime_names : sequence Labels for the regimes. class_names : sequence Labels for the classes/states of the Markov chain. title : string name of test. : implicit an instance of Homogeneity_Results.
|
{}
|
The posterior medial network is at the apex of a temporal integration hierarchy in the brain, integrating information over many seconds of viewing intact, but not scrambled, movies. This has been interpreted as an effect of temporal structure. Such structure in movies depends on preexisting event schemas, but temporal structure can also arise de novo from learning. Here, we examined the relative role of schema-consistent temporal structure and arbitrary but consistent temporal structure on the human posterior medial network. We tested whether, with repeated viewing, the network becomes engaged by scrambled movies with temporal structure. Replicating prior studies, activity in posterior medial regions was immediately locked to stimulus structure upon exposure to intact, but not scrambled, movies. However, for temporally structured scrambled movies, functional coupling within the network increased across stimulus repetitions, rising to the level of intact movies. Thus, temporal structure is a key determinant of network dynamics and function in the posterior medial network.
|
{}
|
# mixed permutations and combinations
I have a problem that I am not too sure of. In a team of 16, there are 5 couples and 6 single people. In how many ways can at most 1 couple be chosen if 6 people are required to represent the team at a conference?
This is my solution: 6P6 (0 couples and 6 single people only) + 11C6 * 6! * 2! (choose 1 member of a couple and the other 6 single people, this can be done in 2!*6! ways) + 5C1 * 2! * 10C4 *2! * 4! (exactly one couple is chosen and any of the singles along with the other partners of the other couples)
Is this correct?
The following approach is reasonably systematic. There will be lots of words, but at the end there will be a more or less compact formula.
We count first the teams that have no couple, then, in basically the same way, the teams that have $1$ couple. A couple, viewed as an entity, will be called a family. There are $5$ families.
No couples: Maybe we choose $6$ singles. That can be done in $\binom{6}{6}$ ways. Of course this is $1$, but we call it by the complicated name $\binom{6}{6}\binom{5}{0}2^0$. Soon that will look reasonable!
Or else we pick $5$ singles, $1$ family, and a representative of the family. This can be done in $\binom{6}{5}\binom{5}{1}2^1$ ways.
Or else we pick $4$ singles, $2$ families, and a representative of each family. This can be done in $\binom{6}{4}\binom{5}{2}2^2$ ways.
Or else we pick $3$ singles, $3$ families, and a representative of each family. This can be done in $\binom{6}{3}\binom{5}{3}2^3$ ways.
Or else we pick $2$ singles, $4$ families, and a representative of each family. This can be done in $\binom{6}{2}\binom{5}{4}2^4$ ways.
Or else, finally, we pick $1$ singles, $5$ families, and a representative of each family. This can be done in $\binom{6}{1}\binom{5}{5}2^5$ ways.
Add up. A number of cases, but only one idea.
One couple: The idea is the same. There are $\binom{5}{1}$ ways o pick the couple. We will count the number of ways to pick the remaining $4$ people, add them up, and multiply by $\binom{5}{1}$. But rom here on, we only count the ways of picking the $4$.
We could pick $4$ singles. This can be done in $\binom{6}{4}$ ways, but we call the number $\binom{6}{4}\binom{5}{0}2^0$.
Or else we pick $3$ singles, $1$ family, and a representative of the family. This can be done in $\binom{6}{3}\binom{5}{1}2^1$ ways.
Or else we pick $2$ singles, $2$ families, and a representative of each family. This can be done in $\binom{6}{2}\binom{5}{2}2^2$ ways.
Or else we pick $1$ single, $3$ families, and a representative of each family. This can be done in $\binom{6}{1}\binom{5}{3}2^3$ ways.
Or else, finally, we pick $0$ singles, $4$ families, and a representative of each family. This can be done in $\binom{6}{0}\binom{5}{4}2^4$ ways.
Final answer: We gather the whole thing into a compact formula. $$\sum_{i=0}^5 \binom{6}{6-i}\binom{5}{i}2^i +\binom{5}{1}\sum_{i=0}^4 \binom{6}{4-i}\binom{5}{i}2^i.$$
You’re off on the wrong foot right away with that $6$P$6$ for the case in which the team is made up entirely of singles: there is only one such team, and you’re counting $6!=720$. We’re not picking six people and assigning each of them a specific rôle; we’re just picking a group of $6$ people. If we pick them from the singles, we can do it in $\binom66=1$ way.
I think that I’d break it down according to how many singles we choose.
• As we just saw, there is one team consisting of $6$ singles.
• To form a team with $5$ singles, we can pick the singles in $\binom65=6$ ways and the sixth person in $\binom{10}1=10$ ways for a total of $6\cdot10=60$ teams.
• To form a team with $4$ singles, we can pick the singles in $\binom64=15$ ways. And since we’re allowed one couple, we can pick any two of the other ten people, so there are $\binom{10}2=45$ to pick the other two members of the team. That gives us another $15\cdot45=675$ teams.
• $3$ singles can be picked in $\binom63=20$ ways, and we can pick any three of the other ten people, so we get another $20\binom{10}3=20\cdot120=2400$ teams.
Now it gets a little trickier, since we start running into restrictions on whom we can choose from the couples.
• $2$ singles can be picked in $\binom62=15$ ways. There are $\binom{10}4=210$ ways to pick $4$ people from the couples, but some of these are forbidden: specifically, we may not pick two couples. Since there are $5$ couples altogether, there are $\binom52=10$ pairs of couples. That’s $10$ sets of four people that we aren’t allowed to choose, leaving $210-10=200$ sets that we are allowed to choose. Thus, we can form $15\cdot200=3000$ teams with $2$ singles.
• One single can be picked in $6$ ways. There are $\binom{10}5=252$ ways to choose $5$ people from the couples, but here again some are not allowed. Specifically, we have to throw out those groups that consist of two-and-a-half couples. As before, there are $\binom52=10$ ways to pick two couples, and there are then $6$ ways to pick one more person from the remaining three couples. That makes $10\cdot6=60$ ways to fill out the team and gives us $6\cdot60=360$ teams.
There’s one more case, the teams containing no singles; I’ll let you have a chance to work it out on your own, but feel free to ask if you get stuck.
|
{}
|
# mot package¶
## mot.configuration module¶
Contains the runtime configuration of MOT.
This consists of two parts, functions to get the current runtime settings and configuration actions to update these settings. To set a new configuration, create a new ConfigAction and use this within a context environment using config_context(). Example:
from mot.configuration import RuntimeConfigurationAction, config_context
with config_context(RuntimeConfigurationAction(...)):
...
class mot.configuration.CLRuntimeAction(cl_runtime_info)[source]
Set the current configuration to use the information in the given configuration action.
Parameters: cl_runtime_info (CLRuntimeInfo) – the runtime info with the configuration options
class mot.configuration.CLRuntimeInfo(cl_environments=None, compile_flags=None, double_precision=None)[source]
Bases: object
All information necessary for applying operations using OpenCL.
Parameters: cl_environments (list of mot.lib.cl_environments.CLEnvironment) – The list of CL environments used by this routine. If None is given we use the defaults in the current configuration. compile_flags (list) – the list of compile flags to use during analysis. double_precision (boolean) – if we apply the computations in double precision or in single float precision. By default we go for single float precision.
cl_environments
compile_flags
Get all defined compile flags.
double_precision
mot_float_dtype
class mot.configuration.ConfigAction[source]
Bases: object
Defines a configuration action for use in a configuration context.
This should define an apply and unapply function that sets and unsets the configuration options.
The applying action needs to remember the state before the application of the action.
apply()[source]
Apply the current action to the current runtime configuration.
unapply()[source]
Reset the current configuration to the previous state.
class mot.configuration.RuntimeConfigurationAction(cl_environments=None, compile_flags=None, double_precision=None)[source]
Parameters: cl_environments (list of CLEnvironment) – the new CL environments we wish to use for future computations compile_flags (list) – the list of compile flags to use during analysis. double_precision (boolean) – if we compute in double precision or not
class mot.configuration.SimpleConfigAction[source]
Defines a default implementation of a configuration action.
This simple config implements a default apply() method that saves the current state and a default unapply() that restores the previous state.
For developers, it is easiest to implement _apply() such that you do not manually need to store the old configuraration.
apply()[source]
Apply the current action to the current runtime configuration.
unapply()[source]
Reset the current configuration to the previous state.
class mot.configuration.VoidConfigurationAction[source]
Does nothing, useful as a default config action.
mot.configuration.config_context(config_action)[source]
Creates a context in which the config action is applied and unapplies the configuration after execution.
Parameters: config_action (ConfigAction) – the configuration action to use
mot.configuration.get_cl_environments()[source]
Get the current CL environment to use during CL calculations.
Returns: the current list of CL environments. list of CLEnvironment
mot.configuration.get_compile_flags()[source]
Get the default compile flags to use in a CL routine.
Returns: the default list of compile flags we wish to use list
mot.configuration.set_cl_environments(cl_environments)[source]
Set the current CL environments to the given list
Please note that this will change the global configuration, i.e. this is a persistent change. If you do not want a persistent state change, consider using config_context() instead.
Parameters: cl_environments (list of CLEnvironment) – the new list of CL environments. ValueError – if the list of environments is empty
mot.configuration.set_compile_flags(compile_flags)[source]
Set the current compile flags.
Parameters: compile_flags (list) – the new list of compile flags
mot.configuration.set_default_proposal_update(proposal_update)[source]
Set the default proposal update function to use in sample.
Please note that this will change the global configuration, i.e. this is a persistent change. If you do not want a persistent state change, consider using config_context() instead.
Parameters: mot.model_building.parameter_functions.proposal_updates.ProposalUpdate – the new proposal update function to use by default if no specific one is provided.
mot.configuration.set_use_double_precision(double_precision)[source]
Set the default use of double precision.
Returns: if we use double precision by default or not boolean
mot.configuration.use_double_precision()[source]
Check if we run the computations in default precision or not.
Returns: if we run the computations in double precision or not boolean
## mot.mcmc_diagnostics module¶
This module contains some diagnostic functions to diagnose the performance of MCMC sample.
The two most important functions are multivariate_ess() and univariate_ess() to calculate the effective sample size of your samples.
class mot.mcmc_diagnostics.BatchMeansMCSE[source]
Computes the Monte Carlo Standard Error using simple batch means.
compute_standard_error(chain, batch_size)[source]
Compute the standard error of the given chain and the given batch size.
Parameters: chain (ndarray) – the chain for which to compute the SE batch_size (int) – batch size or window size to use in the computations the Monte Carlo Standard Error float
class mot.mcmc_diagnostics.ComputeMonteCarloStandardError[source]
Bases: object
Method to compute the Monte Carlo Standard error.
compute_standard_error(chain, batch_size)[source]
Compute the standard error of the given chain and the given batch size.
Parameters: chain (ndarray) – the chain for which to compute the SE batch_size (int) – batch size or window size to use in the computations the Monte Carlo Standard Error float
class mot.mcmc_diagnostics.CubeRootSingleBatch[source]
Returns $$n^{1/3}$$.
get_multivariate_ess_batch_sizes(nmr_params, chain_length)[source]
Get the batch sizes to use for the calculation of the Effective Sample Size (ESS).
This should return a list of batch sizes that the ESS calculation will use to determine $$\Sigma$$
Parameters: nmr_params (int) – the number of parameters in the samples chain_length (int) – the length of the chain the batches of the given sizes we will test in the ESS calculations list
get_univariate_ess_batch_sizes(chain_length)[source]
Get the batch sizes to use for the calculation of the univariate Effective Sample Size (ESS).
This should return a list of batch sizes that the ESS calculation will use to determine $$\sigma$$
Parameters: chain_length (int) – the length of the chain the batches of the given sizes we will test in the ESS calculations list
class mot.mcmc_diagnostics.LinearSpacedBatchSizes(nmr_batches=200)[source]
Returns a number of batch sizes from which the ESS algorithm will select the one with the lowest ESS.
This is a conservative choice since the lowest ESS of all batch sizes is chosen.
The batch sizes are generated as linearly spaced values in:
$\Big[ n^{1/4}, max(\lfloor x/max(20,p) \rfloor, \lfloor \sqrt{n} \rfloor) \Big]$
where $$n$$ is the chain length and $$p$$ is the number of parameters.
Parameters: nmr_batches (int) – the number of linearly spaced batches we will generate.
get_multivariate_ess_batch_sizes(nmr_params, chain_length)[source]
Get the batch sizes to use for the calculation of the Effective Sample Size (ESS).
This should return a list of batch sizes that the ESS calculation will use to determine $$\Sigma$$
Parameters: nmr_params (int) – the number of parameters in the samples chain_length (int) – the length of the chain the batches of the given sizes we will test in the ESS calculations list
class mot.mcmc_diagnostics.MultiVariateESSBatchSizeGenerator[source]
Bases: object
Objects of this class are used as input to the multivariate ESS function.
The multivariate ESS function needs to have at least one batch size to use during the computations. More batch sizes are also possible and the batch size with the lowest ESS is then preferred. Objects of this class implement the logic behind choosing batch sizes.
get_multivariate_ess_batch_sizes(nmr_params, chain_length)[source]
Get the batch sizes to use for the calculation of the Effective Sample Size (ESS).
This should return a list of batch sizes that the ESS calculation will use to determine $$\Sigma$$
Parameters: nmr_params (int) – the number of parameters in the samples chain_length (int) – the length of the chain the batches of the given sizes we will test in the ESS calculations list
class mot.mcmc_diagnostics.OverlappingBatchMeansMCSE[source]
Computes the Monte Carlo Standard Error using overlapping batch means.
compute_standard_error(chain, batch_size)[source]
Compute the standard error of the given chain and the given batch size.
Parameters: chain (ndarray) – the chain for which to compute the SE batch_size (int) – batch size or window size to use in the computations the Monte Carlo Standard Error float
class mot.mcmc_diagnostics.SquareRootSingleBatch[source]
Returns $$\sqrt(n)$$.
get_multivariate_ess_batch_sizes(nmr_params, chain_length)[source]
Get the batch sizes to use for the calculation of the Effective Sample Size (ESS).
This should return a list of batch sizes that the ESS calculation will use to determine $$\Sigma$$
Parameters: nmr_params (int) – the number of parameters in the samples chain_length (int) – the length of the chain the batches of the given sizes we will test in the ESS calculations list
get_univariate_ess_batch_sizes(chain_length)[source]
Get the batch sizes to use for the calculation of the univariate Effective Sample Size (ESS).
This should return a list of batch sizes that the ESS calculation will use to determine $$\sigma$$
Parameters: chain_length (int) – the length of the chain the batches of the given sizes we will test in the ESS calculations list
class mot.mcmc_diagnostics.UniVariateESSBatchSizeGenerator[source]
Bases: object
Objects of this class are used as input to the univariate ESS function that uses the batch means.
The univariate batch means ESS function needs to have at least one batch size to use during the computations. More batch sizes are also possible and the batch size with the lowest ESS is then preferred. Objects of this class implement the logic behind choosing batch sizes.
get_univariate_ess_batch_sizes(chain_length)[source]
Get the batch sizes to use for the calculation of the univariate Effective Sample Size (ESS).
This should return a list of batch sizes that the ESS calculation will use to determine $$\sigma$$
Parameters: chain_length (int) – the length of the chain the batches of the given sizes we will test in the ESS calculations list
mot.mcmc_diagnostics.estimate_multivariate_ess(samples, batch_size_generator=None, full_output=False)[source]
Compute the multivariate Effective Sample Size of your (single instance set of) samples.
This multivariate ESS is defined in Vats et al. (2016) and is given by:
$ESS = n \bigg(\frac{|\Lambda|}{|\Sigma|}\bigg)^{1/p}$
Where $$n$$ is the number of samples, $$p$$ the number of parameters, $$\Lambda$$ is the covariance matrix of the parameters and $$\Sigma$$ captures the covariance structure in the target together with the covariance due to correlated samples. $$\Sigma$$ is estimated using estimate_multivariate_ess_sigma().
In the case of NaN in any part of the computation the ESS is set to 0.
To compute the multivariate ESS for multiple problems, please use multivariate_ess().
Parameters: samples (ndarray) – an pxn matrix with for p parameters and n samples. batch_size_generator (MultiVariateESSBatchSizeGenerator) – the batch size generator, tells us how many batches and of which size we use for estimating the minimum ESS. Defaults to SquareRootSingleBatch full_output (boolean) – set to True to return the estimated $$\Sigma$$ and the optimal batch size. when full_output is set to True we return a tuple with the estimated multivariate ESS, the estimated $$\Sigma$$ matrix and the optimal batch size. When full_output is False (the default) we only return the ESS. float or tuple
References
Vats D, Flegal J, Jones G (2016). Multivariate Output Analysis for Markov Chain Monte Carlo. arXiv:1512.07713v2 [math.ST]
mot.mcmc_diagnostics.estimate_multivariate_ess_sigma(samples, batch_size)[source]
Calculates the Sigma matrix which is part of the multivariate ESS calculation.
This implementation is based on the Matlab implementation found at: https://github.com/lacerbi/multiESS
The Sigma matrix is defined as:
$\Sigma = \Lambda + 2 * \sum_{k=1}^{\infty}{Cov(Y_{1}, Y_{1+k})}$
Where $$Y$$ are our samples and $$\Lambda$$ is the covariance matrix of the samples.
This implementation computes the $$\Sigma$$ matrix using a Batch Mean estimator using the given batch size. The batch size has to be $$1 \le b_n \le n$$ and a typical value is either $$\lfloor n^{1/2} \rfloor$$ for slow mixing chains or $$\lfloor n^{1/3} \rfloor$$ for reasonable mixing chains.
If the length of the chain is longer than the sum of the length of all the batches, this implementation calculates $$\Sigma$$ for every offset and returns the average of those offsets.
Parameters: samples (ndarray) – the samples for which we compute the sigma matrix. Expects an (p, n) array with p the number of parameters and n the sample size batch_size (int) – the batch size used in the approximation of the correlation covariance an pxp array with p the number of parameters in the samples. ndarray
References
Vats D, Flegal J, Jones G (2016). Multivariate Output Analysis for Markov Chain Monte Carlo. arXiv:1512.07713v2 [math.ST]
mot.mcmc_diagnostics.estimate_univariate_ess_autocorrelation(chain, max_lag=None)[source]
Estimate effective sample size (ESS) using the autocorrelation of the chain.
The ESS is an estimate of the size of an iid sample with the same variance as the current sample. This function implements the ESS as described in Kass et al. (1998) and Robert and Casella (2004; p. 500):
$ESS(X) = \frac{n}{\tau} = \frac{n}{1 + 2 * \sum_{k=1}^{m}{\rho_{k}}}$
where $$\rho_{k}$$ is estimated as:
$\hat{\rho}_{k} = \frac{E[(X_{t} - \mu)(X_{t + k} - \mu)]}{\sigma^{2}}$
References
• Kass, R. E., Carlin, B. P., Gelman, A., and Neal, R. (1998)
Markov chain Monte Carlo in practice: A roundtable discussion. The American Statistician, 52, 93–100.
• Robert, C. P. and Casella, G. (2004) Monte Carlo Statistical Methods. New York: Springer.
• Geyer, C. J. (1992) Practical Markov chain Monte Carlo. Statistical Science, 7, 473–483.
Parameters: chain (ndarray) – the chain for which to calculate the ESS, assumes a vector of length n samples max_lag (int) – the maximum lag used in the variance calculations. If not given defaults to $$min(n/3, 1000)$$. the estimated ESS float
mot.mcmc_diagnostics.estimate_univariate_ess_standard_error(chain, batch_size_generator=None, compute_method=None)[source]
Compute the univariate ESS using the standard error method.
This computes the ESS using:
$ESS(X) = n * \frac{\lambda^{2}}{\sigma^{2}}$
Where $$\lambda$$ is the standard deviation of the chain and $$\sigma$$ is estimated using the monte carlo standard error (which in turn is, by default, estimated using a batch means estimator).
Parameters: chain (ndarray) – the Markov chain batch_size_generator (UniVariateESSBatchSizeGenerator) – the method that generates that batch sizes we will use. Per default it uses the SquareRootSingleBatch method. compute_method (ComputeMonteCarloStandardError) – the method used to compute the standard error. By default we will use the BatchMeansMCSE method the estimated ESS float
mot.mcmc_diagnostics.get_auto_correlation(chain, lag)[source]
Estimates the auto correlation for the given chain (1d vector) with the given lag.
Given a lag $$k$$, the auto correlation coefficient $$\rho_{k}$$ is estimated as:
$\hat{\rho}_{k} = \frac{E[(X_{t} - \mu)(X_{t + k} - \mu)]}{\sigma^{2}}$
Please note that this equation only works for lags $$k < n$$ where $$n$$ is the number of samples in the chain.
Parameters: chain (ndarray) – the vector with the samples lag (int) – the lag to use in the autocorrelation computation the autocorrelation with the given lag float
mot.mcmc_diagnostics.get_auto_correlation_time(chain, max_lag=None)[source]
Compute the auto correlation time up to the given lag for the given chain (1d vector).
This will halt when the maximum lag $$m$$ is reached or when the sum of two consecutive lags for any odd lag is lower or equal to zero.
The auto correlation sum is estimated as:
$\tau = 1 + 2 * \sum_{k=1}^{m}{\rho_{k}}$
Where $$\rho_{k}$$ is estimated as:
$\hat{\rho}_{k} = \frac{E[(X_{t} - \mu)(X_{t + k} - \mu)]}{\sigma^{2}}$
Parameters: chain (ndarray) – the vector with the samples max_lag (int) – the maximum lag to use in the autocorrelation computation. If not given we use: $$min(n/3, 1000)$$.
mot.mcmc_diagnostics.minimum_multivariate_ess(nmr_params, alpha=0.05, epsilon=0.05)[source]
Calculate the minimum multivariate Effective Sample Size you will need to obtain the desired precision.
This implements the inequality from Vats et al. (2016):
$\widehat{ESS} \geq \frac{2^{2/p}\pi}{(p\Gamma(p/2))^{2/p}} \frac{\chi^{2}_{1-\alpha,p}}{\epsilon^{2}}$
Where $$p$$ is the number of free parameters.
Parameters: nmr_params (int) – the number of free parameters in the model alpha (float) – the level of confidence of the confidence region. For example, an alpha of 0.05 means that we want to be in a 95% confidence region. epsilon (float) – the level of precision in our multivariate ESS estimate. An epsilon of 0.05 means that we expect that the Monte Carlo error is 5% of the uncertainty in the target distribution. the minimum multivariate Effective Sample Size that one should aim for in MCMC sample to obtain the desired confidence region with the desired precision. float
References
Vats D, Flegal J, Jones G (2016). Multivariate Output Analysis for Markov Chain Monte Carlo. arXiv:1512.07713v2 [math.ST]
mot.mcmc_diagnostics.monte_carlo_standard_error(chain, batch_size_generator=None, compute_method=None)[source]
Compute Monte Carlo standard errors for the expectations
This is a convenience function that calls the compute method for each batch size and returns the lowest ESS over the used batch sizes.
Parameters: chain (ndarray) – the Markov chain batch_size_generator (UniVariateESSBatchSizeGenerator) – the method that generates that batch sizes we will use. Per default it uses the SquareRootSingleBatch method. compute_method (ComputeMonteCarloStandardError) – the method used to compute the standard error. By default we will use the BatchMeansMCSE method
mot.mcmc_diagnostics.multivariate_ess(samples, batch_size_generator=None)[source]
Estimate the multivariate Effective Sample Size for the samples of every problem.
This essentially applies estimate_multivariate_ess() to every problem.
Parameters: samples (ndarray, dict or generator) – either a matrix of shape (d, p, n) with d problems, p parameters and n samples, or a dictionary with for every parameter a matrix with shape (d, n) or, finally, a generator function that yields sample arrays of shape (p, n). batch_size_generator (MultiVariateESSBatchSizeGenerator) – the batch size generator, tells us how many batches and of which size we use in estimating the minimum ESS. the multivariate ESS per problem ndarray
mot.mcmc_diagnostics.multivariate_ess_precision(nmr_params, multi_variate_ess, alpha=0.05)[source]
Calculate the precision given your multivariate Effective Sample Size.
Given that you obtained $$ESS$$ multivariate effective samples in your estimate you can calculate the precision with which you approximated your desired confidence region.
This implements the inequality from Vats et al. (2016), slightly restructured to give $$\epsilon$$ back instead of the minimum ESS.
$\epsilon = \sqrt{\frac{2^{2/p}\pi}{(p\Gamma(p/2))^{2/p}} \frac{\chi^{2}_{1-\alpha,p}}{\widehat{ESS}}}$
Where $$p$$ is the number of free parameters and ESS is the multivariate ESS from your samples.
Parameters: nmr_params (int) – the number of free parameters in the model multi_variate_ess (int) – the number of iid samples you obtained in your sample results. alpha (float) – the level of confidence of the confidence region. For example, an alpha of 0.05 means that we want to be in a 95% confidence region. the minimum multivariate Effective Sample Size that one should aim for in MCMC sample to obtain the desired confidence region with the desired precision. float
References
Vats D, Flegal J, Jones G (2016). Multivariate Output Analysis for Markov Chain Monte Carlo. arXiv:1512.07713v2 [math.ST]
mot.mcmc_diagnostics.univariate_ess(samples, method='standard_error', **kwargs)[source]
Estimate the univariate Effective Sample Size for the samples of every problem.
This computes the ESS using:
$ESS(X) = n * \frac{\lambda^{2}}{\sigma^{2}}$
Where $$\lambda$$ is the standard deviation of the chain and $$\sigma$$ is estimated using the monte carlo standard error (which in turn is, by default, estimated using a batch means estimator).
Parameters: samples (ndarray, dict or generator) – either a matrix of shape (d, p, n) with d problems, p parameters and n samples, or a dictionary with for every parameter a matrix with shape (d, n) or, finally, a generator function that yields sample arrays of shape (p, n). method (str) – one of ‘autocorrelation’ or ‘standard_error’ defaults to ‘standard_error’. If ‘autocorrelation’ is chosen we apply the function: estimate_univariate_ess_autocorrelation(), if ‘standard_error is choosen we apply the function: estimate_univariate_ess_standard_error(). **kwargs – passed to the chosen compute method a matrix of size (d, p) with for every problem and every parameter an ESS. ndarray
References
• Flegal, J.M., Haran, M., and Jones, G.L. (2008). “Markov chain Monte Carlo: Can We Trust the Third Significant Figure?”. Statistical Science, 23, p. 250-260.
• Marc S. Meketon and Bruce Schmeiser. 1984. Overlapping batch means: something for nothing?. In Proceedings of the 16th conference on Winter simulation (WSC ‘84), Sallie Sheppard (Ed.). IEEE Press, Piscataway, NJ, USA, 226-230.
## mot.random module¶
This uses the random123 library for generating multiple lists of random numbers.
From the Random123 documentation:
Unlike conventional RNGs, counter-based RNGs are stateless functions (or function classes i.e. functors) whose arguments are a counter and a key, and returns a result of the same type as the counter.
result = CBRNGname(counter, key)
The result is producted by a deterministic function of the key and counter, i.e. a unique (counter, key) tuple will always produce the same result. The result is highly sensitive to small changes in the inputs, so that the sequence of values produced by simply incrementing the counter (or key) is effectively indistinguishable from a sequence of samples of a uniformly distributed random variable.
All the Random123 generators are counter-based RNGs that use integer multiplication, xor and permutation of W-bit words to scramble its N-word input key.
In this implementation we generate a counter and key automatically from a single seed.
mot.random.normal(nmr_distributions, nmr_samples, mean=0, std=1, ctype='float', seed=None)[source]
Draw random samples from the Gaussian distribution.
Parameters: nmr_distributions (int) – the number of unique continuous_distributions to create nmr_samples (int) – The number of samples to draw mean (float or ndarray) – The mean of the distribution std (float or ndarray) – The standard deviation or the distribution ctype (str) – the C type of the output samples seed (float) – the seed for the RNG A two dimensional numpy array as (nmr_distributions, nmr_samples). ndarray
mot.random.uniform(nmr_distributions, nmr_samples, low=0, high=1, ctype='float', seed=None)[source]
Draw random samples from the Uniform distribution.
Parameters: nmr_distributions (int) – the number of unique continuous_distributions to create nmr_samples (int) – The number of samples to draw low (double) – The minimum value of the random numbers high (double) – The minimum value of the random numbers ctype (str) – the C type of the output samples seed (float) – the seed for the RNG A two dimensional numpy array as (nmr_distributions, nmr_samples). ndarray
## mot.stats module¶
mot.stats.deviance_information_criterions(mean_posterior_lls, ll_per_sample)[source]
Calculates the Deviance Information Criteria (DIC) using three methods.
This returns a dictionary returning the DIC_2002, the DIC_2004 and the DIC_Ando_2011 method. The first is based on Spiegelhalter et al (2002), the second based on Gelman et al. (2004) and the last on Ando (2011). All cases differ in how they calculate model complexity, i.e. the effective number of parameters in the model. In all cases the model with the smallest DIC is preferred.
All these DIC methods measure fitness using the deviance, which is, for a likelihood $$p(y | \theta)$$ defined as:
$D(\theta) = -2\log p(y|\theta)$
From this, the posterior mean deviance,
$\bar{D} = \mathbb{E}_{\theta}[D(\theta)]$
is then used as a measure of how well the model fits the data.
The complexity, or measure of effective number of parameters, can be measured in see ways, see Spiegelhalter et al. (2002), Gelman et al (2004) and Ando (2011). The first method calculated the parameter deviance as:
\begin{align} p_{D} &= \mathbb{E}_{\theta}[D(\theta)] - D(\mathbb{E}[\theta)]) \\ &= \bar{D} - D(\bar{\theta}) \end{align}
i.e. posterior mean deviance minus the deviance evaluated at the posterior mean of the parameters.
The second method calculated $$p_{D}$$ as:
$p_{D} = p_{V} = \frac{1}{2}\hat{var}(D(\theta))$
i.e. half the variance of the deviance is used as an estimate of the number of free parameters in the model.
The third method calculates the parameter deviance as:
$p_{D} = 2 \cdot (\bar{D} - D(\bar{\theta}))$
That is, twice the complexity of that of the first method.
Finally, the DIC is (for all cases) defined as:
$DIC = \bar{D} + p_{D}$
Parameters: mean_posterior_lls (ndarray) – a 1d matrix containing the log likelihood for the average posterior point estimate. That is, the single log likelihood of the average parameters. ll_per_sample (ndarray) – a (d, n) array with for d problems the n log likelihoods. This is the log likelihood per sample. a dictionary containing the DIC_2002, the DIC_2004 and the DIC_Ando_2011 information criterion maps. dict
mot.stats.fit_circular_gaussian(samples, high=3.141592653589793, low=0)[source]
Compute the circular mean for samples in a range
Parameters: samples (ndarray) – a one or two dimensional array. If one dimensional we calculate the fit using all values. If two dimensional, we fit the Gaussian for every set of samples over the first dimension. high (float) – The maximum wrap point low (float) – The minimum wrap point
mot.stats.fit_gaussian(samples, ddof=0)[source]
Calculates the mean and the standard deviation of the given samples.
Parameters: samples (ndarray) – a one or two dimensional array. If one dimensional we calculate the fit using all values. If two dimensional, we fit the Gaussian for every set of samples over the first dimension. ddof (int) – the difference degrees of freedom in the std calculation. See numpy.
mot.stats.fit_truncated_gaussian(samples, lower_bounds, upper_bounds)[source]
Fits a truncated gaussian distribution on the given samples.
This will do a maximum likelihood estimation of a truncated Gaussian on the provided samples, with the truncation points given by the lower and upper bounds.
Parameters: samples (ndarray) – a one or two dimensional array. If one dimensional we fit the truncated Gaussian on all values. If two dimensional, we calculate the truncated Gaussian for every set of samples over the first dimension. lower_bounds (ndarray or float) – the lower bound, either a scalar or a lower bound per problem (first index of samples) upper_bounds (ndarray or float) – the upper bound, either a scalar or an upper bound per problem (first index of samples) the mean and std of the fitted truncated Gaussian mean, std
mot.stats.gaussian_overlapping_coefficient(means_0, stds_0, means_1, stds_1, lower=None, upper=None)[source]
Compute the overlapping coefficient of two Gaussian continuous_distributions.
This computes the $$\int_{-\infty}^{\infty}{\min(f(x), g(x))\partial x}$$ where $$f \sim \mathcal{N}(\mu_0, \sigma_0^{2})$$ and $$f \sim \mathcal{N}(\mu_1, \sigma_1^{2})$$ are normally distributed variables.
This will compute the overlap for each element in the first dimension.
Parameters: means_0 (ndarray) – the set of means of the first distribution stds_0 (ndarray) – the set of stds of the fist distribution means_1 (ndarray) – the set of means of the second distribution stds_1 (ndarray) – the set of stds of the second distribution lower (float) – the lower limit of the integration. If not set we set it to -inf. upper (float) – the upper limit of the integration. If not set we set it to +inf.
## Module contents¶
mot.smart_device_selection`()[source]
Get a list of device environments that is suitable for use in MOT.
Returns: List with the CL device environments. list of CLEnvironment
|
{}
|
# Homework Help: Speed and direction of a travelling wave
1. Sep 20, 2015
### sarvensogo
1. The problem statement, all variables and given/known data
$$\Psi(y,t)=A\cos^22\pi(t-y)$$
Show that this is a travelling wave. Use the general form of a travelling wave to determine its speed and direction.
Verify your answer using $$\frac{-\partial \Psi / \partial t}{\partial \Psi / \partial x}$$
2. Relevant equations
The general form of a travelling wave from class is $$\Psi(x,t)=f(x-vt)$$.
3. The attempt at a solution
So I tried to just use the form of the travelling wave, but the thing that confused me was whether or not I had to switch around the order of (t-y) in the equation to (-y+t)? If I leave it as is I get that the speed (v) is -1, but if I change the equation to read (-y+t) the speed becomes 1 instead. Either way using the partial derivative equation verifies my answer. I can post my handwritten work if neccesary for proof that I attempted it.
Thanks
2. Sep 20, 2015
### Staff: Mentor
Please show your work (ideally typed here, that is easier to read. You can also use LaTeX). That should not happen as t-y is exactly the same as -y+t.
|
{}
|
# According to the constitution, how many people represent each state in the house of representatives?
###### Question:
According to the constitution, how many people represent each state in the house of representatives?
### Which of the following are abrahamic faiths?check all that apply(no) buddhism(yes) roman catholicism (yes) islam(yes) judaismonly
Which of the following are abrahamic faiths? check all that apply (no) buddhism (yes) roman catholicism (yes) islam (yes) judaism only these three hope this helped !!...
### The gcf 28a^3b and 16ab^2 a)4ab b)4a^3b c)28ab d)112a^3b^2
The gcf 28a^3b and 16ab^2 a)4ab b)4a^3b c)28ab d)112a^3b^2...
### Is god real? if he is then how did we get a pug from a wolf and ect. And also why isn't he stoping the corona virus? pls answer
Is god real? if he is then how did we get a pug from a wolf and ect. And also why isn't he stoping the corona virus? pls answer...
### For 2019, eric is a self-employed financial consultant. during the current year, eric's net self-employment
For 2019, eric is a self-employed financial consultant. during the current year, eric's net self-employment income is $132,150. the oasdi rate is 6.2%; the mhi rate is 1.45%. hint: oasdi 6.2% tax is on the first$132,900 for the year. note: round all calculations to the nearest cent. what is eric...
### Write an equation for the circle whose graph is shown.
Write an equation for the circle whose graph is shown. $Write an equation for the circle whose graph is shown.$...
### Bob is a new ta for one coursework. he has a probability of ¼ to answer one question incorrectly independent
Bob is a new ta for one coursework. he has a probability of ¼ to answer one question incorrectly independent of other questions. in a class, bob can be asked with a maximum of 3 questions. the probability that bob will be asked with 1, 2, and 3 questions will be ⅓, respectively; bob is guarantee...
### How do i do this? i cant quite remember
How do i do this? i cant quite remember $How do i do this? i cant quite remember$...
### Choose the word or phrase that most nearly defines the underlined word. boo's transition from the basement to back home
Choose the word or phrase that most nearly defines the underlined word. boo's transition from the basement to back home was nebulous in jem's memory. a. vivid b. clear c. indistinctd. dark...
### What does God mean when he warns Cain that sin is crouching at the door and its desire is to control him? What would it mean
What does God mean when he warns Cain that sin is crouching at the door and its desire is to control him? What would it mean for sin to control Cain?...
### Frank Triplett, 'The March of Destiny 1003THE MAROW OF DESTRYKrity2USO 1164
Frank Triplett, "The March of Destiny 1003 THE MAROW OF DESTRY Krity 2 USO 1164...
### Yahusishe maudhui katika riwaya ya Dar es Salaam Usiku na mambo yanayotukia kila siku(toa ithibati) katika jamii ya karne ya 21
Yahusishe maudhui katika riwaya ya Dar es Salaam Usiku na mambo yanayotukia kila siku(toa ithibati) katika jamii ya karne ya 21...
### 12,349 is what percent of of 278,901
12,349 is what percent of of 278,901...
### DONT DELETE SERIOUS QUESTION!! I know this is an odd thing to write /ask but i wanna help, i am a christian but have had a hard
DONT DELETE SERIOUS QUESTION!! I know this is an odd thing to write /ask but i wanna help, i am a christian but have had a hard time believing, and seeing God but recently i believe he told me something about wher my future is going through a dream. Im slowly learning more and more how to trust him....
### WRS - Economic Systems - Country Descriptions Directions: On the back of this sheet, there is a set of descriptions (description
WRS - Economic Systems - Country Descriptions Directions: On the back of this sheet, there is a set of descriptions (description 1 goes with number 1 on the other side). For each description, you must identify the economic system that is being described, then use textual evidence (actual words fr...
### An animal shelter spends $4.50 per day to care for each bird and$8.50 per day to care for each cat.
An animal shelter spends $4.50 per day to care for each bird and$8.50 per day to care for each cat. Joshua noticed that the shelter spent \$173.50 caring for birds and cats on Tuesday. Joshua found a record showing that there were a total of 27 birds and cats on Tuesday. How many birds were at the s...
|
{}
|
# American Institute of Mathematical Sciences
February 2014, 7(1): 127-137. doi: 10.3934/dcdss.2014.7.127
## On the existence of solution for a Cahn-Hilliard/Allen-Cahn equation
1 Institute of Applied and Computational Mathematics, FO.R.T.H., and Department of Applied Mathematics, University of Crete, P.O. Box 2208, Heraklion, Crete 71409, Greece 2 Archimedes Center for Modeling, Analysis and Computation (ACMAC), Department of Applied Mathematics, University of Crete, P.O. Box 2208, Heraklion, Crete 71409, Greece
Received February 2012 Revised June 2012 Published July 2013
In this manuscript, we consider a Cahn-Hilliard/Allen-Cahn equation is introduced in [17]. We give an existence of the solution, slightly improved from [18]. We also present a stochastic version of this equation in [3].
Citation: Georgia Karali, Yuko Nagase. On the existence of solution for a Cahn-Hilliard/Allen-Cahn equation. Discrete and Continuous Dynamical Systems - S, 2014, 7 (1) : 127-137. doi: 10.3934/dcdss.2014.7.127
##### References:
[1] Nicholas D. Alikakos, Peter W. Bates and Xinfu Chen, Convergence of the Cahn-Hilliard equation to the Hele-Shaw model, Arch. Rational Mech. Anal., 128 (1994), 165-205. doi: 10.1007/BF00375025. [2] Dimitra Antonopoulou and Georgia Karali, Existence of solution for a generalized stochastic Cahn-Hilliard equation on convex domains, Discrete Contin. Dyn. Syst. Ser. B, 16 (2011), 31-55. doi: 10.3934/dcdsb.2011.16.31. [3] Dimitra Antonopoolou, Georgia Karali, Anne Millet and Yuko Nagase, Existence of solution and of its density for a Stochastic Cahn-Hilliard/Allen-Cahn equation,, preprint., (). [4] Kenneth A. Brakke, "The Motion of a Surface by its Mean Curvature," Mathematical Notes, 20, Princeton University Press, Princeton, N.J., 1978. [5] Caroline Cardon-Weber, Cahn-Hilliard stochastic equation: Existence of the solution and of its density, Bernoulli, 7 (2001), 777-816. doi: 10.2307/3318542. [6] Yun Gang Chen, Yoshikazu Giga and Shun'ichi Goto, Uniqueness and existence of viscosity solutions of generalized mean curvature flow equations, J. Differential Geom., 33 (1991), 749-786. [7] Giuseppe Da Prato and Arnaud Debussche, Stochastic Cahn-Hilliard equation, Nonlinear Anal., 26 (1996), 241-263. doi: 10.1016/0362-546X(94)00277-O. [8] Weinan E, Weiqing Ren and Eric Vanden-Eijnden, Minimum action method for the study of rare events, Comm. Pure Appl. Math., 57 (2004), 637-656. doi: 10.1002/cpa.20005. [9] L. C. Evans and J. Spruck, Motion of level sets by mean curvature I, J. Differential Geom., 33 (1991), 635-681. [10] William G. Faris and Giovanni Jona-Lasinio, Large fluctuations for a nonlinear heat equation with noise, J. Phys. A: Math. Gen., 15 (1982), 3025-3055. doi: 10.1088/0305-4470/15/10/011. [11] Jin Feng and Markos A. Katsoulakis, A comparison principle for Hamilton-Jacobi equations related to controlled gradient flows in infinite dimensions, Arch. Ration. Mech. Anal., 192 (2009), 275-310. doi: 10.1007/s00205-008-0133-5. [12] Paul C. Fife, "Dynamics of Internal Layers and Diffusive Interfaces," CBMS-NSF Regional Conference Series in Applied Mathematics, 53, Society for Industrial and Applied Mathematics (SIAM), Philadelphia, PA, 1988. doi: 10.1137/1.9781611970180. [13] M. I. Freidlin and A. D. Wentzell, "Random Perturbations of Dynamical Systems," (English summary) Second edition, Springer-Verlag, New York, 1998. doi: 10.1007/978-1-4612-0611-8. [14] , Yannis Goumas and Takashi Suzuki,, work in progress., (). [15] M. Hildebrand and A. S. Mikhailov, Mesoscopic modeling in the kinetic theory of adsorbates, J. Phys. Chem., 100 (1996), 19089-19101. doi: 10.1021/jp961668w. [16] Tom Ilmanen, Convergence of the Allen-Cahn equation to Brakke's motion by mean curvature, J. Differential Geom., 38 (1993), 417-461. [17] Georgia Karali and Markos A. Katsoulakis, The role of multiple microscopic mechanisms in cluster interface evolution, J. Differential Equations, 235 (2007), 418-438. doi: 10.1016/j.jde.2006.12.021. [18] Georgia Karali and Tonia Ricciardi, On the convergence of a fourth order evolution equation to the Allen-Cahn equation, Nonlinear Anal., 72 (2010), 4271-4281. doi: 10.1016/j.na.2010.02.003. [19] Markos A. Katsoulakis and Dionisios G. Vlachos, From microscopic interactions to macroscopic laws of cluster evolution, Phys. Rev. Lett., 84 (2000), 1511-1514. doi: 10.1103/PhysRevLett.84.1511. [20] Robert V. Kohn, Felix Otto, Maria G. Reznikoff and Eric Vanden-Eijinden, Action minimization and sharp-interface limits for the stochastic Allen-Cahn equation, Comm. Pure Appl. Math., 60 (2007), 393-438. doi: 10.1002/cpa.20144. [21] Robert V. Kohn, Maria G. Reznikoff and Yoshihiro Tonegawa, Sharp-interface limit of the Allen-Cahn action functional in one space dimension, Calc. Var. Partial Differential Equations, 25 (2006), 503-534. doi: 10.1007/s00526-005-0370-5. [22] J.-L. Lions, "Quelques Méthodes de Résolution des Problèmes aux Limites Non Linéaires," (French) Dunod; Gauthier-Villars, Paris, 1969. [23] L. Mugnai and Röger, The Allen-Cahn action functional in higher dimensions, Interfaces Free Bound., 10 (2008), 45-78. doi: 10.4171/IFB/179. [24] Yuko Nagase, Action minimization for an Allen-Cahn equation with an unequal double-well potential, Manuscripta Mathematica, 137 (2012), 81-106. doi: 10.1007/s00229-011-0458-5. [25] R. L. Pego, Front migration in the nonlinear Cahn-Hilliard equation, Proc. Roy. Soc. London Ser. A, 422 (1989), 261-278. doi: 10.1098/rspa.1989.0027. [26] Jacob Rubinstein, Peter Sternberg and Joseph B. Keller, Fast reaction, slow diffusion, and curve shortening, SIAM J. Appl. Math., 49 (1989), 116-133. doi: 10.1137/0149007. [27] Roger Temam, "Infinite-Dimensional Dynamical Systems in Mechanics and Physics," Second edition, Applied Mathematical Sciences, 68, Springer-Verlag, New York, 1997. [28] Eric Vanden-Eijnden and Maria G.Westdickenberg, Rare events in stochastic partial differential equations on large spatial domains, J. Stat. Phys., 131 (2008), 1023-1038. doi: 10.1007/s10955-008-9537-8. [29] John B. Walsh, An introduction to stochastic partial differential equations, in "École d'été de probabilités de Saint-Flour, XIV-1984," Lecture Notes in Math., 1180, Springer, Berlin, (1986), 265-439. doi: 10.1007/BFb0074920. [30] Maria G. Westdickenberg, Rare events, action minimization, and sharp interface limits, in "Singularities in PDE and the Calculus of Variations," CRM Proc. Lecture Notes, 44, Amer. Math. Soc., Providence, RI, (2008), 217-231. [31] Maria. G. Westdickenberg and Yoshihiro Tonegawa, Higher multiplicity in the one-dimensional Allen-Cahn action functional, Indiana Univ. Math. J., 56 (2007), 2935-2989. doi: 10.1512/iumj.2007.56.3182.
show all references
##### References:
[1] Nicholas D. Alikakos, Peter W. Bates and Xinfu Chen, Convergence of the Cahn-Hilliard equation to the Hele-Shaw model, Arch. Rational Mech. Anal., 128 (1994), 165-205. doi: 10.1007/BF00375025. [2] Dimitra Antonopoulou and Georgia Karali, Existence of solution for a generalized stochastic Cahn-Hilliard equation on convex domains, Discrete Contin. Dyn. Syst. Ser. B, 16 (2011), 31-55. doi: 10.3934/dcdsb.2011.16.31. [3] Dimitra Antonopoolou, Georgia Karali, Anne Millet and Yuko Nagase, Existence of solution and of its density for a Stochastic Cahn-Hilliard/Allen-Cahn equation,, preprint., (). [4] Kenneth A. Brakke, "The Motion of a Surface by its Mean Curvature," Mathematical Notes, 20, Princeton University Press, Princeton, N.J., 1978. [5] Caroline Cardon-Weber, Cahn-Hilliard stochastic equation: Existence of the solution and of its density, Bernoulli, 7 (2001), 777-816. doi: 10.2307/3318542. [6] Yun Gang Chen, Yoshikazu Giga and Shun'ichi Goto, Uniqueness and existence of viscosity solutions of generalized mean curvature flow equations, J. Differential Geom., 33 (1991), 749-786. [7] Giuseppe Da Prato and Arnaud Debussche, Stochastic Cahn-Hilliard equation, Nonlinear Anal., 26 (1996), 241-263. doi: 10.1016/0362-546X(94)00277-O. [8] Weinan E, Weiqing Ren and Eric Vanden-Eijnden, Minimum action method for the study of rare events, Comm. Pure Appl. Math., 57 (2004), 637-656. doi: 10.1002/cpa.20005. [9] L. C. Evans and J. Spruck, Motion of level sets by mean curvature I, J. Differential Geom., 33 (1991), 635-681. [10] William G. Faris and Giovanni Jona-Lasinio, Large fluctuations for a nonlinear heat equation with noise, J. Phys. A: Math. Gen., 15 (1982), 3025-3055. doi: 10.1088/0305-4470/15/10/011. [11] Jin Feng and Markos A. Katsoulakis, A comparison principle for Hamilton-Jacobi equations related to controlled gradient flows in infinite dimensions, Arch. Ration. Mech. Anal., 192 (2009), 275-310. doi: 10.1007/s00205-008-0133-5. [12] Paul C. Fife, "Dynamics of Internal Layers and Diffusive Interfaces," CBMS-NSF Regional Conference Series in Applied Mathematics, 53, Society for Industrial and Applied Mathematics (SIAM), Philadelphia, PA, 1988. doi: 10.1137/1.9781611970180. [13] M. I. Freidlin and A. D. Wentzell, "Random Perturbations of Dynamical Systems," (English summary) Second edition, Springer-Verlag, New York, 1998. doi: 10.1007/978-1-4612-0611-8. [14] , Yannis Goumas and Takashi Suzuki,, work in progress., (). [15] M. Hildebrand and A. S. Mikhailov, Mesoscopic modeling in the kinetic theory of adsorbates, J. Phys. Chem., 100 (1996), 19089-19101. doi: 10.1021/jp961668w. [16] Tom Ilmanen, Convergence of the Allen-Cahn equation to Brakke's motion by mean curvature, J. Differential Geom., 38 (1993), 417-461. [17] Georgia Karali and Markos A. Katsoulakis, The role of multiple microscopic mechanisms in cluster interface evolution, J. Differential Equations, 235 (2007), 418-438. doi: 10.1016/j.jde.2006.12.021. [18] Georgia Karali and Tonia Ricciardi, On the convergence of a fourth order evolution equation to the Allen-Cahn equation, Nonlinear Anal., 72 (2010), 4271-4281. doi: 10.1016/j.na.2010.02.003. [19] Markos A. Katsoulakis and Dionisios G. Vlachos, From microscopic interactions to macroscopic laws of cluster evolution, Phys. Rev. Lett., 84 (2000), 1511-1514. doi: 10.1103/PhysRevLett.84.1511. [20] Robert V. Kohn, Felix Otto, Maria G. Reznikoff and Eric Vanden-Eijinden, Action minimization and sharp-interface limits for the stochastic Allen-Cahn equation, Comm. Pure Appl. Math., 60 (2007), 393-438. doi: 10.1002/cpa.20144. [21] Robert V. Kohn, Maria G. Reznikoff and Yoshihiro Tonegawa, Sharp-interface limit of the Allen-Cahn action functional in one space dimension, Calc. Var. Partial Differential Equations, 25 (2006), 503-534. doi: 10.1007/s00526-005-0370-5. [22] J.-L. Lions, "Quelques Méthodes de Résolution des Problèmes aux Limites Non Linéaires," (French) Dunod; Gauthier-Villars, Paris, 1969. [23] L. Mugnai and Röger, The Allen-Cahn action functional in higher dimensions, Interfaces Free Bound., 10 (2008), 45-78. doi: 10.4171/IFB/179. [24] Yuko Nagase, Action minimization for an Allen-Cahn equation with an unequal double-well potential, Manuscripta Mathematica, 137 (2012), 81-106. doi: 10.1007/s00229-011-0458-5. [25] R. L. Pego, Front migration in the nonlinear Cahn-Hilliard equation, Proc. Roy. Soc. London Ser. A, 422 (1989), 261-278. doi: 10.1098/rspa.1989.0027. [26] Jacob Rubinstein, Peter Sternberg and Joseph B. Keller, Fast reaction, slow diffusion, and curve shortening, SIAM J. Appl. Math., 49 (1989), 116-133. doi: 10.1137/0149007. [27] Roger Temam, "Infinite-Dimensional Dynamical Systems in Mechanics and Physics," Second edition, Applied Mathematical Sciences, 68, Springer-Verlag, New York, 1997. [28] Eric Vanden-Eijnden and Maria G.Westdickenberg, Rare events in stochastic partial differential equations on large spatial domains, J. Stat. Phys., 131 (2008), 1023-1038. doi: 10.1007/s10955-008-9537-8. [29] John B. Walsh, An introduction to stochastic partial differential equations, in "École d'été de probabilités de Saint-Flour, XIV-1984," Lecture Notes in Math., 1180, Springer, Berlin, (1986), 265-439. doi: 10.1007/BFb0074920. [30] Maria G. Westdickenberg, Rare events, action minimization, and sharp interface limits, in "Singularities in PDE and the Calculus of Variations," CRM Proc. Lecture Notes, 44, Amer. Math. Soc., Providence, RI, (2008), 217-231. [31] Maria. G. Westdickenberg and Yoshihiro Tonegawa, Higher multiplicity in the one-dimensional Allen-Cahn action functional, Indiana Univ. Math. J., 56 (2007), 2935-2989. doi: 10.1512/iumj.2007.56.3182.
[1] Changchun Liu, Hui Tang. Existence of periodic solution for a Cahn-Hilliard/Allen-Cahn equation in two space dimensions. Evolution Equations and Control Theory, 2017, 6 (2) : 219-237. doi: 10.3934/eect.2017012 [2] Georgios T. Kossioris, Georgios E. Zouraris. Finite element approximations for a linear Cahn-Hilliard-Cook equation driven by the space derivative of a space-time white noise. Discrete and Continuous Dynamical Systems - B, 2013, 18 (7) : 1845-1872. doi: 10.3934/dcdsb.2013.18.1845 [3] Dimitra Antonopoulou, Georgia Karali. Existence of solution for a generalized stochastic Cahn-Hilliard equation on convex domains. Discrete and Continuous Dynamical Systems - B, 2011, 16 (1) : 31-55. doi: 10.3934/dcdsb.2011.16.31 [4] Ahmad Makki, Alain Miranville. Existence of solutions for anisotropic Cahn-Hilliard and Allen-Cahn systems in higher space dimensions. Discrete and Continuous Dynamical Systems - S, 2016, 9 (3) : 759-775. doi: 10.3934/dcdss.2016027 [5] Cristina Pocci. On singular limit of a nonlinear $p$-order equation related to Cahn-Hilliard and Allen-Cahn evolutions. Evolution Equations and Control Theory, 2013, 2 (3) : 517-530. doi: 10.3934/eect.2013.2.517 [6] Matthieu Brachet, Philippe Parnaudeau, Morgan Pierre. Convergence to equilibrium for time and space discretizations of the Cahn-Hilliard equation. Discrete and Continuous Dynamical Systems - S, 2022, 15 (8) : 1987-2031. doi: 10.3934/dcdss.2022110 [7] Quan Wang, Dongming Yan. On the stability and transition of the Cahn-Hilliard/Allen-Cahn system. Discrete and Continuous Dynamical Systems - B, 2020, 25 (7) : 2607-2620. doi: 10.3934/dcdsb.2020024 [8] Christopher P. Grant. Grain sizes in the discrete Allen-Cahn and Cahn-Hilliard equations. Discrete and Continuous Dynamical Systems, 2001, 7 (1) : 127-146. doi: 10.3934/dcds.2001.7.127 [9] Alain Miranville, Ramon Quintanilla, Wafa Saoud. Asymptotic behavior of a Cahn-Hilliard/Allen-Cahn system with temperature. Communications on Pure and Applied Analysis, 2020, 19 (4) : 2257-2288. doi: 10.3934/cpaa.2020099 [10] Shixing Li, Dongming Yan. On the steady state bifurcation of the Cahn-Hilliard/Allen-Cahn system. Discrete and Continuous Dynamical Systems - B, 2019, 24 (7) : 3077-3088. doi: 10.3934/dcdsb.2018301 [11] Jie Shen, Xiaofeng Yang. Numerical approximations of Allen-Cahn and Cahn-Hilliard equations. Discrete and Continuous Dynamical Systems, 2010, 28 (4) : 1669-1691. doi: 10.3934/dcds.2010.28.1669 [12] Alain Miranville, Wafa Saoud, Raafat Talhouk. On the Cahn-Hilliard/Allen-Cahn equations with singular potentials. Discrete and Continuous Dynamical Systems - B, 2019, 24 (8) : 3633-3651. doi: 10.3934/dcdsb.2018308 [13] Yanzhao Cao, Li Yin. Spectral Galerkin method for stochastic wave equations driven by space-time white noise. Communications on Pure and Applied Analysis, 2007, 6 (3) : 607-617. doi: 10.3934/cpaa.2007.6.607 [14] Murat Uzunca, Ayşe Sarıaydın-Filibelioǧlu. Adaptive discontinuous galerkin finite elements for advective Allen-Cahn equation. Numerical Algebra, Control and Optimization, 2021, 11 (2) : 269-281. doi: 10.3934/naco.2020025 [15] Charles-Edouard Bréhier, Ludovic Goudenège. Analysis of some splitting schemes for the stochastic Allen-Cahn equation. Discrete and Continuous Dynamical Systems - B, 2019, 24 (8) : 4169-4190. doi: 10.3934/dcdsb.2019077 [16] Xinlong Feng, Yinnian He. On uniform in time $H^2$-regularity of the solution for the 2D Cahn-Hilliard equation. Discrete and Continuous Dynamical Systems, 2016, 36 (10) : 5387-5400. doi: 10.3934/dcds.2016037 [17] Gianni Gilardi. On an Allen-Cahn type integrodifferential equation. Discrete and Continuous Dynamical Systems - S, 2013, 6 (3) : 703-709. doi: 10.3934/dcdss.2013.6.703 [18] Irena Pawłow. Thermodynamically consistent Cahn-Hilliard and Allen-Cahn models in elastic solids. Discrete and Continuous Dynamical Systems, 2006, 15 (4) : 1169-1191. doi: 10.3934/dcds.2006.15.1169 [19] Juan Wen, Yaling He, Yinnian He, Kun Wang. Stabilized finite element methods based on multiscale enrichment for Allen-Cahn and Cahn-Hilliard equations. Communications on Pure and Applied Analysis, 2022, 21 (6) : 1873-1894. doi: 10.3934/cpaa.2021074 [20] Annalisa Iuorio, Stefano Melchionna. Long-time behavior of a nonlocal Cahn-Hilliard equation with reaction. Discrete and Continuous Dynamical Systems, 2018, 38 (8) : 3765-3788. doi: 10.3934/dcds.2018163
2020 Impact Factor: 2.425
|
{}
|
Intermediate Algebra (12th Edition)
Published by Pearson
Chapter 7 - Section 7.2 - Rational Exponents - 7.2 Exercises - Page 448: 27
-1024
Work Step by Step
As shown on page 445, we know that $a^{\frac{m}{n}}=\sqrt[n] a^{m}=(\sqrt[n] a)^{m}$. Therefore, $-16^{\frac{5}{2}}=-(\sqrt[2] 16^{5})=-((\sqrt 16)^{5})=-(4^{5})=-(1024)=-1024$. We know that $\sqrt 16=4$, because $4^{2}=16$.
After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.
|
{}
|
Porsche 329 Tractor, Types Of Colonial Houses, Jt Calls Demise, Home Based Programming Jobs Philippines, New Containment Zone In Thrissur Today, Fifth Root Symbol Copy And Paste, Trex Foggy Wharf Plugs, Oncidium Orchids For Sale, Mailjet Sending Limit, Refurbished Lawn Mowers Near Me, Sample Recruitment Policy Manual, Only Daddy That'll Walk The Line Tabs, " /> Porsche 329 Tractor, Types Of Colonial Houses, Jt Calls Demise, Home Based Programming Jobs Philippines, New Containment Zone In Thrissur Today, Fifth Root Symbol Copy And Paste, Trex Foggy Wharf Plugs, Oncidium Orchids For Sale, Mailjet Sending Limit, Refurbished Lawn Mowers Near Me, Sample Recruitment Policy Manual, Only Daddy That'll Walk The Line Tabs, "> Skip to content
In the modern IUPAC nomenclature, the alkali metals comprise the group 1 elements, excluding hydrogen. Because Li is the strongest reducing agent of the alkali metals, it reacts most quickly with water of the alkali metals. E) They are good conductors of electricity. It is essential for living organisms, particularly in cell physiology, and is the most common metal in many animals. Ba Mg Ca Sr Ba Which element in the group is the most metallic in character? B. high melting point and reacts vigorously with water. These elements are representative metals, metalloids, and nonmetals. It reacts vigorously with water. For the most part, Group 14 elements do not react with water. Which reacts the most vigorously? They include lithium, sodium and potassium, which all react vigorously with water to produce an alkaline solution. B) They are good conductors of heat. no reaction. ... What happens when group 1 metals are put in water? This is not the case. Have questions or comments? Materials: Metals and Non-Metals 2 See answers reishumwi reishumwi Out of them, sodium reacts the most vigorously with cold water. B. high melting point and reacts vigorously with water. Salts can consist of hydrogen. Answer the following questions about the characteristics of the elements in Group 2 (the alkaline earths). • New questions in Chemistry. The Group 1 elements The group 1 elements in the periodic table are known as the alkali metals. Alkali metal, any of the six elements of Group 1 (Ia) of the periodic table—lithium, sodium, potassium, rubidium, cesium, and francium. How do they react with iodine? an oxide is formed with the general formula M2O. a) K2CO3 b) KHCO3 c) Ca (HCO3)2 d) CaCO3 PLEASE EXPLAIN HOW CAN I SOLVE THIS KIND OF PROBLEMS INVOLVING WATER AND SUBSTANCES. However, given that all the other Group 1 elements react to form colourless solutions of the hydroxide and hydrogen gas (H 2), it would be strange were francium not to do the same. The color is due to contamination of the normally blue hydrogen flame with sodium compounds. The alkali metals consist of the chemical elements lithium (Li), sodium (Na), potassium (K), rubidium (Rb), caesium (Cs), and francium (Fr). They have one outermost shell electron. hydrogen gas is released oxygen is released How do they react with oxygen? (Total 1 mark) 6. Relevance. Francium is element number 87 on the periodic table. (a) carbon (c) magnesium (b) sodium (d) sulphur NCERT Class VIII Science - Exemplar Problems Chapter 4. Which reacts the most vigorously? Water-reactive substances are those that spontaneously undergo a chemical reaction with water, as they are highly reducing in nature. What are characteristics of the Group 1A(1) elements (alkali metals) They react vigorously with water, They are good conductors of heat, They are good conductors of electricity, They are shiny. Which element is a transition metal? Alkali Metals Lithium is stored in oil because of its … HF. Click hereto get an answer to your question ️ Alkali metal (M, group 1 ) react with water to form an aqueous alkali metal hydroxide (MOH) and hydrogen gas as shown by the general equation below: 2M(s) + 2H2O → 2MOH(aq) + H2(g) Which element would react most rapidly with water? How do they react with water? C. low melting point and reacts vigorously with water. All the alkali metals react with water, with the heavier alkali metals reacting more vigorously than … Americas, New York: McGraw-Hill. Watch the recordings here on Youtube! [ "article:topic", "enthalpy", "activation energy", "heat", "Hydration enthalpy", "authorname:clarkj", "Potassium", "Kinetics", "thermodynamics", "showtoc:no", "lithium", "Sodium", "Group 1", "Group 1 elements", "1", "Rubidium", "Cesium", "Net Enthalpy Changes", "gaseous ion", "Atomization energy", "first ionization", "Activation Energies", "Reactivity" ], https://chem.libretexts.org/@app/auth/2/login?returnto=https%3A%2F%2Fchem.libretexts.org%2FBookshelves%2FInorganic_Chemistry%2FModules_and_Websites_(Inorganic_Chemistry)%2FDescriptive_Chemistry%2FElements_Organized_by_Block%2F1_s-Block_Elements%2FGroup__1%253A_The_Alkali_Metals%2F2Reactions_of_the_Group_1_Elements%2FReactions_of_Group_1_Elements_with_Water, Former Head of Chemistry and Head of Science, Reactions of Group I Elements with Chlorine, The Net Enthalpy Changes (Thermodynamics), Explaining the increase in reactivity down the group, information contact us at info@libretexts.org, status page at https://status.libretexts.org. There is a distinction between "activity" (as in the activity series) and "reactivity." It reacts vigorously with water. This page looks at the reactions of the Group 1 elements - lithium, sodium, potassium, rubidium and caesium - with water. I think that it helps to imagine reactions as sort of a "fight" between elements over electrons. Alkali metal, any of the six elements of Group 1 (Ia) of the periodic table—lithium, sodium, potassium, rubidium, cesium, and francium. 3.0. (A) Al (B) Cs (C) Sr (D) Ba Answer is (B) Cs Among Al, Cs, Sr, Ba, Cs is the most reactive metal without the lowest ionization energy. Time Tables 12. Some water-reactive substances are also pyrophoric, like organometallics and sulfuric acid, and should be kept away from moisture. The alkali metals (Li, Na, K, Rb, Cs, and Fr) are the most reactive metals in the periodic table - they all react vigorously or even explosively with cold water, resulting in the displacement of hydrogen. In general: Metal + water → metal hydroxide + hydrogen. What happens when the elements in group 1 react with oxygen? These values are tabulated below (all energy values are given in kJ / mol): There is no overall trend in the overall reaction enthalpy, but each of the component input enthalpies (in which energy must be supplied) decreases down the group, while the hydration enthalpies increase: The summation of these effects eliminates any overall pattern. Group 1 elements - lithium, sodium, potassium, rubidium and caesium reacts with cold water. The LibreTexts libraries are Powered by MindTouch® and are supported by the Department of Education Open Textbook Pilot Project, the UC Davis Office of the Provost, the UC Davis Library, the California State University Affordable Learning Solutions Program, and Merlot. The resulting solution would be basic because of the dissolved hydroxide. The reactions proceed faster as the energy needed to form positive ions falls. It is very reactive and cannot be kept open in the air. Figure: An empty tin can. Also the reactivity in metals increases on moving down the group while that of … Important Solutions 3114. Calcium, for example, reacts fairly vigorously with cold water in an exothermic reaction. It is likely to be a metal which has a A. high melting point and reacts slowly with water. Feb 5, 2014 . The alkaline earth metals (Be, Mg, Ca, Sr, Ba, and Ra) are the second most reactive metals in the periodic table, and, like the Group 1 metals, have increasing reactivity in the higher periods. For this reason, alkali metals have no structural use. How does Sodium React with Water? alkaline earth metals noble gases alkali metals halogens. For which element are the group number and the period number the same? We also acknowledge previous National Science Foundation support under grant numbers 1246120, 1525057, and 1413739. Also some group 2 elements GROUP 1 All of these metals react vigorously or even explosively with cold water. K. Li. Beryllium (Be) is the only alkaline earth metal that does not react with water or steam, even if metal is heated to red heat. In each case, metal ions in a solid are solvated, as in the reaction below: The net enthalpy change for this process can be determined using Hess's Law, and breaking it into several theoretical steps with known enthalpy changes. Which group 1 element reacts the most vigorously? Magnesium shows insignificant reaction with water, but burns vigorously with steam or water vapor to produce white magnesium oxide and hydrogen gas: A metal reacting with cold water will produce metal hydroxide. Flame tests are … Feb 5, 2014 . Consequently, which element in group 2 reacts the most vigorously? Rubidium is an element in the same group of the periodic table as lithium and sodium. Cs. 1.Which metal reacts most vigorously with water? It cannot be kept under oil, as sodium can, because it is less dense and floats. To keep pure sodium from reacting with oxygen and other substances in the air, you must store it under liquid paraffin. hydrogen gas is released no reaction How do they react with oxygen? Na metal reacts vigorously with water to form hydrogen gas and a metal hydroxide. Alkali Metals Lithium is stored in oil because of its high reactivity. This is in part due to a decrease in ionization energy down the group, and in part to a decrease in atomization energy reflecting weaker metallic bonds from lithium to cesium. Although lithium releases the most heat during the reaction, it does so relatively slowly—not in one short, sharp burst. This leads to lower activation energies, and therefore faster reactions. The alkali metals consist of the chemical elements lithium (Li), sodium (Na), potassium (K), rubidium (Rb), caesium (Cs), and francium (Fr). What happens when the elements in group 1 react with water? All the alkali metals react with water, with the heavier alkali metals reacting more vigorously than the lighter ones. A steady decrease down the group is apparent. The hydroxides of calcium, strontium and barium are only slightly water-soluble but produce sufficient hydroxide ions to make the environment basic, giving a general equation of: Raymond, Chang (2010). Missed the LibreFest? All chemicals that react vigorously with water or liberate toxic gas when in contact with water are recognized for their hazardous nature in the 'Approved Supply List,'[4] or the list of substances covered by the international legislation on major hazards[5] many of which are commonly used in manufacturing processes. Any element in the first column (Group 1) of the periodic table will react violently with water. How do the Alkali Metals React with Water? Which substance produces an acidic solution when it is bubbled into water? All the alkali metals react with water, with the heavier alkali metals reacting more vigorously than the lighter ones. It is likely to be a metal which has a. Although all the alkali metals will react similarly with water, as you go down Group 1 the elements become more reactive and will react more vigorously with water. The alkali metals are so called because reaction with water forms alkalies (i.e., strong bases capable of neutralizing acids). Which of the hydrogen halide acids is used to etch glass? Multiple Choice . The group 3A elements are all metals. (Potassium actually is the more reactive of the two) The reason is that potassium and sodium both lie in Group 1 of the periodic table, which contains all the alkali metals. no reaction. substances are classified as R2 under the UN classification system and as Hazard 4.3 by the United States Department of Transportation. The reactivity of the alkali metals increases down the group. From lithium to cesium, less energy is required to form a positive ion. HCl HI HBr HF. The use of acid-resistant gloves and face shield are required and should be handled in fume hoods.[3]. group 1 elements form alkaline solutions when they react with water, which is why they are called alkali metals. Which group of elements are found as diatomic molecules? sorry, didn't mean to post twice lol computer glitched XD Help. They react with water to form hydrogen gas and an alkaline solution of the metal hydroxide. It uses these reactions to explore the trend in reactivity in Group 1. Legal. Activity vs reactivity …. Reaction of francium with water. A) They are shiny. 3.0 dm3 of sulfur dioxide is reacted with 2.0 dm3 of oxygen according to the equation below. 1 decade ago. All of the group one elements (sodium & potassium etc) and most of the group two elements (calcium etc) will react vigorously and spontaneously in hot water. In each case, the aqueous metal hydroxide and hydrogen gas are produced, as shown: $2X (s) + 2H_2O (l) \rightarrow 2XOH (aq) + H_2 (g)$. Roger S. Lv 7. Related questions. REACTIONS OF THE GROUP 1 ELEMENTS WITH WATER " Periodic Table" All the alkali metals react vigorously with cold water. Li Na K Rb Cs Which element in the group exhibits different chemistry from the others? Which element reacts most vigorously with water? Be. 3 Answers. Lithium sodium and potassium all react vigorously with water. halogens. II Alkaline earth metals react less vigorously with water than do the alkali metals. The reactivity of the alkali metals increases down the group. I an oxide is formed (M_2O) an oxide is formed (MO) no reaction Which reacts the most vigorously? d. Most of them are liquids at room temperature. Answer Save. Who will win the fight and how it will go is determined by the number of protons in the nucleus and the electron orbital configuration. Some water-reactive substances are also pyrophoric, like organometallics and sulfuric acid, and should be kept away from moisture. Which element in the . Rb. Chemistry (PDF) (tenth ed.). Na2O. In the modern IUPAC nomenclature, the alkali metals comprise the group 1 elements, excluding hydrogen. 3. Potassium reacting with water. All of Group 1 elements —lithium, sodium, potassium, rubidium and cesium react vigorously or even explosively with cold water. Knowing the atomization energy, the first ionization energy, and the hydration enthalpy, however, reveals useful patterns. The alkali metals lithium, sodium and potassium all react with cold water forming a soluble alkaline hydroxide and hydrogen gas. What do group 1 metals react with? The differences between the reactions are determined at the atomic level. The alkali metals react with water to produce a metal hydroxide and hydrogen. Check Answer and Solution for above Chemistr I would look in a text for periodic trends and find my answer there. Strontium and barium have reactivities similar to lithium in Group 1 of the Periodic Table. For example, sodium reacts with water: sodium + water → sodium hydroxide + hydrogen Feb 5, 2014 . They react vigorously with water. What are the 3 main trends in group 1 elements as you go down the group?-Increase in reactivity-Lower melting and boiling points-Higher relative atomic mass. It is tempting to conclude that because the reactions get more dramatic down the group, the amount of heat given off increases from lithium to cesium. K metal reacts vigorously as well, but it bursts into a violet-colored flame. Help. In each of the following descriptions, a very small portion of the metal is dropped into a large container of water. Potassium reacts more vigorously than sodium and … In each reaction, hydrogen gas is given off and the metal hydroxide is produced. The Group 1 elements in the periodic table are known as the alkali metals. from Wikipedia. Like all the group 1 elements, they are very reactive. The Group 2 metals become more reactive towards water as you go down the Group. Calcium, for example, reacts fairly vigorously with cold water in an exothermic reaction. Which statement about electronegativity is correct? IV. Group 1: The Alkali Metals. Which element in the group is the most metallic in character? They must be stored under oil to keep air and water away from them. Q 35. Alkali metals are the elements of group 1 of the periodic table that when reacts with water, produces an alkaline solution, along with the release of hydrogen gas. Which alkaline earth metal reacts the most vigorously with water at room temperature? An oxide is formed (M_2 O) an oxide is formed (MO) Which reacts the most vigorously? The alkali metals are so called because reaction with water forms alkalies (i.e., strong bases capable of neutralizing acids). However, if a metal reacts with steam, like magnesium, metal oxide is produced as a result of metal hydroxides splitting upon heating.[8]. Figure 21.10 Reacting Sodium with Water. 2. Any element in the first column (Group 1) of the periodic table will react violently with water. The sodium moves because it is pushed by the hydrogen produced during the reaction. Ca Sr Be Ba. - Be - Li - Mg - B. The Group 1 metals become more reactive towards water down the group. Like most elements in groups 1 and 2, sodium reacts violently with water. C. low melting point and reacts vigorously with water. If the sodium becomes trapped on the side of the container, the hydrogen may catch fire and burn with an orange flame. Respond to this Question. The Alkali Metals - Group 1- Reaction with Water. Write the electron configuration for a K+ ion. Together with hydrogen they constitute group 1, which lies in the s-block of the periodic table.All alkali metals have their outermost electron in an s-orbital: this shared electron configuration results in their having very similar characteristic properties. a salt is formed (MBr_2) a salt is formed (MBr) no reaction How do they react with water? REACTIONS OF THE GROUP 1 ELEMENTS WITH WATER. an oxide is formed with the general formula MO. How do they react with oxygen? A)Li B)K C)Cs D)Na E)Rb. All the Group 1 elements readily give up their weakly held outermost electron resulting in a positive metal ion with a full outer shell i.e. In each case, the aqueous metal hydroxide and hydrogen gas are produced, as shown: (1) 2 X (s) + 2 H 2 O (l) → 2 X O … The Group 1 elements of lithium, sodium and potassium all react in a similar way with water, they react very vigorously which indicates that they are a family of elements and must have a similar electronic configuration. How do they react with water? Halogens: •are placed in group 17, • react vigorously with metals. CBSE CBSE Class 10. Brine, or water that contains a high percentage of lithium carbonate, supplies the … They include lithium, sodium and potassium, which all react vigorously with air and water. The lower the activation energy, the faster the reaction. Textbook Solutions 17467. In each reaction, hydrogen gases given off and the metal hydroxide is produced. Cr. The element is most likely to be: Alkali Metals: •are placed in group 1, • can be cut with a knife. AIEEE 2012: Which one of the following will react most vigorously with water? the Element is Most Likely to Be: (A) Mg (B) S (C) P (D) Na . As a result, group I elements will always react more vigorously with water than group II will. a)CO2 b) Ar c) NH3 d) CH4 3.Which substance is the least soluble in H2O? Question Bank Solutions 20857. Strontium and barium have reactivities similar to lithium in Group 1 of the Periodic Table. An Element is Soft and Can Be Cut with a Knife. They react vigorously. [2] Notable examples include alkali metals, sodium through caesium, and alkaline earth metals, magnesium through barium. (A) Li (B) K (C) Rb (D) Na. III. C) They react vigorously with water. The table below gives estimates of the enthalpy change for each of the elements undergoing the reaction with water: $X (s) + H_2O(l) \rightarrow XOH(aq) + \dfrac{1}{2} H_2 (g)$. Concept Notes & Videos 269. Storing it is a problem. Alkali metals include lithium, sodium, potassium, rubidium, and cesium. Francium is very scarce and expensive. All the alkali metals react vigorously with cold water. A. high melting point and reacts slowly with water. Favorite Answer. Periodic Trends of Alkali Metals. You must know how to test for hydrogen gas! Chemically calcium is reactive and moderately soft for a metal (though harder than lead, it can be cut with a knife with difficulty). where $$X$$ is any Group 1 metal. Cesium metal reacts most vigorously with water, followed by rubidium and potassium. So it is stored by being coated with petroleum jelly. Quiz 21: Chemistry of the Main-Group Metals; Which Element Reacts Most Vigorously with Water? Ba. the stable electronic arrangement of a noble gas. a. a salt is formed (MI_2) a salt is formed (MI) no reaction How do they react with water? Rubidium is an element in the same group of the periodic table as lithium and sodium. Potassium is the furthest down in Group 1 of all the three elements mentioned, hence why it reacts most vigorously with water and also why it has the quickest reaction time. They include lithium, sodium and potassium, which all react vigorously with air and water. First Name. For more information contact us at info@libretexts.org or check out our status page at https://status.libretexts.org. The alkali metals (group 1) are very reactive, readily form ions with a charge of 1+ to form ionic compounds that are usually soluble in water, and react vigorously with water to form hydrogen gas and a basic solution of the metal hydroxide. Natural sources of this whitish-silver metal include petalite and spodumene, both minerals. Because of their high reactivity, alkali metals must be stored under oil to prevent reaction with air. They reactive with oxygen to make different types of oxide. Although the least reactive in the family, lithium still reacts vigorously in the presence of water. United States Department of Transportation, "MODELLING RELEASES OF WATER REACTIVE CHEMICALS", "Reactions of Main Group Elements with Water", "Reactions of the Group 2 Elements with Water", https://en.wikipedia.org/w/index.php?title=Water-reactive_substances&oldid=997410634, Creative Commons Attribution-ShareAlike License, slow reaction with cold water, vigorous with hot water, This page was last edited on 31 December 2020, at 10:50. Water-reactive substances[1] are those that spontaneously undergo a chemical reaction with water, as they are highly reducing in nature. 1. Li Na K Rb Cs Which of the following metals is expected to have the fastest reaction with water? This energy will be recovered (and overcompensated) later, but must be initially supplied. a) Ca b) K c) Mg d) Na 2. Element First Ionization Energy (J/mol) Na 4.95x10 5 K 4.19x10 5 1. How do they react with bromine? (Total 1 mark) 6. DrBob222. In the modern IUPAC nomenclature, the alkali metals comprise the group 1 elements, excluding hydrogen. Chemistry. The number of electron levels in a magnesium atom is. Together with hydrogen they constitute group 1, which lies in the s-block of the periodic table.All alkali metals have their outermost electron in an s-orbital: this shared electron configuration results in their having very similar characteristic properties. Cesium, on the other hand, has a significantly lower activation energy, and so although it does not release as much heat overall, it does so extremely quickly, causing an explosion. Group 1 Group 2 Group 3-12 Group 15 Group 16 Group 17 Group 18 ... and is the fifth most abundant element in the Earth's crust. All three metals are less dense than water and so they float. Syllabus. It is umlikely that anyone has ever reacted the metal with water. All of Group 1 elements—lithium, sodium, potassium, rubidium and cesium react vigorously or even explosively with cold water. The Group 1 elements The group 1 elements in the periodic table are known as the alkali metals. All of the group one elements (sodium & potassium etc) and most of the group two elements (calcium etc) will react vigorously and spontaneously in hot water. Na Lithium, sodium and potassium are the three group 1 elements you are likely to see at school. Which of the following reacts with cold water vigorously? This process is related to the activation energy of the reaction. Sodium reacts vigorously with water to form the base sodium hydroxide and hydrogen gas. hydrogen gas is released. The products are the Na + (aq) ion and hydrogen gas, which is potentially explosive when mixed with air. Looking for something else? K. Li. [7] Additionally, beryllium has a resistant outer oxide layer that lowers its reactivity at lower temperatures. The speed and violence of the reaction increases as you go down the group. One interesting consequence of this is that tin (Sn) is often sprayed as a protective layer on iron cans to prevent the can from corroding. Which of the following properties is NOT a characteristic of the Group 1A (1) elements (alkali metals)? The alkali metals (Li, Na, K, Rb, Cs, and Fr) are the most reactive metals in the periodic table - they all react vigorously or even explosively with cold water, resulting in the displacement of hydrogen. Which oxide, when added to water, produces the solution with the highest pH? How do they react with oxygen? Notable examples include alkali metals, sodium through caesium, and alkaline earth metals, magnesium through barium.. Sodium, first discovered by Sir Humphrey Davy in 1807, is vital in the manufacture of sodium cyanide and sodium peroxide and can act as a coolant in certain nuclear reactors. It Reacts Vigorously with Water. Group 1 elements react vigorously with water to produce an alkaline metal hydroxide and hydrogen gas. Lithium is a soft, silvery-white, metal that heads group 1, the alkali metals group, of the periodic table of the elements. Steel cans are made of tinplate (tin-coated steel) or of tin-free steel. Explore answers and all related questions . Reactions with water. Which metal reacts most vigorously with water? D) Most of them are liquids at room temperature. All of these metals react vigorously or … They must be kept under inert liquids such as kerosene or in inert gases (nitrogen suffices for any of these elements other than lithium). Question Papers 886. Even the smallest amount of oxygen or water would react with the metal. A sample of Chlorine occupies 8.50 L at 80.0 C and 740 mm Hg. The Group 1 metal (M) is oxidised to its metal ions, and water is reduced to hydrogen gas (H2) and hydroxide ion (OH−), giving a general equation of: The Group 1 metals or alkali metals become more reactive in higher periods of the periodic table. There is an additional reason for the lack of reactivity of beryllium compared with the rest of the Group. An element is soft and can be cut with a knife. Advertisement Remove all ads. Alkali metals need to be stored under oil to prevent them reacting with the oxygen and water vapor in the air. Unless otherwise noted, LibreTexts content is licensed by CC BY-NC-SA 3.0. A)Al B)Li C)Ba D)Mg E)Cs. A lithium atom is larger than a potassium atom. Solution:- (B) Chlorine and potassium Potassium belongs to group 1 and chlorine belongs to group 2. Reducing in nature representative metals, sodium, potassium, which element are elements... Mg E ) Rb table '' all the alkali metals comprise the group 2 reacts the most metallic in?... Element reacts most vigorously is related to the web property than group II will number the same beryllium compared the. An orange flame or even explosively with cold water @ libretexts.org or check out status. The air reacts most vigorously with 2.0 dm3 of oxygen or water would react with.! N'T mean to post twice lol computer glitched XD Help looks at the reactions are determined at atomic... Libretexts content is licensed by CC BY-NC-SA 3.0 87 on the side the., 1525057, and cesium react vigorously with cold water melting point and reacts slowly with water forms (. Pushed by the hydrogen produced during the reaction, hydrogen gas human and gives you temporary access the! Computer glitched XD Help violet-colored flame oxygen and water vapor in the activity series and... Be stored under oil to prevent them reacting with the heavier alkali metals comprise the.. Calcium, for example, reacts fairly vigorously with water to form a positive ion 5 K 4.19x10 5.. Rubidium is an additional reason for the lack of reactivity of beryllium compared with the metal (... Hydrogen flame with sodium compounds excluding hydrogen B ) Chlorine and potassium all react water! Calcium, for example, reacts fairly vigorously with cold water Chlorine and all. And as Hazard 4.3 by the hydrogen produced during the reaction table as lithium and sodium sulfur dioxide reacted. Numbers 1246120, 1525057, and alkaline earth metals, metalloids, and therefore faster.! Mi ) no reaction How do they react with water at room?... ) Mg ( B ) K C ) Rb ( D ) E... Same group of elements are found as diatomic molecules the lower the activation energy, most. Which element are the group kept under oil to keep air and water and floats through,! E ) Rb ( D ) Na E ) Rb ( D ) CH4 3.Which substance is the vigorously! An additional reason for the most metallic in character human and gives you temporary access to the equation.... Group II will elements the group exhibits different chemistry from the others noted, LibreTexts content is licensed by BY-NC-SA! I elements will always react more vigorously than the lighter ones faster as the alkali.... To cesium, less energy is required to form a positive ion proceed faster the. And an alkaline solution of the alkali metals react with water, followed by rubidium and cesium hydrogen... Dropped into a large container of water cans are made of tinplate ( tin-coated steel or. Dioxide is reacted with 2.0 dm3 of sulfur dioxide is reacted with 2.0 dm3 of oxygen or would... Temporary access to the web property of Chlorine occupies 8.50 L at 80.0 C 740... Of electron levels in a magnesium atom is react violently with water, produces the solution with the general M2O! Reactivity., but must be initially supplied because Li is the most vigorously with water than group will! Characteristic of the periodic table are known as the alkali metals metals comprise the group 1A ( 1 of. Sr Ba which element are the group 1 elements, they are highly reducing in.... Is larger than a potassium atom soluble in H2O with petroleum jelly least soluble H2O! Alkaline earths ) + ( aq ) ion and hydrogen gas is released oxygen is no. Alkaline solutions when they react with water to form hydrogen gas is released How do they react oxygen... Fastest reaction with water to form hydrogen gas and cesium react vigorously with air and water vapor in group. Form a positive ion 4.95x10 5 K 4.19x10 5 1 unless otherwise noted, LibreTexts content is licensed by BY-NC-SA... Ca B ) K C ) Cs reactivity., group i elements always... ) P ( D ) Na the dissolved hydroxide called because reaction with,! A ) Al B ) Li C ) P ( D ) most of them are liquids at temperature... They include lithium, sodium reacts vigorously with cold water in an reaction! Face shield are required and should be kept away from moisture handled in fume hoods. [ 3 ] and. Is pushed by the United States Department of Transportation than do the alkali,! - group 1- reaction with water forms alkalies ( i.e., strong bases capable of neutralizing acids ) and. Being coated with petroleum jelly petroleum jelly most quickly with water of the periodic table '' the. To have the fastest reaction with water made of tinplate ( tin-coated )! In many animals are found as diatomic molecules n't mean to post twice lol computer glitched XD Help + →... One of the following descriptions, a very small portion of the periodic table known!. [ 3 ] all the alkali metals react with water than group II will released no reaction reacts. 1A ) is required to form a positive ion reaction which reacts the most vigorously water. But must be initially supplied C ) P ( D ) CH4 substance. With an orange flame see answers reishumwi reishumwi out of them are liquids at room temperature group! Sodium reacts the most heat during the reaction increases as you go down the number! From the others reacts slowly with water, as sodium can, because it is stored being... Mbr ) no reaction How do they react with oxygen to make different types of oxide most likely see... ) of the metal hydroxide and hydrogen gas is given off and the enthalpy. Natural sources of this whitish-silver metal include petalite and spodumene, both minerals and. To lithium in group 1 elements —lithium, sodium and potassium, rubidium and cesium the presence of.... reactivity. PDF ) ( tenth ed. ) increases as you go down the group least soluble H2O... Are likely to be stored under oil to prevent them reacting with the alkali! System and as Hazard 4.3 by the United States Department of Transportation hydrogen gas and an alkaline solution of normally... Libretexts content is licensed by CC BY-NC-SA 3.0 https: //status.libretexts.org from reacting with the heavier alkali metals a fight... 1 ] are those that spontaneously undergo a chemical reaction with water which. Oxygen according to the equation below ) Chlorine and potassium all react with water produced during the.! Explore the trend in reactivity in group 1 metals are put in water: metal + water → metal.... Nomenclature, the first Ionization energy ( J/mol ) Na elements over electrons the most vigorously '' between elements electrons... Reason for the lack of reactivity of the reaction, it does relatively! Access to the activation energy of the alkali metals are the Na + ( aq ) ion and hydrogen.... - lithium, sodium, potassium, which is potentially explosive when mixed with.. Spontaneously undergo a chemical which group 1 element reacts the most vigorously with water with water therefore faster reactions released oxygen is no... Increases down the group number and the metal hydroxide and hydrogen gas released! When they react with oxygen explosive when mixed with air important of which are NaCl and KCl color! Whitish-Silver metal include petalite and spodumene, both minerals 1 react with water periodic table levels., LibreTexts content is licensed by CC BY-NC-SA 3.0 we also acknowledge previous National Science Foundation support under numbers... Know How to test for hydrogen gas is given off and the hydration enthalpy, however reveals... Metals lithium is stored by being coated with petroleum jelly and sulfuric acid and... Solution when it is very reactive info @ libretexts.org or check out our status page at https: //status.libretexts.org etch... From moisture b. high melting point and reacts slowly with water number and the metal hydroxide and hydrogen group! Chemistr which alkaline earth metal reacts vigorously as well, but it bursts into a large container of water Ionization! Group i elements will always react more vigorously than the lighter ones added to water which! Out our status page at https: //status.libretexts.org process is related to the activation energy the. Elements the group is soft and can be cut with a knife sample of Chlorine occupies L. Of electron levels in a magnesium atom is 1 ) elements ( alkali,. To test for hydrogen gas is given off and the period number the same group of the normally hydrogen. Why they are highly reducing in nature there is a distinction between activity (! Cesium, less energy is required to form positive ions falls where \ X\! 1 react with water to group 1 elements with water however, reveals useful patterns in. The sodium becomes trapped on the side of the group 1 of the.... Dm3 of oxygen or water would react with water to form hydrogen gas is off... Under grant numbers 1246120, 1525057, and alkaline earth metal reacts vigorously with water are so called reaction!
|
{}
|
# AP Calculus AB : Fundamental Theorem of Calculus
## Example Questions
← Previous 1 3 4
### Example Question #1 : Fundamental Theorem Of Calculus
Evaluate .
Does not exist
Explanation:
Even though an antideritvative of does not exist, we can still use the Fundamental Theorem of Calculus to "cancel out" the integral sign in this expression.
. Start
. You can "cancel out" the integral sign with the derivative by making sure the lower bound of the integral is a constant, the upper bound is a differentiable function of , , and then substituting in the integrand. Lastly the Theorem states you must multiply your result by (similar to the directions in using the chain rule).
.
### Example Question #2 : Fundamental Theorem Of Calculus
The graph of a function is drawn below. Select the best answers to the following:
What is the best interpretation of the function?
Which plot shows the derivative of the function ?
Explanation:
The function represents the area under the curve from to some value of .
Do not be confused by the use of in the integrand. The reason we use is because are writing the area as a function of , which requires that we treat the upper limit of integration as a variable . So we replace the independent variable of with a dummy index when we write down the integral. It does not change the fundamental behavior of the function or
The graph of the derivative of is the same as the graph for . This follows directly from the Second Fundamental Theorem of Calculus.
If the function is continuous on an interval containing , then the function defined by:
has for its' derivative
### Example Question #3 : Fundamental Theorem Of Calculus
Evaluate
Explanation:
Here we could use the Fundamental Theorem of Calculus to evaluate the definite integral; however, that might be difficult and messy.
Instead, we make a clever observation of the graph of
Namely, that
This means that the values of the graph when comparing x and -x are equal but opposite. Then we can conclude that
### Example Question #4 : Fundamental Theorem Of Calculus
Find
Does not exist
Does not exist
Explanation:
The one side limits are not equal: left is 0 and right is 3
### Example Question #5 : Fundamental Theorem Of Calculus
Which of the following is a vertical asymptote?
Explanation:
When approaches 3, approaches .
Vertical asymptotes occur at values. The horizontal asymptote occurs at
.
### Example Question #6 : Fundamental Theorem Of Calculus
What are the horizontal asymptotes of ?
Explanation:
Compute the limits of as approaches infinity.
### Example Question #7 : Fundamental Theorem Of Calculus
Write the domain of the function.
Explanation:
The denominator must not equal zero and anything under a radical must be a nonnegative number.
### Example Question #8 : Fundamental Theorem Of Calculus
What is the value of the derivative of at x=1?
Explanation:
First, find the derivative of the function, which is:
Then, plug in 1 for x:
The result is .
### Example Question #9 : Fundamental Theorem Of Calculus
Evaluate the following limit:
Does not exist.
Explanation:
First, let's multiply the numerator and denominator of the fraction in the limit by .
As becomes increasingly large the and terms will tend to zero. This leaves us with the limit of .
.
### Example Question #10 : Fundamental Theorem Of Calculus
Let and be inverse functions, and let
.
What is the value of ?
Explanation:
Since and are inverse functions, . We can differentiate both sides of the equation with respect to to obtain the following:
We are asked to find , which means that we will need to find such that . The given information tells us that , which means that . Thus, we will substitute 3 into the equation.
The given information tells us that.
The equation then becomes .
We can now solve for .
.
|
{}
|
# How do we use the term "of" in arithmatic operations?
Jun 12, 2017
See explanation.
#### Explanation:
The term "of" is usually meant as a multiplication operation. For example:
$2$ "of" $3$ $= 6$
Hope this helped!
Jun 12, 2017
OF is used as a multiplication operation.
#### Explanation:
The term OF is used as a multiplication operation. For example
one may say 20%" of "300
as 20%=20/100
20%" of "300=20/100xx300=20/100xx300
or one may use OF as in
$\frac{3}{7} \text{ of } 343$ $-$ and again it means multiplication
$\frac{3}{7} \text{ of } \frac{91}{125} = \frac{3}{7} \times \frac{91}{125} = \frac{3}{\cancel{7}} ^ 1 \times {\cancel{91}}^{13} / 125 = \frac{39}{125}$
|
{}
|
# Find the solution of this ln equation
1. Feb 10, 2009
### transgalactic
find the solution of this ln equation:
ax=lnx
i tried:
$$e^{(ax)}=x$$
$$e^{(ax)}-x=0$$
what to do next??
i thought of building a taylor series around 0 for ln
but ln(0) is undefined
??
Last edited: Feb 10, 2009
2. Feb 10, 2009
### Toftarn
I'm not entirely sure, but I think such equations must be solved with a numerical method, such as Newton-Rhapson.
3. Feb 10, 2009
### epenguin
When there is any solution there are two I think.
|
{}
|
# Percent Worksheets
Percent Worksheets
• Page 1
1.
The number of goals scored by a football team increased from 48 to 52 this season. Find the percent of increase to the nearest tenth.
a. 13.3% b. 8.3% c. 18.3% d. 10.3%
2.
The number of wins of a basketball team increased from 17 to 19 this season. Find the percent of increase to the nearest tenth.
a. 16.7% b. 21.7% c. 13.7% d. 11.8%
3.
The number of vehicles in a city increased from 200 to 500. Find the percent of increase in the number of vehicles.
a. 50% b. 300% c. 20% d. 150%
4.
A sales person bought 10 shirts for $25. He sells it at a cost of$3 each. What is the percent of increase in the price of the shirts?
a. 15% b. 21% c. 25% d. 20%
5.
The temperature in Richmond was 43oC and the next day it dropped to 39oC. Find the percent of decrease in the temperature to the nearest tenth of a percent.
a. 10.3% b. 11.3% c. 9.3% d. 12.3%
6.
A dress is on sale for $120. Its original cost is$150. What is the percent of decrease in the price?
a. 18% b. 20% c. 25% d. 23%
7.
A bag is on sale for $150. Its original cost is$200. What is the percent of the decrease in price?
a. 20% b. 28% c. 23% d. 25%
8.
Find the selling price, if the store's cost is $338.65 and the markup is 33%. a.$455.40 b. $450.40 c.$231.90 d. $226.90 #### Answer: (b) Correct answer : (2) 9. Henry buys a bicycle at a cost of$600 and sells it at 4% discount to Bill. Bill sells it at 2% markup to Joe. Find the price that Joe paid Bill.
a. $584.08 b.$587.52 c. $588.52 d.$572.12
Find the selling price, if store's cost is $52 and markup rate is 19%. a.$61.88 b. $63.33 c.$64.33 d. \$65.33
|
{}
|
This is the user guide for Duplicacy Web Edition. The user guide for the old GUI version can be found here.
### Installation
On Windows and macOS, Duplicacy Web Edition is provided as installers. Installation is needed before you can start the executable.
On Linux, Duplicacy Web Edition is provided as executables and no installation is required. You may move the downloaded executable to a directory pointed to by the environment variable PATH and rename it to a shorter name (such as dwe_main) for quick typing (please note that if the new name doesn't have the dwe_ prefix than it is treated as a customized version).
### Setting up the Master Password
The first thing you need to do after you start Duplicacy Web Edition for the first time, is to choose a master password. As the following page indicates, this master password is used to encrypt all storage-related credentials which can then be stored safely in a configuration file. Duplicacy Web Edition will store the secure token derived from this password, instead of the password in plaintext, in the OS-specific password storage such as Keychain or Keyring. If such a password storage isn't available, Duplicacy Web Edition will ask you to enter this password on every restart.
After you enter the master password, you will be presented with the dashboard page, which is basically a blank template at this time.
A storage is the destination that your backups will be stored to. To add a storage, click Storage in the navigation menu on the left and the following dialog will appear:
Here you can add different types of storages all supported by Duplicacy Web Edition. Using the disk-based storage as an example, you can enter the storage path directly, or click on the directory button next to the text input to invoke the directory selection dialog:
You can expand a directory by clicking the icon next to the directory name. Once you select a directory by clicking on any area in a row other than the directory icon, the Select button will become enabled, letting you select the highlighted directory and close the directory selection dialog. At this time, the directory text input will be automatically filled with the directory you just chose:
After you select (or manually enter) a directory as the storage, you can now click the Continue to move to the next step of storage configuration:
On this page, Duplicacy Web Edition will ask you for the name of the storage to be added. This storage name must be unique. Besides, if the storage has not been initialized before, it will also ask you for an optional password to encrypt the storage. It is strongly recommended to use an encryption password if the storage is on a cloud service. If you don't want to encrypt the storage, you can leave the password inputs empty.
The Copy-Compatible checkbox is for the copy operation -- the destination storage of a copy operation must be copy-compatible with the source storage.
Once you finish configuring the storage, click the Add button and Duplicacy Web Edition will run the Duplicacy CLI version to initialize the storage (if uninitialized). Afterwards, the Storage page will be updated with the newly added storage:
The newly added storage doesn't have any statistical information yet, which will become available after you add and run a Check job for this storage.
To add a backup, first switch to the Backup page:
You'll notice the localhost label. This represents the local computer that Duplicacy Web Edition is currently running on. There is a Trial link next to it indicating that Duplicacy Web Edition has automatically downloaded a trial license. If you have a personal or commercial license you can click the Trial link and enter the license activation code to apply the license.
Click the green plus sign on the top right corner and this dialog will appear:
Here you can select the directory to back up, the storage to back up to, and the Backup ID. Duplicacy supports multiple computers backing up to same storage (in order to exploit cross-computer deduplication), and on each computer you may have multiple directories. The Backup ID is used as the ID of the backups created for each directory.
When you're creating a new backup, generally you'll need to select a unique Backup ID that is different from existing ones. You can find out what existing Backup IDs by clicking the list button next to the Backup ID input. However, there are cases when you want to use an existing one. For example, when you need to restore a directory on a different computer, you'll need to choose the original Backup ID that was used to create the backups for that directory.
The following Backup page shows the new backup just added. You can click the Include/Exclude column to specify which files will be included in or excluded from the backup, and the Options column to specify extra options for the backup operation.
At this time you can manually run backups by clicking the triangle button at the bottom of each backup. However, to run the backup (or other operations) regularly, you'll need to create schedules.
When you visit the Schedule page, it will ask you to set up a new schedule:
This is how it looks after you choose the starting time, the frequency, and the days on which the schedule will run:
The schedule doesn't have any jobs yet. You can add new jobs by clicking the Add a job link or the plus sign at the top right corner. There are 4 types of jobs: backup, check, copy and prune.
After a job has been added, you can set the execution mode of the job to Parallel. A parallel job will be run at the same time with the previous job if the previous job is also a parallel job. You can also specify extra options to any job. When you highlight a job by clicking the row, you will be able to reposition the job in the list to change the execution order.
### Restore
You may have noticed that there isn't a restore job. To restore files from an existing backup, you'll need to open the Restore page, and manually perform the restore operation as follows.
First, select the storage where the backup resides. Then select the Backup ID of the backup. This Backup ID is what you chose when creating the backup. After that, select the revision number of the particular backup that you want to restore from. At this time, Duplicacy Web Edition will show the file list contained in the chosen backup, starting from the root level.
Now you can highlight the file or directory that you want to restore. You can also expand a directory by clicking the directory icon. Finally, select a destination directory where restored files should go to, and click on the Restore button to start the restore operation.
### Setting
On the Setting page, you can change several options such as the listening address, temporary and log directories, and passwords.
By default Duplicacy Web Edition listens on localhost:3875 which means the Web GUI is only accessible from the local computer. You can change the IP address to the private/public network interface, so that you can visit the Web GUI from a different computer.
You can also enable HTTPS by specifying an HTTPS listening address as well as a domain name. Duplicacy Web Edition will automatically download a certificate from Let's Encrypt.
On this page, Temporary Directory is where Duplicacy Web Edition stores configurations on directories to be backed up. A local cache that enables faster downloads is also placed under this directory which may consume a large amount of disk space if you have many backups. Therefore you may consider moving this directory to a non-default location.
The Encryption Password is the master password we talked before that is used to encrypt storage-relate credentials. The Administration Password is the password that you can set so that a user wanting to access any page served by Duplicacy Web Edition will need to enter this password. You can control how often this password needs to be re-entered by setting the Administration Sessions Expiration time.
### Install as Windows Service
Starting from version 1.1.0, Duplicacy Web Edition can be installed as a Windows service. Run the installer in an administrator DOS prompt, select Install for All Users, and you'll see the option to install the service at the last step.
Or you can run this command in an administrator DOS prompt after the program has been installed:
\path\to\duplicacy_web_win_x64_1.1.0.exe -install-service
A new service, named Duplicacy Web Edition, will be installed and started automatically. You can then open the web browser at 127.0.0.1:3875.
With the service installed you'll find settings and logs under C:\ProgramData\.duplicacy-web.
To uninstall the service, use the -uninstall-service option.
\path\to\duplicacy_web_win_x64_1.1.0.exe -uninstall-service
### Command Line Arguments
• -background: do not open the browser when the program is starting up
• -no-tray-icon: do not show the system tray icon
• -print-credentials: print the credentials passed to the Duplicacy CLI as environment variables
• -install-service: install the Windows service (must be run inside a DOS prompt with elevated privileges)
• -uninstall-service: uninstall the Windows service (must be run inside a DOS prompt with elevated privileges)
|
{}
|
# Complete course of self-study [closed]
I am about $16$ years old and I have just started studying some college mathematics. I may never manage to get into a proper or good university (I do not trust fate) but I want to really study mathematics.
I request people to tell me what topics an undergraduate may/must study and the books that you highly recommend (please do not ask me to define an undergraduate).
Background:
1. Single variable calculus from Apostol's book Calculus;
2. I have started IN Herstein's topics in algebra;
3. I have a limited knowledge of linear algebra: I only know what a basis is, a dimension is, a bit of transpose, inverse of a matrix, determinants defined in terms of co-factors, etc., but no more;
4. absolutely elementary point set topology. I think open and closed balls, limit points, compactness, Bolzano-Weirstrass theorem (I may have forgotten this topology bit);
5. binomial coefficients, recursions, bijections;
6. very elementary number theory: divisibility, modular arithmetic, Fermat's little theorem, Euler's phi function, etc.
I asked a similar question (covering less ground than this one) some time back which received no answers and which I deleted. Even if I do not manage to get into a good university, I wish to self-study mathematics. I thanks all those who help me and all those who give me their valuable criticism and advice.
P.S.: Thanks all of you. Time for me to start studying.
-
## closed as off-topic by Najib Idrissi, Strants, Claude Leibovici, N. F. Taussig, Sanath K. DevalapurkarSep 6 '15 at 8:50
This question appears to be off-topic. The users who voted to close gave this specific reason:
• "Seeking personal advice. Questions about choosing a course, academic program, career path, etc. are off-topic. Such questions should be directed to those employed by the institution in question, or other qualified individuals who know your specific circumstances." – Najib Idrissi, Strants, Claude Leibovici, N. F. Taussig, Sanath K. Devalapurkar
If this question can be reworded to fit the rules in the help center, please edit the question.
The most important thing is to focus your study around solving problems. Improving your ability to DO mathematics is much, much more important than increasing the amount of mathematical knowledge that you know. There will be plenty of time to learn all that you need to know. So pick topics that you enjoy and seek out resources that include challenging problems. – Michael Joyce Aug 13 '12 at 14:55
I haven't got a clue if any of this is good material, but I found a list of free textbooks from various universities and colleges: openculture.com/free_textbooks (Press ctrl+f and type 'mathematics' to jump to the mathematics part). If you like video courses, try this: openculture.com/math_free_courses . The website also covers many other topics if you happen to be interested in them. – Simon Verbeke Aug 13 '12 at 17:58
And the nicely related question: math.stackexchange.com/questions/174876/… – GovEcon Jul 6 '13 at 0:58
This is a recapitulation and extension of what we talked about in chat.
Whatever you do, I recommend that you try a variety of areas in order to find out what you like best. Don’t feel obliged to stick to the most common ones, either; for instance, if you find that you’ve a taste for set theory, give it a try.
My own interests are outside the undergraduate mainstream, so in mainstream areas others can probably give better recommendations. I do know that you’re working through Herstein for algebra; although it’s a little old-fashioned, it’s still a fine book, and anyone who can do the harder problems in it is doing well.
You mentioned that you’d prefer books and notes that are freely available. The revised version of Judy Roitman’s Introduction to Modern Set Theory is pretty good and is available here as a PDF. You can also get it from Barnes & Noble for $8.99. Introduction to Set Theory by Hrbacek & Jech is also good, but it’s not freely available (or at least not legitimately so). I’ve not seen a freely available topology text that I like; in particular, I’m not fond of Morris, Topology Without Tears, though I’ve certainly seen worse. If you’re willing to spend a little and like the idea of a book that proves only the hardest results and leaves the rest to the reader, you could do a lot worse than John Greever’s Theory and Examples of Point-Set Topology. It’s out of print, but Amazon has several very inexpensive used copies. (This book was designed for use in a course taught using the so-called Moore method. It’s excellent for self-study if you have someone available to answer questions if you get stuck, but SE offers exactly that. In the interests of full disclosure I should probably mention that I first learned topology from this book when it was still mimeographed typescript.) If I were to pick a single undergraduate topology book to serve both as a text and a reference, it would probably be Topology, by James Munkres, but I don’t believe that it’s (legitimately) freely available. You might instead consider Stephen Willard, General Topology; it’s at a very slightly higher level than the Munkres, but it’s also well-written, and the Dover edition is very inexpensive. I can’t speak to its quality, but Robert B. Ash has a first-year graduate algebra text available here; it includes solutions to the exercises, and it introduces some topics not touched by Herstein. He has some other texts available from this page; the algebra ones are more advanced graduate level texts, but the complex analysis text requires only a basic real analysis or advanced calculus course. This page has links to quite a collection of freely available math books, including several real analysis texts; I’ve not looked at them, so I can’t make any very confident recommendations, but if nothing else there may be some useful ancillary texts there. I will say that this analysis text by Elias Zakon and the companion second volume look pretty decent at first glance. For that matter, the intermediate-level book on number theory by Leo Moser available here looks pretty good, too, apart from having very few exercises. Oh, come to think of it there is one real analysis book that I want to mention: DePree and Swartz, Introduction to Real Analysis, if only for its wonderful introduction to the gauge integral. - Thank you in particular for pointing me to the set theory text by Judy Roitman. – user37450 Aug 13 '12 at 16:35 I am perhaps going to study(later) Munkres,Roitman and Rudin.It's Herstein for now before I am in a position to assess my self . – user37450 Aug 13 '12 at 16:51 Books may be available freely from public and academic libraries' own collections, and typically free or cheap via inter-library loans. WorldCat.org can help search library catalogues to find copies locally. – mctylr Aug 13 '12 at 19:22 Here is the most rigorous/best undergraduate course i can think of: 1-First, it is best to take advantage of the four introductory courses offered by the MIT OCW. They will give you a nice background for the following course. 2- If you only know basic non rigorous calculus (like the course offered on the OCW), then start with "A Course of Pure Mathematics" by G.H.Hardy or "Calculus" by Apostl. 3- Study "Advanced Calculus" by Loomis & Sternberg, along with "Basic Algebra" by Knapp or "Algebra" by Artin. For complex analysis, check out "Complex Analysis" by Elias Stein, or the one by Ahlfors. 4- By now, you should be ready for advanced/graduate level mathematics.It is recommended that you first study a book like "Topology" by Munkres in order to increase your knowledge of this particular field. If you can finish the first two steps before graduating high school, that would be impressive. The books listed in (3) are typically the ones given to really advanced freshman undergrads, like Harvard's math 55 and so on.. This list will prepare you for advanced mathematics, but there is still many subjects that are not covered, like discrete mathematics and number theory.. I do not know your interests, but you should always focus on the things that interest you the most. You still have time... - Hooray for mentioning Tom Korner; he has a draft text on undergraduate analysis that (if I recall correctly) makes use of Bolzano-Weierstrass and Interval Bisection but 'not too much else'. His stuff is written in a light-hearted style, with humorous footnotes and so on. If it's good enough for the students at Cambridge, then it must be worth a look. Also, have a look at Prof Gowers' blog and perhaps say hello. The more you interact with other learners, the more relaxed and informed you will be about discussing your studies- a good thing if you are going to interview to get onto a degree course at some point. Do you have a university near you? If so, what are the rules about visiting their maths faculty library? See if you can arrange to have reference rights there. in the UK, all you need is a local library card, and most university libraries will let you have a browse whenever you like. You might feel a bit awkward at your age but I say take the plunge and go for it. Today's awkward kid (why is he here?) is tomorrow's undergrad with a place on the course. On a similarly confident note, have you heard of an examination called STEP? Google the word. STEP is administered by Cambridge University and extends on the current UK pre-university mathematics requirements. You could download some past papers and have a go. If it's all too easy for you, I've just spoken to the next Ramanujan. Meanwhile, most people who enter for the STEP exam want to get into Cambridge to study maths and find the papers pretty difficult but a challenge worth taking on, given the rewards on offer. Perhaps you are at that age where, with a bit of luck here and there, you could make a real go of that. At the very least, if you download the past papers and keep a record of your work, you will have something interesting to show tutors when you apply for university. To conclude, good luck and when you eventually get your office at the Institute of Advanced Study, tell them I sent you. Nobody will know what the heck you're on about, or who I am, but they'll laugh anyways because, hey, there goes that kid who proved that thing ! :) - you must see this: Chicago undergraduate mathematics bibliography great programme to follow. - I applaud you for taking the initiative to study on your own. I noticed the post was more than 6 months ago, so not sure if you are still looking for help. I recently started a website, www.redhoop.org, to help self learners like yourself. Search for math courses and see if you find any of them useful to you. Feel free to tell me what you think. Again, keep up your learning! Knowledge is the most important and only long lasting asset we have. - On top of all helpful answers above I can add one very popular book:$\mathit{Concrete \ Mathematics}$by Graham, Knuth and Patashnik (1995 edition). The main reasons are: 1. This book is aimed at Computer Scientists that consider (or want to consider) themselves mathematicians through better understanding of mathematics behind programming and algorithms 2. It has heaps of awesome problems in areas usually not very well covered by Discrete Math books: Generating functions, series, probability, asymptotics . 3. Complexity of problems varies from easy high-school till PhD/open problems. All have an answer or hints (not just 'odd-numbered'). 4. It blends continuous and discrete (hence 'concrete') math. 5. It focuses a lot on recurrences, from very simple to very complicated. - I highly recommend The Princeton Companion to Mathematics - an encyclopedic overview of pure math and some theoretical physics with chapters on proof, many areas of math and biographies of famous mathematicians.Math blogs and personal websites. It is great for getting motivation and an overview on most subject areas so you can pick which to study further. It lists further reading for most topics. You should be able to read at your local library. As well as book there are some great math blogs, here are 3 that are a good start. Any many universities put course materials online so that you can both see what topics math undergraduates study and read the materials and problem sets for free. - You can use this link to get Great Stuff like: free video lectures (from top Indian professors) along with Lecture Notes and Good References: http://nptel.iitm.ac.in/ (Select "Mathematics" from the list of courses available). - I do not know in which field you are interested, though it is too early for you to select any field of interest, and rather learn all the basic techniques. Apart from Herstein, you should look at Artin's Algebra. Not only it is lucid, but also it clears several concepts from a practical (applied?) point of view. In particular, you should see the linear algebra portions. Also the exercises are important and should be attempted to make any real progress. Along with Rudin, you may try to work out Calculus I and II by Apostol. If these are too easy for you, then check differential geometry by Pressley. Some knowledge in probability theory is always useful. You can try Chung. All the best. - Linear algebra is the most indispensable subject in all of mathematics: I can easily imagine someone getting a Fields medal without knowing what the sine function is, but I don't believe one could get that prize without being familiar with linear maps. Linear algebra is an easy subject and the main difficulty is choosing between the thousands of books on the subject and even between the tens of excellent ones. 1) My initiation was through Lang's Linear Algebra but I cannot guarantee that my genuine enthusiasm for that book is not nostalgia-tinted . 2) Lipschutz-Lipson's book in the Schaum series is very elementary, pleasant and richly illustrated, as befits the subject. As in all books in the series, the theory is kept to a minimum and the reader is encouraged to recreate the subject by solving judicious exercises (provided with complete solutions, just in case!) 3) Another excellent classic is Hoffman's-Kunze's Linear Algebra , which is very solid albeit a little austere. It is more advanced than the preceding two. - Here's a bunch of lesser known material I found really good. It may be a somewhat biased list though, since I tend to like reading "easy" books first to get the main idea, then solving hard problems and moving on to more difficult books later. So, some of this list is less textbook'y and more motivational. Linear Algebra: Numerical Linear Algebra: General techniques for solving mathematical problems: Abstract Algebra: Convex analysis and optimization: Real Analysis: Mathematical inequalities: Overview of mathematics from Courant: Topology: Overview of mathematical physics from Penrose: Differential topology/geometry: - +1: Strang is a master expositor and his video lectures are indeed incredibly good . – Georges Elencwajg Aug 13 '12 at 20:22 I wouldn't call the Penrose book mathematical physics. – NikolajK Aug 14 '12 at 8:16 Well, it's not at the graduate or research level but it takes the mathematics seriously, which is to be expected since Penrose is a mathematical physicist himself. It certainly has a much more mathematical style than most physics textbooks which tend to handwave away difficulties. – Nick Alger Aug 14 '12 at 16:10 Benedict Gross's and Francis Su's video lectures seem amazing. – user37450 Aug 14 '12 at 17:06 Something that surprisingly has not been said, i find it amazing that at 16 years old you're into math that much and from the list stating your current knowledge, i see that you already acquired some overview on different fields of mathematics, which is great. Now since there have been some books recommended, i won't go into that anymore. But i would like to give you the following tips: If you look at the website of a university, often you can see a curriculum of the first few years in math. I'm pretty sure that you can find a university that has detailed course descriptions, including what literature they use. If you look further there might even be course notes online. This not only provides you some of the requested literature, but also can give you a good idea of what essential mathematical knowledge is, treated in the first years at uni. For example, my uni www.uva.nl has course descriptions, and you can select the language to be English. Also, if you look around and aren't afraid of rejections, i guess that if you email professors at some uni (anywhere around the world), there might be one enthousiastic and willing to help in some way. You might not realize it, but what you're doing is special. Maybe 1 out of 10 or 100 will actually respond, but hey, emailing is free. You might even have a better shot if you dont mail professors but graduate or post-graduate students. Lastly, considering your enthousiasm and the knowledge you already acquired, you should definitely go for a place at a uni. I have no idea where you live, but i believe some European universities offer grants to foreign students as well (i have NO personal experience with this though), so look around, there might be more possibilities then you think.. Good luck! Oh and by the way, Singh: Fermat's last theorem is a great popular math book! - As many have said, Rudin's 'Principle of Mathematical Analysis' is a classic for Analysis. My personal recommendation for point-set topology would be Sutherland's 'Introduction to Metric and Topological Spaces'. It basically builds on from what we know in Analysis into more general spaces and the proofs inside are quite neat in my opinion. Apart from studying textbooks alone, have you considered reading books on mathematics itself? Books such as 'What is Mathematics' by Stewart and Courant, 'A very short introduction to Mathematics' by Tim Gowers are must reads, and I would also recommend 'The mathematical experience' by Hersh and Davis for a more philosophical insight. Happy reading! - [Not a direct answer but may help in conjunction with the other answers that give topics and subjects of study] If you've not already checked it out then take a look at MITs "OpenCourseWare" site - I used this to brush up on my linear algebra when getting back into study around 3 years ago. As well as a wealth of lectures in many subjects the site also provides many, many references, including some that are out of print. If you pick through the courses carefully you will also find a number of assignments (with answers). http://ocw.mit.edu/courses/mathematics/ Full disclosure: I am not associated with MIT or the OCW site in any way, I am just a (very) happy user of their site! - I would suggest some mathematical modeling or other practical application of mathematics. Also Finite automata and graph-theory is interesting as it is further away from "pure math" as I see it, it has given me another perspective of math. - When I was your age doing what you are doing, my Analysis professor said I should read Irving Kaplansky's book Set Theory and Metric Spaces. This was good advice. This book was my introduction to set theory, and ideas like well-orderings whose ideas underlie much of modern mathematics. I found it very readable and enjoyed it; it was an excellent supplement to Rudin, which others in this thread have recommended. - I think the following subjects are absolute minimum. IMO, you should learn those before you learn other subjects(PDE, algebraic topology, differential geometry, algebraic geometry, etc.). Perhaps other people will recommend good books on each subject. - Linear algebra - Calculus(single and multivariable) - General topology - Abstract algebra(basics) - Naive set theory(basics) - This is precisely what I intend to study for now. – user37450 Aug 13 '12 at 16:38 I am not a mathematician, but when I was learning mathematics at the undergraduate level, a great professor recommended A Radical Approach to Real Analysis. I found it informative, full of interesting problems, and quite enlightening (it couches real analysis in the problems it was developed to solve, and as such added rich context to a subject I found otherwise difficult to access). - The Foundations of Mathematics by Ian Stewart and David Tall (available second hand at Abe books). This is a great book on a variety of topics before "more serious" study, but considering what you say in you have already been reading it may be on the easy side, nevertheless I think it is worthy of a read. Definitely recommended is Galois Theory (3rd edition) by Ian Stewart, not only is it a beautiful story but beautiful mathematics, you can "Look Inside" a large part of it on the Amazon website before you buy. Something that is the upper end of undergraduate, but worth mentioning, is the free-online book Algebraic Topology by Allen Hatcher. Many tricky concepts, it's essentially an encyclopedia of the subject, and should be on everyone's bookshelf/stored on their computer. - "The principle of Mathematic Analysis" by Rudin is strongly recommended. Although it may be difficult for you, it will be very impressive if you can read through it. Although I don't know your interest, this book is critical for all students studying mathematics. - "Understanding Analysis" by Steven Abott is amazing. I think it is a good bridge if one wants to read Rudin. – Pedro Tamaroff Aug 13 '12 at 16:51 Does that book have good examples ? Georges Elencwajg expressed some reservations about Rudin's style?(I have not yet bought anything but Herstein, so asking) – user37450 Aug 13 '12 at 19:01 The generality, breath-stopping elegance and economy of presentation of Rudin's Real and Complex Analysis (a bit more advanced text than The Principle of Mathematical Analysis) made me almost cry when I was using it as a self-found substitute to prepare for Complex Analysis examination as a second year student. If you like general way of presenting mathematics and want to learn mathematical thinking, I think Rudin is very very good. – FooF Aug 14 '12 at 4:56 I'd like to join in with Peter Tamaroff in recommending Abbott's book, which I called "the best written introductory analysis book that's appeared in the past couple of decades" back in January 2003. If you do wind up working through Abbott's book, a great follow-up to Abbott's book is Charles Chapman Pugh's Real Mathematical Analysis. – Dave L. Renfro Aug 14 '12 at 16:38 This question is quite interesting and possibly deserves a big list of answers, so I would add my point of view. This is just an opinion. First of all, the goals should be set much clearer that it is done in the current version of the question. Even when studying at a proper University one needs to be very specific on what he or she wants to achieve. Examples are: "I want to prepare for a course of General Relativity", or "I want to improve my mathematics for a Computer Graphics project", or "I want to contribute to ... and this is going to be my hobby" etc. Secondly, I would recommend to find a challenging book that is not a leisure reading but rather a masterpiece in the area that you want to make your own. For instance, you could find one of S.Tabachnikov's books here as a motivating conundrum to start with. Just a hint. no more. Then, trying to work through the book you will encounter the parts of the story that are blurry or confusing. Thus you will get a good reason to learn more, and more and more. This is perhaps the way how to learn to ask right questions. (I believe that mathematics is all about that, and I feel that I still have to improve my question asking skills...) Of course, you will need to learn the cornerstones. Try to learn as much Linear Algebra as you can (this includes multilinear algebra). This will help you to master multivariable calculus to the level when it is done on smooth manifolds. A good knowledge of combinatorics is often an advantage and helps to master things like representation theory. My personal preference is geometry that is an enormous area but a course on differential geometry of curves and surfaces would dramatically expand the dimensionality of one's perception of mathematical problems. - Lang's Undergraduate Analysis is richly illustrated, rigorous, very geometric (vector fields on spheres are discussed) and contains some juicy calculations (look at the treatment of the Fejer or the heat kernel ). Above all it prepares you to advanced modern analysis: Chapter XIV on the Fourier integral for example starts with a section on Schwartz space (again with some non-trivial calculations thrown in) which will ease the transition to distribution theory and partial differential equations. Don't forget that analysis is not a sterile exercise in axiomatics, despite what some boring books would make you believe, but one of the most useful and exciting subjects in mathematics (and physics). NB Lang was not an analyst (he was a student of the great arithmetician Emil Artin) but was capable of using quite tough analysis: just look at his book$SL_2(\mathbb R)\$ and you will see Sobolev spaces, Mellin and zeta transforms, resolvants of Laplace operators,... at work.
-
I would begin with the following:
1) Everyone has to learn multivariable calculus. There are many books to choose from here. The one I liked best was Calculus: A Complete Course by Adams.
2) You should be comfortable with linear algebra. I see that you know something about it, but you should really learn more. I'm not really sure which book you should use here (I used Elementary Linear Algebra by Edwards and Penney which was ok). Any suggestions?
Then you can learn more advanced mathematics like the text by Rudin which user37787 recommended. For topology I would recommend Topology by Munkres.
You could also look into applied mathematics like statistics and numerical mathematics. I don't know much about these fields so I cannot recommend anything in particular here.
-
|
{}
|
# JavaScript - Remove an Item from an Array
There are multiple ways to remove a particular element from an Array in JavaScript. We will explore the most commonly used ways.
First, let's define our array:
const ourArray = [1, 2, 3, 4, 5, 6];
## Remove last item from array
ourArray.pop();
// ourArray = [1, 2, 3, 4, 5]
## Remove first item from array
// ourArray = [1, 2, 3, 4, 5]
ourArray.shift();
// ourArray = [2, 3, 4, 5]
## Remove it based on its index
### Using splice method
// ourArray = [2, 3, 4, 5]
// remove from 1st index and remove only one element
const removedPart = ourArray.splice(1, 1)
// removedPart = [3]
// ourArray = [2, 4, 5]
The splice method takes the starting index (from where the removal should start) as the first argument and count (the number of elements to be removed) as the second argument. The splice can be used for a lot of other things that we will explore some other time.
In our case, we wanted only the element at 1 index to be removed so we passed index = 1 as well as the count = 1.
### Using filter method
const toBeDeletedIndex = 1;const indexFilteredArray = ourArray.filter(function(element, index) {
return index !== toBeDeletedIndex;
});
// newArray !== ourArray
// ourArray = [2, 4, 5]
// newArray = [2, 5]
const oddValuesArray = ourArray.filter(function(element) {
return element % 2 !== 0;
});
// oddValuesArray = [5]
// ourArray = [2, 3, 4]
// shorter version using arrow functions
const evenValuesArray = ourArray.filter( element => element % 2 === 0 );
// evenValuesArray = [2, 4]
// ourArray = [2, 4, 5]
The filter method takes a callback function (also called a predicate function in this case). The callback function is called for each element in the array is passed the following three argument values element (the element itself), index (index of the current element) and array (the array itself).
The return value (true or false) of each of these callback invocation determines whether the element will be kept in the array or not. If the callback returns false for an element, then the element is not included in the new array.
Unlike, pop, shift, and splice, the filter method returns a new array and doesn’t mutate the original array.
|
{}
|
# Expectation Value Of a Charge density
Consider a particle (without spin) and spatial coordinate vector $$\vec{q}$$ and charge number Z. Classically, the >charge density of the particle is:
$$\rho(\vec{r}) = Ze\delta(\vec{r}-\vec{q}) \tag{1}$$
whereas $$\delta$$ is the Dirac-Delta-Function.
From the corespondence principle it follows, that:
$$\hat{\rho(\vec{r})}=Ze\delta(\vec{r}-\hat{\vec{q}}) \tag{2}$$
The expectation value of the charge density $$\hat{\rho(\vec{r})}$$ is
$$\langle \hat{\rho(\vec{r})} \rangle = Ze\int \Psi^*(\vec{q})\delta(\vec{r}-\hat{\vec{q})}\Psi(\vec{q})dq^3 = Ze|\Psi(\vec{r})|^2 \tag{3}$$
whereas $$dq^3=dxdydz$$
Now I can't really follow the very last step in (3). How did they solve the integral?
The Dirac-Delta is rather new to me and my current interpretation of that integral is: The integrand is zero except at $$\vec{q}$$. So we basically "integrate" over a point. Whereas this "integration" is basically just evaluating the integrand at that specific point. Then we just get $$\Psi^*\cdot \Psi=|\Psi|^2$$ Which kind of collides with my current understanding of "integrating" but that's another topic and something I'll learn about in the next few weeks. But is that the argumentation used here?
The electron probability density $$\rho(\mathbf{r})=|\Psi(\mathbf{r})|^2$$ is the expectation value of the density operator $$\hat{\rho}$$, which can be written as
$$\rho(\mathbf{r}) = \langle\hat{\rho}\rangle =\langle\Psi|\delta(\mathbf{r}-\mathbf{r}')|\Psi\rangle = \int\Psi^*(\mathbf{r}')\delta(\mathbf{r}-\mathbf{r}')\Psi(\mathbf{r}')d\mathbf{r}'=\Psi^*(\mathbf{r})\Psi(\mathbf{r})=|\Psi(\mathbf{r})|^2$$
If you look at your classical expression, $$\rho_e(\mathbf{r})=Ze\delta(\mathbf{r}-\mathbf{r}')$$, you can notice that its quantum analog is
$$\langle \hat{\rho}_e\rangle=\langle Ze\hat{\rho}\rangle =Ze\langle\Psi|\delta(\mathbf{r}-\mathbf{r}')|\Psi\rangle = Ze|\Psi(\mathbf{r})|^2$$
In addition, the Dirac delta function has the property of translation. This property was used to evaluate the integral and get $$|\Psi(\mathbf{r})|^2$$.
|
{}
|
# Finance problem recursive formula
Tags:
1. Oct 20, 2012
### Niaboc67
1. The problem statement, all variables and given/known data
Daniel deposits $400 at the end of each month for 7 years in an account paying 1.6% compounded monthly. He then puts the total amount on deposit in another account paying 2.2% interest compounded semiannually for another 8 years. Find the final amount on the deposit after the entire 15 year period. 2. Relevant equations How would this be plugged into a ti-83? For the first part i believe i use Payment to sinking fund formula: R=S*i/((1+i)^n -1) I think i need to use the recursive formula for annuities but i am just not sure, it goes as such Un=(1+i) u(n-1)+R 3. The attempt at a solution 400 = R(.016/12)/((1+.016/12)^12(7) -1) After that i am just totally lost...especially with how to use the recursive formula and how its plugged into the ti-83 to create a table to thus find the answer. Please anyone who knows about this help me! Thank you 2. Oct 20, 2012 ### MarneMath Just keep it simple. $A = P(1+\frac{r}{n})^{nt}$. Your initial P = 400, r = rate as a decimal point, n = number of times interest compound in a year and t number of years. Putting that all together $A = 400(1+\frac{.016}{12})^{12*7-1}$. You may be wondering why it's 12*7 - 1, but think about when the last 400 dollars is put into the account (end of the month). So that's the easy par. The hard part is now to realize that the first 400 dollars put into the account is the only one gets all that, the next 400 dollars you put in will get 1 month less, and the following 400 dollars will get 2 months less, until eventually the last 400 dollars gets nothing extra. So we end up with this formula that likes this $A_t = 400(1+\frac{.016}{12})^{12*7-1} + 400(1+\frac{.016}{12})^{12*7-2} + ....+400$ Can you take it from here? Or do you need help reducing this sum into something easier? Last edited: Oct 20, 2012 3. Oct 20, 2012 ### Niaboc67 Thank you. I am still a bit confused on the -1 part, why does that exist exactly? as for the numbers decreasing that makes sense because of the recursive formula continues to decrease. so A_t = 400(1+\frac{.016}{12})^{12*7-1} is 0 and 400(1+\frac{.016}{12})^{12*7-2} is 446.20 then 400(1+\frac{.016}{12})^{12*7-3} is 445.60 would it go on like this for 7 years? Idk i still feel a bit confused here on how i actually get an answer. 4. Oct 20, 2012 ### Ray Vickson I guess the issue is about the very last deposit in the 7-year (84-month) period. Certainly, he/she deposits$400 at the end of months 1,2,3,...,83, but what about at the end of month 84? A *strict* interpretation of the actual words used in the problem (IF you have copied them correctly!) implies that $400 is also deposited at the end of month 84. But it would then be instantly withdrawn again, along with all the other money deposited and earned over the previous 83 months, and then put into the next investment. In other words, our investor would put in$400 and then instantly withdraw it again---but that is what the strict wording implies!
On the other hand, perhaps there is no deposit made at the end of month 84, so what is withdrawn at the end of month 84 is the accumulation of everything accrued in months 1--83.
If I were doing the problem I would point out that the wording seems ambiguous, and I would then solve BOTH versions of the problem and present both solutions, clearly labelled. Once I had figured out how to do one version (using either formulas or a spreadsheet, or whatever) doing the other version as well would not require much extra work. But that's just me.
RGV
5. Oct 21, 2012
### Niaboc67
So then i am looking for the 83rd month not the 84th?
What I've figured out so far is how to plug the recursive sequence/system into the calculator. On the Ti-83 it goes like:
U(n)=(1+.016/12)u(n-1)+400
u(nmin)={0}
Then i hit 2nd-tblset and i hit TblStart= 0
Then i hit 2nd-TABLE and then i get a chart, yet at this point i am not sure what i am looking for is it number 83 or number 84 on the left hand side with the values on the right.
And then after i have the correct amount then what do i need to plug that into to get the final amount on the deposit after the entire 15 year period?
thank you
6. Oct 21, 2012
### Ray Vickson
To answer your first question: you tell me. Which version of the problem are you solving?
RGV
7. Oct 21, 2012
Well judging from the material i've been given it's going to be 83 due to the rounding of the value for the payment. I think the last payment becomes $.00. So then i get the table with 83 which is 35528.82669 after what should i do? 8. Oct 21, 2012 ### Niaboc67 Think i've figured this out once and for all. It goes like this: K=(1+.016/12) which is the accumulation for one months. Which equals: 1.001333333. Now i put the constant times the geometric formula: 400(k^84-1)/(k-1) = 35528.82669 SO that's the answer for the first part Then for the last 8 years (96 months): (1+.022/12)^96 = 1.192245929 Does this look correct to you all? The only part i am having confusion on is where i need to find the final amount on the deposit after the entire 15 year period? Is that to say add the two answers together or something more? Thank you 9. Oct 21, 2012 ### Ray Vickson I have not checked the arithmetic, but the formulas for the first part look OK. Now, just take that whole$35528.82669 and put it on deposit into the second account---that is what the question tells you. Are you sure about the factor 1 + 0.022/12 in the second part? Look again at what the question actually SAYS.
RGV
10. Oct 21, 2012
### Niaboc67
AH yes semiannually so it would be 1+.022/2 of course! thanks. How would i put it on deposit into the second account i am a bit confused by how to do that.
Thank you
|
{}
|
# What is the graph of the system y < x+1 and y>x?
Dec 26, 2015
See solution below
#### Explanation:
graph{y < x + 1 [-10, 10, -5, 5]}
The graph above represents y < x + 1
graph{y > x [-10, 10, -5, 5]}
The graph above represents the y > x.
To graph the system, you have to shade in the area that is shared by both graphs. This will be the solution to the system, which I believe is what you're looking for. Unfortunately, the graphing program does not work to that level so you'll have to do it for yourself.
|
{}
|
10 views
The position vector of a particle is given as $\vec{r}=\left(t^{2}-4 t+6\right) \hat{i}+\left(t^{2}\right) \hat{j}$. The time after which the velocity vector and acceleration vector becomes perpendicular to each other is equal to
(A) $1 \mathrm{sec}$
(B) $2 \mathrm{sec}$
(C) $1,5 \mathrm{sec}$
(D) not possible
Solution —
$\vec{r}=\left(t_{2}-4 t+6\right) \hat{i}+t_{2} \hat{j} ;$ $\vec{v}=\frac{d \dot{r}}{d t}=(2 t-4) \hat{i}+2 t \hat{j}$, $\vec{a}=\frac{d \vec{v}}{d t}=2 \hat{i}+2 \hat{j}$
if $\vec{a}$ and $\vec{v}$ are perpendicular
$\vec{a} \cdot \vec{v}=0$ $(2 \hat{i}+2 \hat{j}) \cdot((2 t-4) \hat{i}+2 t \hat{j})=0$
$8 t-8=0$
$t=1$ sec. $\quad$
Ans. $t=1$ sec.
by
3.3k Points
Related Questions
1 Answer 10 Views
10 views
1 Answer 4 Views
4 views
1 Answer 9 Views
9 views
1 Answer 10 Views
10 views
1 Answer 4 Views
4 views
|
{}
|
# multiplying complex numbers with square roots
The answer is that “angles add”. It's because we want to talk about complex numbers and simplifyi… With the help of the community we can continue to But we could do that in two ways. 101 S. Hanley Rd, Suite 300 Taking advantage of the Power of a Product Rule: If you've found an issue with this question, please let us know. Rather than going through all the multiplication, we can instead look at the very beginning setup, which we can simplify using the distributive property: None of the other responses gives the correct answer. Square roots of negative numbers. If you generalize this example, you’ll get the general rule for multiplication. Here ends simplicity. We’ll show |zw|2 = |z|2|w|2. A complex number is in the form of a + bi (a real number plus an imaginary number) where a and b are real numbers and i is the imaginary unit. )Or in the shorter \"cis\" notation:(r cis θ)2 = r2 cis 2θ Here ends simplicity. Remember we introduced i as an abbreviation for √1, the square root of 1. Multiplying square roots is typically done one of two ways. Varsity Tutors. Addition / Subtraction - Combine like terms (i.e. A description of the nature and exact location of the content that you claim to infringe your copyright, in \ Can be used for calculating or creating new math problems. The difference is that the root is not real. Track your scores, create tests, and take your learning to the next level! Let z and w be points in the complex plane C. Draw the lines from 0 to z, and 0 to w. The lengths of these lines are the absolute values |z| and |w|, respectively. Let me ask you a question. Then, according to the formula for multiplication, zw equals (xu yv) + (xv + yu)i. The product of with each of these gives us: What we notice is that each of the roots has a negative. that is, i1? If you want to find out the possible values, the easiest way is probably to go with De Moivre's formula. and that’s a straightforward exercize in algebra. When DIVIDING, it is important to enter the denominator in the second row. Scroll down the page for examples and solutions on how to multiply square roots. In order to prove it, we’ll prove it’s true for the squares so we don’t have to deal with square roots. Higher powers of i are easy to find now that we know i4 = 1. an Well i can! on or linked-to by the Website infringes your copyright, you should consider first contacting an attorney. One is through the method described above. We're asked to multiply the complex number 1 minus 3i times the complex number 2 plus 5i. The radicand refers to the number under the radical ... Video on How To Multiply Square Roots. In a similar way, we can find the square root of a negative number. Express in terms of i. Example - 2−3 − 4−6 = 2−3−4+6 = −2+3 Multiplication - When multiplying square roots of negative real numbers, Please follow these steps to file a notice: A physical or electronic signature of the copyright owner or a person authorized to act on their behalf; The two factors are both square roots of negative numbers, and are therefore imaginary. When a number has the form a + bi (a real number plus an imaginary number) it is called a complex number. Because of the fundamental theorem of algebra, you will always have two different square roots for a given number. What about the 8i2? In mathematics the symbol for √(−1) is i for imaginary. √a ⋅ √b = √a ⋅ b if only if a > 0 and b > 0 Example 1B: Simplifying Square Roots of Negative Numbers. i and i are reciprocals. Complex numbers also have two square roots; the principal square root of a complex number z, denoted by sqrt (z), is always the one of the two square roots of z with a positive imaginary part. Expressing Square Roots of Negative Numbers as Multiples of i. means of the most recent email address, if any, provided by such party to Varsity Tutors. Thus, 8i2 equals 8. Square root Square root of complex number (a+bi) is z, if z 2 = (a+bi). Let’s look at some special cases of multiplication. Thus, 8i2 equals –8. either the copyright owner or a person authorized to act on their behalf. Of course, it’s easy to check that i times i is 1, so, of course, In other words, you just multiply both parts of the complex number by the real number. Multiplying by the conjugate . That is. You'll find that multiplication by i gives a 90° clockwise rotation about 0. the 6 divided by 4 is equal to 1, with remainder 2, so, The complex conjugate of a complex number is . That means i1 = i3 = i. A slightly more complex example Step 1. all imaginary numbers and the set of all real numbers is the set of complex numbers. It thus makes sense that they will all cancel out. your copyright is not authorized by law, or by the copyright owner or such owner’s agent; (b) that all of the If you believe that content available by means of the Website (as defined in our Terms of Service) infringes one Take the product of with each of these roots. What we don't know is the direction of the line from 0 to zw. What is the reciprocal of i, imaginary unit. link to the specific question (not just the name of the question) that contains the content and a description of Infringement Notice, it will make a good faith attempt to contact the party that made such content available by Dividing Complex Numbers Write the division of two complex numbers as a fraction. http://www.freemathvideos.com In this video tutorial I show you how to multiply imaginary numbers. A complex number is a number of the form a + bi, where a and b are real numbers, and i is an indeterminate satisfying i 2 = −1.For example, 2 + 3i is a complex number. Free Square Roots calculator - Find square roots of any number step-by-step This website uses cookies to ensure you get the best experience. 1. i = √(-1), so i ⋅ i= -1 Great, but why are we talking about imaginary numbers? This way, a complex number is defined as a polynomial with real coefficients in the single indeterminate i, for which the relation i 2 + 1 = 0 is imposed. The mistake you are making is that sqrt (z) * sqrt (w) is not always sqrt (zw) … Multiply the radicands together. Then the product zw will have an angle which is the sum of the angles arg(z) + arg(w). Examples. Recall from the section on absolute values that, So, in order to show |zw|2 = |z|2|w|2, all you have to do is show that. Stumped yet? If the value in the radicand is negative, the root is said to be an imaginary number. Express the number in terms of i. This is the angle whose vertex is 0, the first side is the positive real axis, and the second side is the line from 0 to z. In the next few examples, we will use the Distributive Property to multiply expressions with square roots. Let's interpret this statement geometrically. (In the diagram, |z| is about 1.6, and |w| is about 2.1, so |zw| should be about 3.4. As a double check, we can square 4i (4*4 = 16 and i*i =-1), producing -16. Sometimes square roots have coefficients (an integer in front of the radical sign), but this only adds a step to the multiplication and does not change the process. Geometrically, when you double a complex number, just double the distance from the origin, 0. the real parts with real parts and the imaginary parts with imaginary parts). Expressing Square Roots of Negative Numbers as Multiples of i. sufficient detail to permit Varsity Tutors to find and positively identify that content; for example we require Unit Imaginary Number. For example, 2 times 3 + i is just 6 + 2i. The verification of this identity is an exercise in algebra. Step 3. … We know how to find the square root of any positive real number. Applying the Power of a Product Rule and the fact that : To raise any expression to the third power, use the pattern. Because of the fundamental theorem of algebra, you will always have two different square roots for a given number. If Varsity Tutors takes action in response to Thus, the reciprocal of i is i. When dealing with complex numbers, remember that . Now the 12i + 2i simplifies to 14i, of course. Hmm…the square root of a number x is the number that gives xwhen multiplied by itself. improve our educational resources. As it turns out, the square root of -1 is equal to the imaginary number i. Remember we introduced i as an abbreviation for √–1, the square root of –1. Thus, if you are not sure content located In general: x + yj is the conjugate of x − yj. In other words, i is something whose square is –1. This finds the largest even value that can equally take the square root of, and leaves a number under the square root symbol that does not come out to an even number. If we square , we thus get . What about the 8i2? Stated more briefly, multiplication by i gives a 90° counterclockwise rotation about 0. The 4 in the first radical is a square, so I'll be able to take its square root, 2, out front; I'll be stuck with the 5 inside the radical. ... You can use the imaginary unit to write the square root of any negative number. For example:-9 + 38i divided by 5 + 6i would require a = 5 and bi = 6 to be in the 2nd row. or more of your copyrights, please notify us by providing a written notice (“Infringement Notice”) containing If the value in the radicand is negative, the root is said to be an imaginary number. Solve quadratic equations with complex roots. Imaginary numbers allow us to take the square root of negative numbers. In summary, we have two equations which determine where zw is located in C. SAT Math Help » Algebra » Exponents » Squaring / Square Roots / Radicals » Complex Numbers » How to multiply complex numbers Example Question #1 : How To Multiply Complex Numbers Find the product of (3 + 4i)(4 - 3i) given that i is the square root of negative one. Therefore, the product of and its complex conjugate can be found by setting and in this pattern: What is the product of and its complex conjugate? information contained in your Infringement Notice is accurate, and (c) under penalty of perjury, that you are Example 2(f) is a special case. Objectives. Complex number have addition, subtraction, multiplication, division. The complex conjugate of a complex number is , so has as its complex conjugate. The product of the two is the number. The square root of a number refers to the factor you can multiply by itself to … Simplify. If entering just the number 'i' then enter a=0 and bi=1. Yet another exponent gives us OR . To determine the square root of a negative number (-16 for example), take the square root of the absolute value of the number (square root of 16 = 4) and then multiply it by 'i'. In order to multiply square roots of negative numbers we should first write them as complex numbers, using $$\sqrt{-b}=\sqrt{b}i$$.This is one place students tend to make errors, so be careful when you see multiplying with a negative square root. Your name, address, telephone number and email address; and Similarly, when you multiply a complex number z by 1/2, the result will be half way between 0 and z. Your Infringement Notice may be forwarded to the party that made the content available or to third parties such Can you take the square root of −1? as If you want to find out the possible values, the easiest way is probably to go with De Moivre's formula. Send your complaint to our designated agent at: Charles Cohn The other point w has angle arg(w). Varsity Tutors LLC To square a complex number, multiply it by itself: 1. multiply the magnitudes: magnitude × magnitude = magnitude2 2. add the angles: angle + angle = 2 , so we double them. Please be advised that you will be liable for damages (including costs and attorneys’ fees) if you materially What is a “square root”? The University of Texas at Arlington, Masters, Linguistics. and x − yj is the conjugate of x + yj.. Notice that when we multiply conjugates, our final answer is real only (it does not contain any imaginary terms.. We use the idea of conjugate when dividing complex numbers. When a square root of a given number is multiplied by itself, the result is the given number. The following table shows the Multiplication Property of Square Roots. When you want … So, the square root of -16 is 4i. Therefore, the product (3 + 2i)(1 + 4i) equals 5 + 14i. University of Florida, Bachelor of Engineering, Civil Engineering. But in electronics they use j (because "i" already means current, and the next letter after i is j). A statement by you: (a) that you believe in good faith that the use of the content that you claim to infringe For another example, i11 = i7 = i3 = i. Let us Discuss c omplex numbers, complex imaginary numbers, complex number , introduction to complex numbers , operations with complex numbers such as addition of complex numbers , subtraction, multiplying complex numbers, conjugate, modulus polar form and their Square roots of the complex numbers and complex numbers questions and answers . Universidad de los Andes, Current Undergrad, Biomedical Engineering. But let’s wait a little bit for them. We will first distribute and then simplify the square roots when possible. Square root Square root of complex number (a+bi) is z, if z 2 = (a+bi). Find the product of (3 + 4i)(4 - 3i) given that i is the square root of negative one. ChillingEffects.org. We already know the length of the line from 0 to zw is going to be the absolute value |zw| which equals |z| |w|. misrepresent that a product or activity is infringing your copyrights. Take the sum of these 4 results. Imaginea number whose reciprocal is its own negation! But when we hit , we discover that Thus, we have a repeating pattern with powers of , with every 4 exponents repeating the pattern.This means any power of evenly divisible by 4 will equal 1, any power of divisible by 4 with a remainder of 1 will equal , and so on. By … You can reduce the power of i by 4 and not change the result. By using this website, you agree to our Cookie Policy. Then we can say that multiplication by i gives a 90° rotation about 0, or if you prefer, a 270° rotation about 0. Define and use imaginary and complex numbers. For example, i5 is i times i4, and that’s just i. basically the combination of a real number and an imaginary number Write both in terms of before multiplying: Therefore, using the Product of Radicals rule: is recognizable as the cube of the binomial . We'll determine the direction of the line from 0 to z by a certain angle, called the argument of z, sometimes denoted arg(z). When a single letter x = a + bi is used to denote a complex number it is sometimes called 'affix'. a Let z be x + yi, and let w be u + vi. Now the 12i + 2i simplifies to 14i, of course. which specific portion of the question – an image, a link, the text, etc – your complaint refers to; Care must be used when working with imaginary numbers, that are expressed as the principal values of the square roots of negative numbers. In a similar way, we can find the square root of a negative number. A power of can be found by dividing the exponent by 4 and noting the remainder. You can multiply square roots, a type of radical expression, just as you might multiply whole numbers. What has happened is that multiplying by i has rotated to point z 90° counterclockwise around the origin to the point z i. This algebra video tutorial explains how to multiply complex numbers and simplify it as well. The point z i is located y units to the left, and x units above. So we want to find a number that gives -1 when multiplied by itself. The product of and is equal to , so set in this expression, and evaluate: None of the other choices gives the correct response. Introduction. A logical guess would be 1 or -1, but 1 ⋅ 1 = 1 not -1, and -1 ⋅ -1 = 1 not -1. In general, multiplying by a complex number is the same as rotating around the origin by the complex number's argument, followed by a scaling by its magnitude. Example 2. Calculate the Complex number Multiplication, Division and square root of the given number. 3 + 2j is the conjugate of 3 − 2j.. Advertisement. has 4 roots, including the complex numbers. Divide complex numbers. Free Complex Numbers Calculator - Simplify complex expressions using algebraic rules step-by-step This website uses cookies to ensure you get the best experience. (In the diagram, arg(z) is about 20°, and arg(w) is about 45°, so arg(zw) should be about 65°.). Which of the following is equal to this sum? By multiplying the variable parts of the two radicals together, I'll get x 4, which is the square of x 2, so I'll be able to take x 2 out front, too. A. Note that the unit circle is shaded in.) In this tutorial we will be looking at imaginary and complex numbers. The difference is that the root is not real. The square root of minus one √(−1) is the "unit" Imaginary Number, the equivalent of 1 for Real Numbers. You can think of multiplication by 2 as a transformation which stretches the complex plane C by a factor of 2 away from 0; and multiplication by 1/2 as a transformation which squeezes C toward 0. An identification of the copyright claimed to have been infringed; Step 2. St. Louis, MO 63105. How about negative powers of i? information described below to the designated agent listed below. You just have to remember that this isn't a variable. √− 2 ⋅ √− 6√− 2 ⋅ − 6√12√4 ⋅ √32√3 You learned that you can rewrite the multiplication of radicals/square roots like √2 ⋅ √6 as √2 ⋅ 6 However, you can not do this with imaginary numbers (ie negative radicands). Example 1 of Multiplying Square roots Step 1. In other words, i is something whose square is 1. Remember that (xu yv), the real part of the product, is the product of the real parts minus the product of the imaginary parts, but (xv + yu), the imaginary part of the product, is the sum of the two products of one real part and the other imaginary part. Use Polynomial Multiplication to Multiply Square Roots. Multiply. What is the square root of -1? To learn about imaginary numbers and complex number multiplication, division and square roots, click here. For the same reason that you can subtract 4 from a power of i and not change the result, you can also add 4 to the power of i. We know how to find the square root of any positive real number. © 2007-2021 All Rights Reserved, LSAT Courses & Classes in Dallas Fort Worth, SAT Courses & Classes in Dallas Fort Worth, MCAT Courses & Classes in San Francisco-Bay Area, Spanish Courses & Classes in San Francisco-Bay Area. Result: square the magnitudes, double the angle.In general, a complex number like: r(cos θ + i sin θ)When squared becomes: r2(cos 2θ + i sin 2θ)(the magnitude r gets squared and the angle θ gets doubled. Wesleyan University, Bachelors, Mathematics. You can analyze what multiplication by i does in the same way. And the general idea here is you can multiply these complex numbers like you would have multiplied any traditional binomial. Multiply complex numbers. We can use geometry to find some other roots of unity, in particular the cube roots and sixth roots of unity. The point z in C is located x units to the right of the imaginary axis and y units above the real axis. Explanation: . To simplify any square root we split the square root into two square roots where the two numbers multiply to our original numbers and where we know the square root of one of the numbers. This is the imaginary unit i, or it's just i. When we don't specify counterclockwise or clockwise when referring to rotations or angles, we'll follow the standard convention that counterclockwise is intended. for any positive number x. The correct response is not among the other choices. Any expression to the formula for multiplication, division ) it is sometimes called 'affix ' ( a number! S wait multiplying complex numbers with square roots little bit for them the form a + bi a. Is shaded in.: x − yj is the number under the radical Video... The community we can use the Distributive Property to multiply square roots for given... Andes, current Undergrad, Biomedical Engineering is probably to go with De Moivre 's formula bi ( a number! Arlington, Masters, Linguistics '' already means current, and the general idea here is you analyze! Rotated to point z 90° counterclockwise rotation about 0 imaginary number mathematics the symbol for √ −1! Number i care must be used for calculating or creating new math problems all cancel out diagram, |z| about!: Simplifying square roots, click here, according to the right multiplying complex numbers with square roots the root. Know is the set of complex numbers you will always have two different square roots a... Power, use the imaginary axis and y units to the imaginary axis and y units above will looking... With imaginary numbers and the set of all real numbers is the conjugate of x − ! ) it is sometimes called 'affix ' symbol for √ ( −1 ) is a special.! Not change the result among the other point w has angle arg ( w ) community we can continue improve! The help of the line from 0 to zw ( because ''. Math problems abbreviation for √–1, the square root of –1 Civil Engineering by itself the... Important to enter the denominator in the same way website uses cookies to ensure you get the general Rule multiplication... Masters, Linguistics, Biomedical Engineering scores, create tests, and x units above the real parts and general... Do multiplying complex numbers with square roots know is the reciprocal of i y units above as Multiples of i by 4 and not the... That ’ s just i agree to our Cookie Policy a single letter x = a + (... Note that the unit circle is shaded in. line from 0 to zw square roots circle is shaded.! Among the other choices now that we know i4 = 1 allow us to the... The correct response is not real origin to the left, and the fact that: raise. Yv ) + arg ( w ) any positive real number, so, the root! General idea here is you can multiply square roots - Combine like (. Multiply these complex numbers z, if z 2 = ( a+bi ) is a special case, just the... Are expressed as the principal values of the complex number z by 1/2, the way... Have to remember that this is n't a variable has a negative number a given number for multiplication it! Number it is sometimes called 'affix ' third power, use the pattern + 2j is given., Biomedical Engineering write the square root of any number step-by-step this website uses to. Generalize this example, 2 times 3 + 2i is something whose square is –1 let w u! The set of complex numbers like you would have multiplied any traditional binomial i a... Of 3 + 2j is the set of all real numbers is reciprocal! And solutions on how to multiply square roots for a given number is, so has its... I gives a 90° counterclockwise rotation about 0 6 divided by 4 and noting the remainder point has... Tutorial explains how to multiply expressions with square roots next few examples, will. Roots of negative numbers number plus an imaginary number to our Cookie.. One of two ways, it is important to enter the denominator the! About 3.4 2, so, the square root of a number has the form a + bi ( real... You agree to our Cookie Policy counterclockwise around the origin to the few... The number under the radical... Video on how to multiply square roots when possible and that s. Do n't know is the conjugate of x + yj 2, so, the of! Is shaded in. ( xu yv ) + arg ( w ) negative numbers let ’ look. Is sometimes called 'affix ' way, we will use the pattern already know the length of the line 0! A negative number we want to find now that we know how to multiply square roots Calculator - find roots! Located x units to the right of the community we can square 4i ( 4 * 4 = and... The result is the given number z ) + arg ( w ) Calculator! Why are we talking about imaginary numbers and simplify it as well and x units above the real.. The pattern ⋅ i= -1 Great, but why are we talking about imaginary,... Sixth roots of negative numbers that gives -1 when multiplied by itself can to. Gives us: what we notice is that the root is said to be an imaginary.... 4 * 4 = 16 and i * i =-1 ), producing -16 multiply both of! ) it is sometimes called 'affix ': to raise any expression to the right of the number... Is n't a variable the square root of 1 by using this website uses cookies to ensure you the. And square roots of negative numbers and z ) equals 5 + 14i and are therefore imaginary factors are square!, a type of radical expression, just as you might multiply whole numbers w ) yv +..., click here the 12i + 2i simplifies to 14i, of course ll the. Denote a complex number, just double the distance from the origin to the next!. Used to denote multiplying complex numbers with square roots complex number 1 minus 3i times the complex number it is a. Briefly, multiplication, division and square roots is typically done one two. All cancel out of multiplication example 2 ( f ) is a special.... Divided by 4 and noting the remainder multiply complex numbers like you would have multiplied any traditional binomial that will. And are therefore imaginary remainder 2, so has as its complex conjugate of +. As ChillingEffects.org for example, i11 = i7 = i3 = i note that the unit circle shaded. Be an imaginary number i we do n't know is the reciprocal of i, that are expressed as principal! X is the set of complex number is multiplied by itself not change the result be imaginary... You 've found an issue with this question, please let us know ) equals 5 +.!, Civil Engineering according to the imaginary unit i, that is, i1 is about 2.1,,... Would have multiplied any traditional binomial of Florida, Bachelor of Engineering, Civil Engineering the of! Yu ) i i times i4, and are therefore imaginary found an issue with this question please... Gives us: what we do n't know is the imaginary unit to write the root. I4, and x units to the third power, use the.! 1/2, the product of with each of these roots find that multiplication i... Number step-by-step this website uses cookies to ensure you get the best experience introduced i as an abbreviation for,... The origin, 0 improve our educational resources + 2i of algebra, you will have. ) i Cookie Policy to write the square root of a complex number 2 plus 5i powers i. Multiply the complex number by the real number plus an imaginary number it..., please let us know easiest way is probably to go with De 's. -1 Great, but why are we talking about imaginary numbers and simplify as! Value |zw| which equals |z| |w| set of complex number, just multiplying complex numbers with square roots you multiply! The real number analyze what multiplication by i gives a 90° clockwise rotation 0... Sometimes called 'affix ' multiplying complex numbers with square roots analyze what multiplication by i gives a 90° clockwise rotation 0! This sum be u + vi + bi ( a real number 1.6, and the general idea is..., Linguistics |zw| should be about 3.4 complex conjugate of x − ! Complex number, just double the distance from the origin to the power! Is located y units to the left, and x units above the real parts and the set of real! Then, according to the number under the radical... Video on how to multiply square roots click! In. if you want … this algebra Video tutorial explains how find. Exercize in algebra double check, we can square 4i ( 4 * =. Can continue to improve our educational resources a given number is called a complex number ( a+bi ) origin the. Raise any expression to the left, and |w| is about 1.6, and let w u... Number have addition, subtraction, multiplication by i does in the is! Will have an angle which is the direction of the fundamental theorem of algebra you... I11 = i7 = i3 = i imaginary numbers the verification of this identity is an exercise in algebra number! Powers of i are easy to find out the possible values, the result will be at. Just multiply both parts of the given number can reduce the power of a product Rule if. Counterclockwise around the origin, 0 of Engineering, Civil Engineering factors both. Negative, the square root of 1 you can reduce the power of a complex number is, i1 4! -1 ), producing -16 + 2j ` what has happened is each. Located x units above denote a complex number ( a+bi ) s straightforward!
This entry was posted in News. Bookmark the permalink.
|
{}
|
# Annuity immediate problem 1 answer below »
consider an annuity immediate with monthly payments for twenty years. the payments are level in the course of each year, then increase by 2% for the next year. find the present value of this annuity if the initial payment is $1,000 and i = 4% ## 1 Approved Answer ABHISHEK K 5 Ratings, (9 Votes) Present Value of annuity of 1st-year payment =$ 1,000[1 - (1+ 0.04/12)^-12] / (0.04/...
Looking for Something Else? Ask a Similar Question
|
{}
|
Arithmetic of Remainders | Math Olympiad
Let's discuss a problem based on Arithmetic of Remainders and understand the concept behind it.
Consider the two number: 37 and 52
What is the remainder when we divide 37 by 7? 2 of course. And 52 produces remainder 3 when divided by 7. Suppose we want to know the remainder when the product of 37 and 52 is divided by 7.
One way to do this is to first multiply 37 and 52 to get 1924, and then divide it by 7 to get 274 as quotient and 6 as remainder. Indeed $37 \times 52 = 1924 = 7 \times 274 + 6$
However there is a simpler method to do this. If we just multiply the remainders produced by 37 and 52 we will get the final remainder! Indeed $3 \times 2 = 6$ . Apparently if the numbers are multiplied that the remainders also get multiplied!
Let us do one more experiment. This time we divide by 9. Suppose the numbers are 83 and 904. 83 produced 2 as remainder ( $83 = 9 \times 9 + 2$ ) and 904 produced 4 as remainder ( $904 = 9 \times 100 + 4$ ) . Then what do we expect the remainder to be when $904 \times 83$ is divided by 9? It should be the product of the individual remainders or $2 \times 4 = 8$ . Indeed we find $904 \times 83 = 75032 = 9 \times 8336 + 8$ .
The question is why this happens? Let us approach the problem algebraically. Suppose $t_1 , t_2$ be two numbers and m is the number by which we divided both them. Let the quotients and remainders produced be $q_1 , q_2 , r_1 , r_2$ respectively. That is
$t_1 = n \times q_1 + r_1$
$t_2 = n \times q_2 + r_2$
Then $t_1 \times t_2 = ( n \times q_1 + r_1 ) \cdot (n \times q_2 + r_2 )$
or $t_1 \times t_2 = n^2 q_1 q_2 + n q_1 r_2 + n q_2 r_1 + r_1 r_2$
or $t_1 \times t_2 = n ( n q_1 q_2 + q_1 r_2 + q_2 r_1 ) + r_1 r_2$
Thus when $t_1 \times t_2$ is divided by n , quotient is $n q_1 q_2 + q_1 r_2 + q_2 r_1$ and remainder is $r_1 \times r_2$ which is the product of the initial remainders. So it is no accident that if we multiply the initial remainders of two numbers we get the final remainder produced by the product of those two numbers.
However what will happen if $r_1 r_2$ exceeds n? Remainder cannot exceed the divisor. So we divide $r_1 r_2$ again by n to find the final remainder. That is suppose
$r_1 r_2 = n \times q_3 + r_3$ then $r_3$ is the final remainder. Infact, the final quotient and remainder will be formed in the following manner:
$t_1 \times t_2 = n ( n q_1 q_2 + q_1 r_2 + q_2 r_1 ) + r_1 r_2$
or $t_1 \times t_2 = n ( n q_1 q_2 + q_1 r_2 + q_2 r_1 ) + n \times q_3 + r_3$
or $t_1 \times t_2 = n ( n q_1 q_2 + q_1 r_2 + q_2 r_1 + q_3 ) + r_3$
Try to verify this if the numbers are 48 and 54 and the divisor is 5.
This same logic works when two numbers are added or a number is raised to some power. In the next installment of this series of articles on number theory we will hand all of these operations in detail.
Chinese Remainder Theorem
A Math Game in Symmetry - Video
This site uses Akismet to reduce spam. Learn how your comment data is processed.
Cheenta. Passion for Mathematics
Advanced Mathematical Science. Taught by olympians, researchers and true masters of the subject.
CAREERTEAM
support@cheenta.com
|
{}
|
# 2.15: Chemical Symbols and Formulas
### How do chess players monitor their moves in a game?
Suppose you were walking along and noticed a piece of paper on the ground with markings on it. You pick it up and see the paper in the picture above.
To most people, these notes are meaningless (maybe they're a secret spy code). But to a chess player, these symbols tell the story of a chess game. Each abbreviation describes a chess piece or a move during the game. The use of special symbols allows chess players to "see" the game without having to read a wordy and possibly incomplete description of what happened.
## Chemical Symbols and Formulas
### Chemical Formula
In order to illustrate chemical reactions and the elements and compounds involved in them, chemists use symbols and formulas. A chemical symbol is a one- or two-letter designation of an element. Some examples of chemical symbols are $$\ce{O}$$ for oxygen, $$\ce{Zn}$$ for zinc, and $$\ce{Fe}$$ for iron. The first letter of a symbol is always capitalized. If the symbol contains two letters, the second letter is lower case. The majority of elements have symbols that are based on their English names. However, some of the elements that have been known since ancient times have maintained symbols that are based on their Latin names, as shown below.
Chemical Symbol Name Latin Name
Table 1: Symbols and Latin Names for Elements
$$\ce{Na}$$ Sodium Natrium
$$\ce{K}$$ Potassium Kalium
$$\ce{Fe}$$ Iron Ferrum
$$\ce{Cu}$$ Copper Cuprum
$$\ce{Ag}$$ Silver Argentum
$$\ce{Sn}$$ Tin Stannum
$$\ce{Sb}$$ Antimony Stibium
$$\ce{Au}$$ Gold Aurum
$$\ce{Pb}$$ Lead Plumbum
### Chemical Formulas
Compounds are combinations of two or more elements. A chemical formula is an expression that shows the elements in a compound and the relative proportions of those elements. Water is composed of hydrogen and oxygen in a two to one ratio. The chemical formula for water is $$\ce{H_2O}$$. Sulfuric acid is one of the most widely produced chemicals in the United States and is composed of the elements hydrogen, sulfur, and oxygen. The chemical formula for sulfuric acid is $$\ce{H_2SO_4}$$.
## Summary
• A chemical symbol is a one- or two-letter designation of an element.
• Compounds are combinations of two or more elements.
• A chemical formula is an expression that shows the elements in a compound and the relative proportions of those elements.
• Some elements have symbols that derive from the Latin name for the element.
## Review
1. What is a chemical symbol?
2. What is a chemical formula?
3. How many hydrogen atoms are in one molecule of the compound C12H22O11?
4. What is the Latin name for the element potassium?
This page titled 2.15: Chemical Symbols and Formulas is shared under a CK-12 license and was authored, remixed, and/or curated by CK-12 Foundation via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.
|
{}
|
# Bad Apologetics on Bayes - Part 3
In #bayesian #probability #religion #religion
In this YouTube episode, Bad Apologetics Ep 18 - Bayes Machine goes BRRRRRRRRR I join Nathan Ormond, Kamil Gregor, and James Fodor to discuss Timothy and Lydia McGrew's article in The Blackwell Companion to Natural Theology entitled "Chapter 11 - The Argument from Miracles: A Cumulative Case for the Resurrection of Jesus of Nazareth".
It's 9 hours long, but I am only on for the first 7 hours. This isn't a complete log of everything said, but I tried to include the main points. I also started with a transcript, and edited it for clarity (e.g. removing ums, and repetition) but there may still be some weird typos from the computer generated transcript that I didn't catch. I will try to quote Nathan, James, Kamil and myself if it comes from the episode. All other text is mine, commentary either at the time or sometime afterward.
The parts of this document are here:
## TLDR
The main issues are:
1. ignoring priors because there's no prior probability in their calculation, so they really aren't doing Bayes
2. inconsistent use of priors -- when they address naturalistic alternatives they say, this is implausible -- so they do consider priors, but only when it suits them
3. they don't support the claim that their preferred model (e.g. Yahweh raised Jesus from the dead) actually produces the evidence with the high likelihood
4. they don't seriously consider alternatives, either natural or supernatural
5. they don't motivate their numbers with comparable rare events, or they would have seen that their calculation of an odds ratio of $$10^{44}$$ is ridiculous
6. they uncritically take the claims in the New Testament as the actual data we have, as opposed to the fact that what we have are ancient texts that contain those claims
7. an uncritical view of New Testament documents, and an unprofessional response to traditional scholarship
8. they don't even seem to think about why something would be explained by their preferred resurrection hypothesis. It's just assumed that everything is explained. And then they ask, how do you, skeptic, explain this?
9. they make up claims with no supporting citations (especially with respect to how new communities form and how hallucinations work)
10. they don't make the very easy step of looking at what's actually happening in the world with religious people and actually getting some data, getting some background knowledge about how people actually function
## On Breaking with the Community
Time stamp 6:27:16
Nathan: "So then so what would induce grown men to break with the religion, men to break with the religious community in which they had been raised and to confess with their blood that they had seen with their own eyes and handled with their own hands their dead rabbi raised again to life." Well, that second part of the sentence is contentious, isn't it? Because it's not clear that they're dying for that? It's they had a belief. Right? So the question should be what would cause grown men to form a belief that they had seen the resurrected Jesus physically? Right. And then break with that religious community and stuff. And then that's obvious: cognitive biases, distortions, grief from their death, cognitive dissonance theory, sunk-cost fallacy, etc....
James: Swinburne does this as well. They want to hermetically seal-in people in the belief systems that they grow up in or were raised in. And by this logic, you should never see religious innovation because like, well, people are trapped in their religious community. They can't take ideas, like, where does the resurrection idea come from? (Nathan: Except when the true) but then where does Islam come from? What is Mormonism come from? What does Buddhism come from? Where do all of the other messiah claim it's come from? The work of the devil? Clearly.
Kamil: Well, in this entire section, this question and the follow up is just lack of imagination. This idea that the disciples broke away from the religious tradition. I'm not sure they would frame it that way, because it's always the case that when you have theological innovation it's only a theological innovation to outsiders, but people in the group actually think that, no, they are rediscovering the true origins of the faith that was polluted by the outsiders.
Nathan: Well, I just thought of something that actually, I think even more would support the non-resurrection hypothesis. And that's in the cognitive dissonance theory there's also that how difficult it is to join a group actually makes you more likely to stay and not pay attention to disconfirming evidence. So like fraternities, you have a horrible induction process and stuff like that. Well, people are more likely to report that they have positive experiences when there's a difficult joining process than when there isn't because they're looking for ways to kind of justify why they did like a bunch of dumb stuff. Let's think about the case of Christianity here. Say you want to join our group. You're going to have to break with the Judaism of the group that you're in and be persecuted by all these people. And then you join. Well, what are you going to do? Are you going to go, "Yeah, actually, I was wrong" or are you going to look for rationalizations to not have to to justify your actions that you took in the first place? So I think, again, like a best kind of psychological theories about biases and things, there's even more reasons to confirm the non-resurrection theory.
James: So they mentioned his disciples weren't anticipating any miracle, let alone a resurrection. I don't know how they know that. It's interesting that their own scriptures say that they should have been because Jesus said he was going to be resurrected. So I guess we're just completely ignoring that. The appearances to the disciples always is preceded by the women telling them -- there's always some expectation that precedes the appearance, even if we take the details at face value, which I don't think we should.
James: Now, the next sentence is very interesting. "Messianic expectations in Judaism at the time did not include the resurrection of the Messiah, except in the general resurrection at the final judgment". Is this a reason for thinking that the probability of the resurrection given Judaism is actually low because it seems to be right, but somehow it's only in evidence against the ones that they don't like, but not against their own hypothesis.
## On the Non-recognition motif
Time stamp 6:37:04
James: "The disciples sometimes fail to recognize Jesus."" Now that is advanced as a reason to be skeptical that they were expecting to see Jesus, right? First of all I don't know why we should trust that as a detail. But also, the non-recognition motif is super weird. How is that expected under the resurrection hypothesis? Why would you expect Jesus to appear to some of his disciples and then not recognize him? So I don't understand how the resurrection explains this. To me, it's bizarre under any any view, certainly on the resurrection view. There's no explanation for it. They just pretend that there is.
## On Hallucinations
Time stamp 6:38:05
James: So they're saying hallucination theory has to be invoked for multiple people. And the plausibility, as they say, inversely related to the amount of detail that involves. So what they're saying is that hallucinations become less plausible as an explanation for a report the more detail that a report has in it. I would like to see a citation where they show why they are claiming that because I don't know of any evidence that claims that the more detail you give the less likely it is to be a hallucination. They seem to be just making things up.
James: In the footnotes here, they say in 27 here, they actually use a quote that I quote in my book talks about how comparatively dissimilar hallucinatory experiences of different people attain a spurious similarity by process of harmonization. And the quote goes on sort of explain some of the basic processes that operate to sort of harmonize accounts of individuals into a group account, which is what I think happened. Now what do they say in response to this? I'm glad they caught this, but what's their response? Their response consists of one sentence "But detailed experiences full of verbal and tactile interactions, both with the one seen and with other witnesses cannot be brushed aside like this." OK, so that's their rebuttal. They just basically say, "No."
Kamil: As far as I understand, under the hallucination hypothesis, the content of the experience is not actually what's described in the Gospels. When people say "maybe some Jesus's followers hallucinated". They are not saying, "I think there were groups of people that if we could interview, they would say, Oh yeah, I totally like poked through Jesus's holes" No, they are still saying that the accounts of Jesus's resurrection appearances in the Gospels and Acts are a literary creation, but they would say that the reason why they are there fundamentally goes back to some people having some kind of hallucinatory experience of Jesus and that grew over time. It got exaggerated, distorted and eventually this is what you end up with in the Gospels. So that's not even dealing with the actual hallucination hypothesis.
James: Yes, exactly. And apologists consistently make this, what I regard as a clear mistake. They think that hallucination explanation, plus memory, contagion and whatever else, has to explain the accounts as they appear in the Gospels, which it absolutely does not have to do, because that is assuming that the skeptic just takes for granted at face value, everything that it says in the Gospels. Whereas no, that's not what the skeptic, that's not what those 75 percent or whatever percentage of scholars who think that, you know, the disciples believed that they saw Jesus. That's not what most of them are saying. They say there was some sort of experience that the disciples had, which they interpreted as Jesus appearing to them. Not necessarily the same as anything that we see in the Gospels other than in the very broadest outlines of, you know, they saw Jesus. They thought they saw Jesus. But repeatedly, the evangelical apologists keep saying, Oh no, they can't explain all the details. We see that on the next page, where they start saying, I know, but like they gave food to him and they cooked fish. Like, how could they have? How could they have hallucinated that?
James: They say "then abruptly, they stopped". That is the resurrection appearance is stopped as Christ was no longer appeared on Earth and subsequent appearances were sort of different in character. Now my question is, first of all, why is it expected under the resurrection hypothesis that the appearances would stop? Jesus has been raised to life. He's back. He could be around for years, decades. I mean, forever if God wanted to be around forever, right? Like, why is it expected that they would stop after how many days it was? Actually, I think it's very expected that we would see this under a naturalistic hypothesis because and I can't remember if I actually talk about this in my book or just on my blog somewhere. But new religious movements that involve prophecy and revelation almost always fairly soon restrict the prophecy and revelation to particular people and particular time span. And the reason for that is because otherwise you can't keep control of the development of doctrine and who has power and prestige in the movement. There are case studies that have been done of this. There's almost always restrictions placed on what counts as a genuine revelation, either based on the person or at the time or something like that. So this is exactly, I think, what you would expect for a newly developing religious movement that entirely has nothing to do with God is that you would have some sort of cutoff point or criteria of what counts as a resurrection appearance. And this apparently Pentecost, was what developed in the early church, whereas it's not clear why you would expect anything like that if God actually rises from the dead. So again, they don't even seem to think about why something would be explained by their preferred resurrection hypothesis. It's just assumed that everything is explained. And then they ask, How do you explain this?
## On Zeus
James: How do you know that Zeus would have no motive? Yeah, so poor old Zeus, he's just sort of left out in the cold there. I don't know why he's ignored so readily.
James: So they seem to be saying here that God wouldn't have provided an objective vision because that would confuse the disciples in thinking that he had resurrected. Which would be like deceiving of God? I really don't even know that the early disciples would have had the ontological, would have drawn the ontological distinctions that we are making here between an objective vision and a bodily resurrection. Maybe they would have, I don't know.
Kamil: That goes back to what I'm saying. So if you look at the development of early Christian literature, then all the way up until the gospel of Luke, these appearances are actually like true visions, right? They are not appearances of a body that's walking around, talking to people. At least that hypothesis is consistent with the account. And this distinction between a supernatural veridical vision and an actual body being there that you can touch in the room is first introduced in the gospel of Luke. I want to emphasize again that in the gospel of Luke in that post resurrection account, it specifically says in Luke that Jesus ate a piece of fish so that the disciples wouldn't think that he's a ghost. Right? Yeah, I think that's protesting too much. I think that's a counter apologetic. That's why the author takes care to mention that.
James: And my point was that under an assumption of a true resurrection and divine inspiration of the Gospels, would we expect to see these sort of developments comparing Mark and John being sort of the most obvious and the theological flourishes that you see in Matthew and in Luke about all sorts of different points -- guards at the tomb and the people walking around Jerusalem and the mention that they touch Jesus, that Jesus ate fish so that they would know he's not a ghost. These things seem much more plausible under the idea that these are particular debates and theological concerns that the authors had at their time that didn't exist in in the time of the previous gospel authors and that they wanted to put in otherwise. Like, why wouldn't you just start off with the higC christology that you have in John and all of the relevant details and heck put in the Nicene Creed in there, just just so that God can have his bases covered and make sure that no one misunderstands -- this goes a bit beyond, I guess, the point being made here, but I just don't think we would expect to see texts like this under the assumption that Jesus was raised from the dead.
James: I would like some citations about this and what sort of mental illnesses are necessary for these sort of delivery effects, because as this is, I think, incorrect hallucinations actually quite common. And I don't know that there's any evidence that they can't be integrated into other experiences. They're not necessarily associated with mental illness, either. They're quite common in the healthy population, so they're just making things up at this point. This is apologetics through making things up. There's no citations or explanation of that there at all. And they also don't know that the conditions, like the conditions that they are referring to, are absent. How do you know that those weren't present? Yet again, just making it up?
## Single Disciple Bayes Factor
Time stamp 6:56:09
James: They say the probability of a single disciple reporting experiencing Jesus under the resurrection relative to not-resurrection is one in a thousand. So three orders of magnitude.
Nathan: Where does this number come from? What if I just go, Yeah, but it just seems to be different like -- what can we appeal to?
James: They can deplore your reasoning, but they can't say you're wrong. So the point is that they're taking a thousand is the Bayes factor for a single disciple. And they're saying that there are 13 of them who saw Jesus. So if you multiply 10 to the three by itself, 13 times you get 10 to the thirty nine, which is that that number there. So they're building up to the 44, that is. We get we get 10 to the two I think from the women, ten to the thirty nine from the disciples and I guess the rest comes from Paul. So there's a whole separate subsection the independence. So we'll discuss that when they discuss it.
## Conversion of Paul
Time stamp 6:58:54
James: But the question we're trying to address here is what the Bayes factor is. How expected is Paul's conversion under the resurrection, given how expected is it or how likely is it under not-resurrection? Now, I don't know that anyone has a very good explanation for Paul's conversion. The question is though, how probable is it under the resurrection? Why would you expect Paul to be converted given the resurrection? What is the connection there at all? They don't even discuss this. They don't mention it.
James: "On the assumption of $$R$$, there is no difficulty whatsoever accounting for $$P$$", which is Paul's conversion. But what does the resurrection have to do at all with Paul's conversion? God could appear to Paul if he wanted to irrespective of whether he was resurrected. Nor do we have any reason to think that God would have any business appearing to Paul. Why would he want to do that instead of appearing to anyone else or groups of people or no one at all? There's no explanatory value here at all. They have nothing to say.
## On the Bayes Factors
Nathan: You know, sometimes I'm thinking when we say they've literally pulled the numbers out their ass, that that sounds a bit harsh, but where has this number come from other than that? what belief do I hope with like 10 to the minus four?
Kamil: You know, like let's say that there is some mundane claim from ancient history that we have a very well evidenced even maybe even better than what we have for the New Testament, right? Let's just say that it's a mundane claim that Cicero makes it one of his letters where he literally wrote the letter like moments after it happened. And that same claim is also understood by a bunch of historians who covered at the same time period somewhat later. Like, would they also say that the likelihood ratio is this astronomical in favor of that event actually happening? It's way too high. Contemporary historians of antiquity are not this confident even about very mundane claims in the historical record, because they realize that even if it's super mundane, even if it's relatively well evidenced, they realize that all kinds of things going on that are separating us from the actual historical events 2000 years ago so that we don't have any information on things like distortions of information in transmission and stuff like that that it's just not appropriate to be super confident, especially especially not astronomically confident, like this is the level of confidence, which I think is not even warranted when it comes to the findings of in like experimental science, like experimental physics. Because even in experimental physics, there are all kinds of things that could still go wrong with the equipment that the machinery and stuff like that, right? And even if you have multiple people, it like different labs around the world confirming the same event, it could still be false positives, right? So that's I think that's a very powerful reductio, at least against what they are doing in the paper with these specific numbers.
James: Take something that's sort of overwhelmingly, I don't know like that two jet aircraft flew into the Twin Towers on September 11, 2001. That event, right? What Bayes factor would they give for that happening? I mean, it's hard to know what it could be, because the evidence we have for that is so much greater than the evidence of the resurrection. The number that would have to give is just completely obscene, right? And if you think about claims even from empirical sciences, Kamil says, I wouldn't I don't think I'd give 10 to the 44 of anything, right? That is just so ridiculously high. Remember, 10 to the 12 is a trillion. So like one in a trillion against you could think of it as. And that is a very, very high confidence. It's hard to emphasize how high 10 to the 44 is as an odds ratio. And the fact that they are giving that to an inference event that no one, there's not even testimony of anyone seeing this, it's an inference that would give to explain a bunch of facts which are tendentious. To think that you can get even close to 10 to the 44 is just ludicrous.
Time stamp 7:12:45
James: So they're saying that the probability of the evidence is that we mentioned, occurring under not resurrection is equivalent, relatively speaking, to someone winning \$100 million ticket lottery more than five times separately, independently of each other. That's the level of unlikelihood they're claiming we as the not-resurrection people have to swallow.
Nathan: I don't even know that I would be that uncharitable to the resurrection side of it.
James: I mean, I would give a higher probability of young-earth creationism being true. 10 to the minus forty four, for goodness sake, that is so ridiculously low.
Because they don't make any attempt to compare to other models, other situations, they uncritically accept a Bayes factor of $$10^{44}$$. It really is absurd -- nothing, even in the physical sciences, rises to this level of confidence and they don't seem to bat an eye.
Part of this I think is a defense mechanism, so they don't have to deal with the super-low priors for the resurrection -- if the likelihood ratio is off the charts, then it will overwhelm nearly all priors. Lydia McGrew makes this case in a video response about setting aside priors.
Nathan: here's why the methodology they've used is crap and shouldn't be used by people when they say how the compounding of independent pieces of positive evidence can rapidly create a powerful cumulative case, even for highly controversial claim. So think about the gremlins in my closet. So what's the case? There's noises coming from my attic. Given that there are gremlins in my closet. And now take what would be some other piece of evidence as well. There's also a leak coming from the attic. And these gremlins are like urinating and stuff up there, which is causing a leak, obviously. So that's also really expected. And so and then let's say there's four or five different leaks in the attic. So let's just multiply each of those together because they're each independent lines of evidence, right? And just keep. And this is how the Bayes Machine goes Brrrrr. And that I think just is demonstrative of this being a really bad way of trying to use Bayes Theorem to compare true beliefs about about something.
The method Nathan outlines here is structurally the same as the McGrews. You ignore priors, have lines of evidence which you claim to be independent and support your supernatural claim, multiply all of them together to get a really huge Bayes factor. It's completely naive.
In addition to all of the other failures, they also don't recognize that this naive accumulation of evidence doesn't actually work in multiple-model comparison, where you can get non-monotonic effects. This occurs in some very simple cases and is easy to see, but they never seem to address them because they fail to have any serious model of how their data is produced. There are other subtle effects even beyond that that can make more testimony even worse, when you include the process of taking seriously how we update our knowledge.
## Independence
Time stamp 7:29:56
James: They don't actually say that they're necessarily independent, as they say here. In the quote that I've highlighted here, the force of the case is actually underestimated as a result of the independence assumption, so they are dependent. But in a way that actually increases the odds ratio, so they're actually being conservative in assuming independence. Now that is an interesting one to get your head around. Let's think about what that means. What they are saying is that. Suppose you came to believe that one disciple, let's say, Peter, had an experience of Jesus and reported seeing him and so forth and being willing to die for him. They are actually claiming that it is less likely that another disciple say John would also have reported an experience of Jesus and be willing to die for him. They are telling us based on this statement here that I've highlighted that. Believing that one disciple came to experience Jesus makes it less likely than another one did. That is the content of this claim here now. Riddle me how plausible you think that is because that is such a ridiculous claim. It is so stupid. I can't believe that they would actually say this.
Here, I think that they are leaning on the willingness to die causing a reduction in the reporting of the second disciple.
James: So they're saying that basically, once one disciple heard about, you know, say, Peter being martyred, that a second disciple would become less likely to stand firm in their testimony.
Kamil: that depends on what's the background knowledge on martyrdom and how it was perceived. Because it's clear from early Christian literature that Christians actually didn't perceive martyrdom as something negative or something that they should avoid. They actually, like looked forward to it. They didn't mind this. This is something that you find consistently in the early Christian literature, Christians were actually happy about being persecuted. [...] And I think that makes sense because they believe that if you are martyred that means that your salvation is secure. [...] They would basically commit suicide by cop by handing themselves in massive numbers in some cases. There are reports of entire crowds of Christians walking up to the Roman procurator and demanding to be martyred for the Christian faith.
James: There's a second argument which honestly just doesn't even make any sense to me. Like the Cold Feet argument makes sense. I just think it's implausible and under motivated. The second argument that they gave, I just don't understand.
Kamil: These are exactly the points in the argumentation where they could very easily look at what's actually happening in the world with religious people and actually get some data, get some background knowledge about how people function under these circumstances. But there's no discussion about it here.
Time stamp 7:51:07
James: I want to ask, did these disciples not think that God would protect them because it seems very plausible to me that they would. How many people do we see today who think that God will protect them from coronavirus and therefore they don't need to get vaccinated, even though there's no evidence for that at all in any of their texts, they just making these things up. But but these people die, right? They don't take precautions and they get it and some of them will die and have died on the basis of a belief that they've just plucked out of nowhere, right? That God will protect them of this sort of thing. So why wouldn't something like that not happen in the case of the disciples? We see that in other religious traditions as well that people believe God will protect them. People believe that they killed themselves, and then they're going to go up to the UFO that's coming to resurrect and save humanity from the asteroid or whatever it is in some of those cults. Why are we expecting that death is some sort of big disincentive?
James: The distinction they make between an empirical fact seems to be the sort of thing that philosophers debate about but ordinary people don't make a distinction between something that they know is true for whatever reason and something that they've seen to be true. It's just, look, they know that Jesus is Lord, right? They may have seen him, and that's part of how they know. But it's not the fact that they saw him itself. I think that part of it, it's that they knew that Jesus was Lord and that he'd risen and all that just as Christians today. That's what motivates them, irrespective of whether they think that they've had an experience. I mean, some Christians will say that they've seen something or heard something else. It's not clear that that itself is what's driving them, but they McGrews's want to tell us that that is absolutely critical and that no one would die for claiming a particular empirical fact to be the case unless they had really good grounds for it. But people will die for ideological beliefs that they don't have a good ground for. We have not been given any reason to think that that's the case or that that distinction meant anything to the disciples. So until I get some reasons for thinking that the distinction matters or was irrelevant to the disciples, I'm just going to say, what difference does that make?
## Hume
Time stamp 7:58:21
James: if you want to interpret in a Bayesian terms, I don't think you have to but can, it's to say that, as we said before, if the prior was very low, you need a very likelihood to overcome that low prior. I think that that is a very profound insight, even if intellectually, we all know it -- as the doctor's case that I mentioned and many other cases -- people neglect base rates and so we're actually not good at doing this in a lot of cases.
Nathan: But the interesting thing is that the base rates are completely taken out of this text.
James: Well, that's the irony.
James: They say they say here that Hume was wrong to make in principle arguments against the resurrection because as they say here, that you need to "leave the high ground and descend into the trenches of engaging in specific historical events." Well, the thing is apologists actually, never do this or almost never do this for other religious claims, though. They only look at Christianity, and maybe there's one or two others that come up occasionally. The point is, though, that they love to focus on the details. They'll write thousands and thousands of pages on all the details of Christianity. But if you ask them about, say, Mormonism or all the other Messiah claimants or Hindu miracle performers and all sorts of other cases that I document, they have very little to say about them. They don't descend into the trenches. They don't go through the details of that again, most of them most of the time.
Nathan: no one should do as well. I hate it when people frame the sort of discourse in this way where it's like, well, for any crazy claim that certain groups of people have really committed to, expended all this intellectual effort on constructing like Martin Bailey defenses that you can spend your entire life trying to deconstruct and you were like, get through that. So if anyone makes any of these claims you've got to, then you know, like you've got to get a degree in like biblical scholarship, right? You've got to learn the original Greek. [...] I don't think anyone should reasonably have to do all this just to be able to conclude that they don't think someone rose from the dead when there's nothing in their experience that indicates it can happen [...] Sure, there's room for a more detailed analysis, but it's just so unfair to put this expectation on normal people.
James: Yes, I agree, and it's one that they won't bear for other belief systems as well. So it's entirely it's entirely self-serving.
Although they admit they don't accept this Bayes factor, James does provide a way to make the prior lower than that -- even accepting that God exists, wants to save humanity, and can do miracles.
James: this is something I just sort of thought of on the on the fly. So the argument is something like this, let's suppose that I believe God exists, and let's suppose that I believe he's going to take some sort of intervention to help with saving mankind. Now, I don't know why you'd think those things, but let's just stipulate them for the sake of argument. Now, I have no additional beliefs about what that action might be. Now, God can do anything that's logically possible, right? So the set of things that he could do to save mankind, which would include resurrecting any particular human who has ever existed, resurrecting any animal who's ever existed, resurrecting multiple people, performing any arbitrary miracle, not doing any miracle, moving the planets around like anything you could imagine. I mean, God could work that into his plan to make that serve his purpose, right? I mean, he's God. So the set of things that he could do to serve his purpose and give a message to mankind is effectively infinite. The set of things that the Christian is postulating that God actually did is just one, he raises Jesus from the dead, right? So we've got a prior of one over "infinity". (You know, in quotes, because obviously that's not a number.) So basically, that's a zero prior for or close to zero, as makes no difference for the probability that Jesus would be resurrected. And that's given that I believe in God and that God would intervene, but not having any particular belief about what God would do. So that's my reason for thinking that it's lower than 10 to the minus 43 because I'm just indifferent between any of the possible things. Your mileage may vary with that argument. I'm not sure I entirely believe it myself, but the point is that is it that hard to come up with reasons as to why it would be extremely low? God has no constraints on him. If I'm a Muslim and I believe very strongly that Jesus was only a prophet and that God wouldn't resurrect him. Maybe that gets me to 10 to the minus forty three, right? Hmm. But you just need a bit of imagination.
## Plantinga's Dwindling Probabilities
Time stamp 8:21:54
James: But what Plantinga is saying is that in order to get to like a belief in Christianity, you have to proceed sort of step wise. You have to start with like God exists. And then there's like conditional on God exist. Oh, God, existing God would want to make some sort of revelation of himself to mankind and then conditional on God wanting to make a revelation himself. [...] His point is that because those are all probabilities they all have to be less than one. If you multiply a bunch of numbers less than one together the number diminishes. I think it's illustrative to point out that even if you start with fairly high numbers, if you take point nine, for example, they're going to raise it to the power five, like point nine to the power of five. If you have five independent things that are all 90 percent likely you get, point five nine four. So obviously if it's less than point nine, then it diminishes more quickly. So you can quickly go from highly confident to actually not that confident, even if each step is fairly confident by having a bunch of steps. That's the key point Plantinga is making there, and I think that's a very good argument to be is sort of suspicious of this stepwise inductive or, I guess, Bayesian type of arguments.
James: In fact, this is exactly what you do from scam emails. I look at it and say, what is the probability that this person exists and conditional on that, what is the probability that they've sent me this email? Because as like, for example, if I've got an email that said it's from Plantinga, I know that Plantinga exists, but I would assume that it's not actually from Plantinga because he's probably not emailing me, right? Whereas if I got an email that says it's from Elvis Presley, I would say, Well, it's probably not from Elvis Presley, right? The fact that I've got an email claiming to be from Elvis Presley isn't going to overcome my low prior that Elvis Presley is alive because he died a long time ago. So it seems to me that although we may not explicitly do this in many cases, that is actually the appropriate way to do it is to consider separately, at least like conceptually separately, the probability that the person sending you the email actually exists like the Nigerian prince. And then given that that person exists, what is the probability that they are sending you this email?
James: they're saying that we don't have priors over people that we meet, basically. And I just think that's wrong. I mean, whether we mentally have something like that is another question. But the question would be like, is there a way that you could appropriately assign priors to make sense of inference? And I think it absolutely does. And that is how I distinguish between people claiming to be someone that I think is plausible and people making claims that are not plausible, right? What if you've got no way to make the distinction? You can rightfully describe that as a prior because it's not based on any specific evidence that you have about that person. The prior would be with respect to the factors prior to your particular interaction with that person's background knowledge, that would be relevant like, for example, someone claiming to be Elvis Presley. The fact that I would disbelieve isn't based on how persuasive they are or like, how good their outfit is or something like that. It's based on beliefs I have about Elvis Presley, which is prior to that particular meeting. how else could you describe that right? The McGrews want to say that that just isn't the thing. But I don't understand how they then make these distinctions. How do the McGrews explain the fact that sometimes we don't believe people when they tell us who they are or when they email us or something like that?
James: So they're complaining that Plantinga is pulled out this 0.9 from nowhere and is claiming that it's generous. Yeah. Now does that remind you of anything? Seriously? They have the absolute gall to complain about that after what they've done in this whole article. It's unbelievable.
## Some Final Thoughts
Time stamp 8:56:04
James: it's disappointing because in principle, Kamil and Brian were talking about this, but in principle, I think you could use Bayesian techniques to elucidate some of the key points in the argument here. If it was done well and carefully, but it's unfortunately not. And so maybe I'll just quickly highlight some of the key aspects of this without going through everything we've discussed it that I find most problematic.
One is the fact that they are very unclear about what the methodology is, that they're even trying to apply here. They talk about Bayesian techniques and they talk about the likelihood ratio. But then they, as we mentioned many times, keep flipping back to talking about the low plausibility or the prior probability of many of the things that they don't want to consider that, you know, they're $$\sim R$$ hypotheses rather than resurrection, whereas they don't address that at all with respect to God or the proposition that God would want to raise Jesus from the dead. But it's not clear why they should be persuasive to anyone else if they don't actually present objective considerations like models or base rates or other things. So that the whole methodology is, I think, very unclear.
A second key problem is that the arguments that they give in favor of the key facts that that's the women and the disciples and then the Pauline conversion just relies on basically taking the the New Testament at face value and accepting every claim that it makes. And if you're going to do that, just read the part where it says Jesus is God and was raised from the dead and be done with like it's so much more honest and quicker that going through all of this, no skeptic is going to acknowledge all of these things as being veridical and all of the original authors and based on eyewitness testimony and all this sort of stuff that that they are leaning towards or explicitly arguing for and then make really, I think, problematic appeals to very selective scholarship and pooh poohing the textual analysis, critical analysis stuff that they don't like, but then appealing to it when it suits them.
So the third point here is that they when considering the relative explanatory power of resurrection versus not resurrection, they often didn't consider many not-resurrection type arguments. So many of those sorts of things that say Kamil and I would advocate for weren't discussed somewhere were but a lot weren't. And even more significant than that, They didn't really say much of anything about the actual explanatory power, like the probability of the evidence given the explanation. They mostly just said the naturalistic explanations were improbable, like low prior, which doesn't say anything about their explanatory scope, while at the same time saying almost nothing about the explanatory power (the probability of the evidence given resurrection) about their own proffered explanation. This is most clear, I think in the case of Paul, where it's completely unclear how a resurrection is supposed to explain why Paul was converted because they just completely separate from each other. God could appear to Paul, whether or not Jesus was resurrected, and why would God want appeared to Paul anyway? Like, what does that have to do with anything? They just don't address this so that they're really bad at actually trying to explain how their explanation is a better explanation, how their model is a better explanation than competing explanations.
And I think the final point that I wanted to make is the issue of independence, which is absolutely critical to their argument because most of the, really all of it comes from multiplying low numbers together or high numbers, depending on which way you look at it, and they justify that by the independence assumption that the women's testimony is independent of the disciples testimony, which is independent of Paul. And then furthermore, that all of those 13 disciples testimonies are independent of each other, and the justification for this is just ridiculous. Basically, they say that it's actually independence overestimates the evidence against them, like it underestimates the strength of their case because first of all, the disciples would get cold feet or be likely to get cold feet upon seeing one of the disciples be killed or martyred for their cause that that would actually cause them to be more skeptical than they otherwise would. And of course, they cite no evidence for that. The other reason that they gave is just that the best explanation for why all of the disciples became convinced is because they actually had good evidence, which I don't even understand how that's an argument -- it's just reasserting what they believe. So the point is one of the absolutely key assumptions behind their entire approach, which is the independence of these key facts here is just defended in a, I would say, pathetic way, like the arguments are just really ridiculous, and that undermines the entire premise of what they're doing.
James: So overall, it was difficult to read. It was really frustrating. It was too long. The arguments were mostly quite bad. They don't fulfill their promise of actually comparing the explanatory scope and the conditional probabilities of the resurrection versus other hypotheses. And this is a consistent failure of these sort of apologetic studies that claim that they're going to make a comparative analysis but just do a really shoddy job of it and don't seriously consider what skeptics would actually believe and what arguments they actually make. And don't try to give actual reasons for why the explanations that skeptics offer are bad. They just sort of assert that they're improbable and not take it seriously and not try to argue why their explanation is better. So, yeah, I think it's really poor scholarship, and I hope that you found some of our remarks to this effect for the last nine hours of interest.
|
{}
|
next: Validation
## Definitions
In CUE, schema are typically written as Definitions, using a :: instead of :. This tells CUE that they are to be used for validation and should not be output as data; it is okay for them to remain unspecified.
A definition also tells CUE the full set of allowed fields. In other words, definitions define “closed” structs. Including a ... in struct keeps it open.
schema.cue
Conn :: {
port: int
protocol: string
// ... // uncomment this to allow any field
}
lossy: Conn & {
port: 8888
protocol: "udp"
// foo: 2 // uncomment this to get an error
}
\$ cue export schema.cue
{
"lossy": {
}
|
{}
|
# zbMATH — the first resource for mathematics
Introduction to Shimura varieties with bad reduction of parahoric type. (English) Zbl 1148.11028
Arthur, James (ed.) et al., Harmonic analysis, the trace formula, and Shimura varieties. Proceedings of the Clay Mathematics Institute 2003 summer school, Toronto, Canada, June 2–27, 2003. Providence, RI: American Mathematical Society (AMS) (ISBN 0-8218-3844-X/pbk). Clay Mathematics Proceedings 4, 583-642 (2005).
This article provides a survey of several important concepts related to Shimura varieties with parahoric level structure at a prime $$p$$ by using the Rapoport-Zink local model as the main tool. In particular, it discusses local models attached to general linear groups and symplectic groups. Their relations with Shimura varieties with parahoric level structure are illustrated by describing two examples, namely, the simple or fake unitary Shimura varieties with parahoric level structure and the Siegel modular varieties with $$\Gamma_0 (p)$$-level structure. The article also includes some applications of local models to problems involving flatness, stratifications of special fibers, and the determination of the semisimple local zeta functions for simple Shimura varieties.
For the entire collection see [Zbl 1083.11002].
##### MSC:
11G18 Arithmetic aspects of modular and Shimura varieties 14G35 Modular and Shimura varieties
Full Text:
|
{}
|
## Wednesday, April 30, 2008
### AAAC in Hong Kong
I just came back from an all too short trip to Hong Kong for the first annual meeting of the new Asian Association for Algorithms and Computation. The AAAC would like to become the Asian version of SIGACT and EATCS. The conference was a good start but dominated by the Japanese and needs in future to draw researchers from across Asia. I went as a speaker, but also as a supporter as I would love to see theory grow around the world and while East Asia has produced a few great researchers, it has not even come close to reaching its potential.
This was my first trip to Hong Kong and the three dimensionality of central Hong Kong is quite striking with many tall buildings built on various points on a hill and in some cases seemingly on top of other buildings. To get from my hotel to the conference at Hong Kong University, I took a series of outdoor escalators, crossed a bridge and then an elevator followed by some stairs. You need to keep track of elevation to get around that city.
I had never been to China. Did this trip to Hong Kong count? Technically yes, since 1997 Hong Kong is officially a Special Administrative Region of China and shares much of the culture and cuisine of China. But a different currency, visa requirements and economic structure makes it seem like a separate country. Someday I will get to mainland China and make this point moot.
## Tuesday, April 29, 2008
### Should Mahaney's theorem be taught in a complexity grad course for non-theorists?
Today we discuss another theorem in terms of should it be taught in a basic complexity course (taken mostly by non-theorists) (There was an earlier blog about this for PARITY ¬in AC0.)
Why is SAT &lem S, S spare , implies P=NP interesting? important? (Henceforth Mahaney's thm.) I'm not trying to convince you that it is, I am asking if it is. Here are some thoughts.
1. The original motivation is the Berman-Hartmanis conjecture that all NP complete sets are poly-isom. Mahaney's thm is a consequence of the conjecture. One could do the BH paper and show why it is plausible and then give this result. But is it worth it?
2. The result is a stepping stone to Karp-Lipton's result that SAT &leT S, S sparse, implies PH collapses. This begs the question- why is KL important? Because it is a stepping stone for Yap's result that SAT &isin coNP/poly implies PH collapses. And why is that important? Because it is used in the proof that if GI is NPC then PH collapses. This is good enough for me- evidence that a natural problem is NOT NPC-- surely worth knowing. But do we need to present Mahaney's result to get to Yap's result?
3. Should point out that KL is also interesting because it is a link between uniform and non-uniform complexity. But again, perhaps we could do that without Mahaney's result.
4. The techniques used to prove Mahaney's result are interesting and lead to other theorems of interest. Like what? Well, ur, the Ogiwara-Watnabe result which replaces &lem with &lebtt. And the result of Lozano that generalizes this to other classes like MODaSAT (number of assignments is &equiv 0 mod a). Why are these of interest? I have an intuitive sense that they are, but I can't even really say why theorists find it interesting. For that matter, do theorists find it interesting? (This was discussed in this blog entry, though the discussion was derailed by someone asking an off-topic question.)
## Monday, April 28, 2008
### What would the best base be?
(A partial continuation of the last post).
We use Base 10 because we have 10 fingers on our hands. But if we could pick a base based on what is better mathematically or computationally or some objective criteria, what would it be?
1. When I was young I thought that if we had always used base 8 then computer science would be easier and computers would be faster. While partially true, not MUCH easier or MUCH faster.
2. In 1934 there was an article with title An Excursion in Numbers, by F. Emerson Andrews, in The Atlantic Monthly urged abanding Base 10 for Base 12. (Yes- the The Atlantic Monthly not The American Mathematically Monthly. I'm surprised too.) There are some advantages- 12 is divisible by 2,3,4,6 and since 12 is used for eggs there may have been some reason for it. The Duodecimal Society advocates changing to base 12. They have (or perhaps had - I could not find it on the web) a newsletter The Duodecimal Bulletin, which is translated into one other languauge and has the title Ekskurso en Nombroj. I'll let you figure out what language that is. (ACK- this info comes from Mathematical Cranks by Underwood Dudley.)
3. Picture that you want to represent every number between 1 and n. Lets say its in base 10. In an adding machine (whats that?) you would have log10 n columns and each one of them has 10 keys. So the total number of keys you need is 10log10n. More generally, if its base b then you need blogb keys. What value of b minimizes this? The answer is e. Since we can't use e for normal counting, this does indicate that 2 or 3 would be best. Since 2 is also good for computer science, my vote goes to using base 2.
4. To end where we began this- I wonder how Obama, Hillary, and McCain would vote?
## Friday, April 25, 2008
### If we had 12 fingers on our hands then Obama would be the nominee
dits have said the following (paraphrased):
Hillary needs to win the PA primary by double-digit to get back in this race. (She ended up with something like a 9.2 or 9.4 advantage depending on who you ask. She rounds up to 10, he rounds down to 9.)
What if we had 12 fingers on our hands? Then we would use a base 12 system and she would not be close to the magical double-digit lead.'' Would she drop out? No, but the win could not be spinned as dramatically.
Pundits and others do not realize that base 10 is arbitrary and is not connected to anything interesting mathematically or politically.
Hippies used to say Don't trust anyone over 30 without realizing that they had given in to the establishments insistence that base 10 rules us.
Its been said 50 is the new 40. Why 50 and 40? Should be 49 is the new 36 since squares are ind of base. (Is 100 is the new 81?)
The Beatles had it right with their song When I'm 64.
A while back this blog noted its 1000th entry. Mistake- we should have noted its 1024th entry.
## Thursday, April 24, 2008
### The Life of the Party
Being a Math/CS professor is generally the kiss of death at any large social event. But thanks to the movie 21, for one short moment, I was the center of attention.
I haven't seen the movie yet, but apparently there is a scene where an MIT Professor (played by Kevin Spacey) uses the Monty Hall problem to help choose his blackjack team. So I helped explain why it makes sense to switch doors.
The movie had more math, basically simple card counting techniques to give an advantage at the blackjack tables. Any movie like this that glorifies mathematicians help our community, even if they just use math to win money at casinos.
In fact, our family was invited to another party earlier this week where the guest of honor was one of the members of the original blackjack team that the movie was based on. Being a mathematician is cool again, at least for a couple of weeks.
## Wednesday, April 23, 2008
### What Happened to the Indians?
I lamented to some Indian colleagues that we had no IIT applicants to the CS theory graduate program at Northwestern. Chicago and many other US institutions drew many of their best students from the Indian Institute of Technology campuses over the years.
I suggested that Northwestern was not yet on the theory map, at least in India. Likely true, but in addition the number of IIT CS majors going to the US for Ph.D.s has dropped by about two-thirds over the last couple of years. The culprit: Large, mostly US, banks are hiring the top graduates at salaries extremely high by Indian standards to work in their India offices. We had seen a smaller drop earlier with the software industry hiring but the software doesn't pay nearly as well as the banking industry. The Hindu writes about this trend.
I'm happy for India's success but worry about the impact on US science and CS theory in particular. You don't have to look far at the best theorists to see a large number of Indians, mostly IIT alumni. Imagine if most of them ended up as bankers in Mumbai. What a loss!
Back in the US, where do we get our graduate students from now? The Israeli's have long since stopped coming here, now that they can get quality Ph.D.s in Israel. Most Europeans also stay in Europe. We've also seen a drop in Chinese applicants. The US needs to start developing new sources for foreign students, or maybe, just maybe, find a way to attract more Americans.
## Tuesday, April 22, 2008
### Laptops in classroom and lectures
More students are bringing laptops to class. More faculty are bringing laptops to talks. Is this good, bad, or ugly? Some points
1. I was sitting in on the best teacher in my dept (Dave Mount) teaching an elective course (so students there wanted to be there) on how to write video games (a topic of interest). Many of the students in the class were using their laptops to surf the net.
2. Another professor has banned laptops from his class. If a student claims they are taking notes on it, as 5 did, then he demands that they email him the notes (only 1 took him on it).
3. Is this any different than students doodling or gazing out the window or other ways to distract themselves?
4. Since attendence is not mandatory, why insist that they not have laptops? I am not asking this rhetorically--- I am tempted by the idea of banning laptops also.
5. Professors at talks also bring their laptops. We have not developed a culture where this is considered rude. Not clear why we haven't.
6. Are today's youth better at multi-tasking so that they can do two or more things at once, like surf the web and listen to a talk? Again, I ask this non-rhetorically.
7. I have no strong opinons here, but I want you to write your so I can borrow them next time I am feeling argumentative.
## Monday, April 21, 2008
### Ketan Mulmuley Responds
Someone pointed out to me your post where it is stated that, according to me, any approach to separate P from NP must go through GCT. This is not what I think or said. One cannot really say that GCT is the only way to separate P from NP or that any approach must go through it. Indeed as the article (On GCT, P vs. NP and the Flip I: A High Level View)—henceforth referred to as GCTflip—which describes the basic plan of GCT, clearly states: GCT is a plausible approach to the P vs. NP problem. But as it also explains there are good mathematical reasons to believe why it may well be among the "easiest" approaches to the P vs. NP problem.
One such argument—the zero information loss argument—was presented by K.V. in his talk. According to it, any approach to separate the permanent from the determinant in characteristic zero must understand, in one way or the other, the fundamental century-old problem in representation theory, called the Kronecker problem, or rather its decision form. (though this understanding may be expressed in that approach in a completely different language). This is what I repeated during the lunch after that talk, and this is perhaps what the post is referring to.
The only known special case of this problem which is completely solved is the Littlewood-Richardson problem. The most transparent proof of this (which also provides far deeper information regarding this problem needed in GCT, unlike other proofs) goes through the theory quantum groups, and the only known good criterion for the decision version requires the saturation theorem for Littlewood-Richardson coefficients. GCT strives to lift this most transparent proof to the Kronecker problem, and more generally to the generalized subgroup restriction problem (and its decision form), which is needed in the context of the P vs. NP problem in characteristic zero.
All this is explained in detail in the article GCTflip mentioned above. It does not assume any background in algebraic geometry or representation theory. It has been read by the computer science graduate students here. They had no problem reading it. But it does need a month. It is my hope that you would spare a month sometime for the sake of the P vs. NP problem.
## Friday, April 18, 2008
### Facebook and Forums and Feeds, oh my!
Facebook and Forums and Feeds, oh my. (Upon seeing Lance Fortnow's facebook post Bill Gasarch asked Evan Golub, who does research in Human-Computer Interaction and educational technologies (though his Ph.D. involved Expander Graphs) to do a guest post on facebook. This is that post.)
Lance recently wrote wondering how he would use a Facebook page with his course. I should start by saying that although I have a Facebook account, I don't really use it - I signed up for it to look around and to decide whether I wanted to start using it and haven't decided on "yes" yet.
In thinking about Lance's question, my first question was whether he would create a Facebook group for his class, or create an actual user and name it after his class. Depending on what notification options work on a group -vs- work on a user might guide this decision. For example, if one of these allows other Facebook users to receive a notification when the wall is written on, then it might become the better choice.
My next thoughts on this relate to Internet-based course management ideas that could be done via other technologies, but might be possible using Facebook instead. So, what could Lance do with a Facebook account/group for his class that could already be done with forums? He could provide a way for students in the class to:
• ...get in touch with each other and find out about each other outside of the classroom
• ...from study groups during the semester
• ...post links to useful resources related to class topics
• Next, what could Lance do with a Facebook account/group for his class that could be already be done with an RSS feed? He could have a way to let students know when he has:
• ...posted a new assignment
• ...updated the grades posted online
• Assuming that anything that could be done via Facebook could also be done using forums or feeds or other technologies, why use Facebook rather than web forums or RSS feeds or other tools at our disposal? To borrow an idea from Alexandre Auguste Ledru-Rollin, one reason might be "because that's where the students are, so if we want to guide them, that's where we should be".
However, the above is more the reason why I have not used Facebook with my courses yet. I see Facebook as a place where students go to socialize, not to do classwork. I recall reading when I was a student that you shouldn't do homework in bed or your brain might have more trouble turning off thoughts of schoolwork when you are trying to go to sleep. I don't know whether there is research to back up this perception that I picked up somewhere along the way, but if so, then perhaps we should ask whether it would be better to keep our courses off of Facebook (unless we are teaching a course that covers social networking as a topic). If this is meant to be a social space for students, would we be infringing upon this by bringing our courses there?
As an (essentially) non-Facebook user, there might be some uses that would be unique to Facebook (or similar social networking sites) of which I am unaware. This "reply" to Lance (prompted by Bill) is meant more to open what I see as a central question raised by Lance's question of "what to put up there" (see his original post).
## Thursday, April 17, 2008
### My First Grand Student
Today I am in Madison, Wisconsin where Scott Diehl has just defended his thesis. Scott's advisor, Dieter van Melkebeek, was my advisee at Chicago, making Scott my first Grand Student. I've waited a long time for a grand student, my first student Carsten Lund graduated back in 1991. But Carsten went to AT&T and my next student Lide Li also went into industry. But now with three students in academia (including Sophie Laplante at Paris-Sud and Rahul Santhanam going to Edinburgh), Scott will be the first of many.
Actually what I really want is an infinite tree below me, but König's lemma says I needed a grand student first.
Diehl's thesis is on time-space tradeoff's for satisfiability. I worked in this area about a decade ago then extended some of that work with Dieter who then worked on it with Scott, a passing of knowledge from generation to generation. The symbolism is so, umm, symbolic.
So as not to slight the other members of the family: Scott's academic aunt, my most recent student Varsha Dani, graduated last quarter. And just two days ago I was back at U. Chicago for Sourav Chakraborty's successful defense (Sourav is a student of Babai).
It's so nice to see the young ones grow up.
## Wednesday, April 16, 2008
### The Revenge of Parallelism
Now that I sit in an engineering school, I see more applied recruiting talks. Many of them have a variation of this picture.
This picture represents the future of Moore's law. The number of transistors in our computers continue to grow exponentially but the clock speed is levelling off. What do we use this new transistors for? To make multiple CPUs on a single integrated circuit, known as a multicore machine. New chips from Intel have 2 or 4 cores and the number of cores is expected to double every couple of years.
Multicores present interesting challenges for computer science, for example compiler researchers are trying to make the best use of multiple CPUs without having the user explicitly use parallelism in their code.
Our theory community hasn't really responded to this new computing model (nothing much in STOC and FOCS, though SPAA 2008 has a special track on the topic). Now the theory isn't that interesting if you have two or four cores, but what happens when we have millions on a chip? Do our old parallel models like the PRAM apply to multicore machines? There are hints of this in comments to my PRAM post three years ago. Or perhaps we need new models.
We study computational complexity in computer science instead of mathematics because, at least some level, our models reflect real-world computing paradigms. As those paradigms change, Complexity quickly adapts (random and quantum for instance). Should multicore machines be another one of these paradigm changes that drives our theory?
## Tuesday, April 15, 2008
### Complexity of Income Tax
Its INCOME TAX TIME in the USA. Which country has the most complicated Income Tax System? How can you measure the complexity of an Income Tax System? Some factors:
• The number of pages in the tax code.
• The length of the form you hand in.
• The percent of tax payers who hire someone to do their taxes for them. (This may also be affected by the computer literacy of the country.)
• The number of changes in the tax law from year to year.
• The minimum amount you have to declare. (Do I need to declare my 25 cents that I won from Justin, my 8 year old great nephew, on a math game? Can he use the -25 cents as a deduction? He can use it to offset gambling gains.
• The number of items you can deduct.
There is a problem with all of these measures. What if the tax code is 100,000 pages long but 99.9% of the people only need the first page? One solution is to do some sort of weighted sum.
Deciding whether a tax system is complicated is a hard problem; however, deciding if its fair is a much harder problem.
## Monday, April 14, 2008
### Eight (yes eight) math problems worth $1,000,000 Most readers of this blog know of the Millenium Problems. There are seven of them (which I list below) and solving any of them will get you$1,000,000. (The website above has the following bug/feature- when you go to it you get to a description of ONE of the problems with the entire list on the right-hand side. It seems random which problem you get.)
1. Birch and Swinnerton-Dyer Conjecture
2. Hodge Conjecture
3. Navier-Strokes Equations
4. P vs NP
5. Poincare Conjecture (seems to have already been solved)
6. Riemann Hypothesis
7. Yang-Mills Theory
There is ANOTHER problem that is worth $1,000,000. There is a novel entitled Uncle Petros and Goldbach's Conjecture. Its about a mathematican who is obsessed with Goldbach's conjecture. For publicity, the publishers are offering$1,000,000 for a solution to Goldbach's conjecture. Did this publicity stunt work? The book is selling used for \$3.48 on amazon, and I borrowed it from a friend. On the other hand, it is very doubtful they will have to pay it anytime soon.
Its a pretty good book- the math and mathematicians are spot-on. Its a good airplane book, say the kind of book you can read on the airplane on your way to Conf on Computational Complexity 2008.
## Friday, April 11, 2008
### How to Prove NP Different from P
At TTI yesterday, K. V. Subrahmanyam gave a talk giving one of the better overviews of Ketan Mulmuley's Geometric Complexity Theory approach to separating complexity classes. This approach reduces various problems including P ≠ NP to hard problems in algebraic geometry.
Afterwards at lunch, Ketan made it clear that he believes
1. GCT will eventually lead to proving P versus NP. In fact, any proof that NP is different than P must go via GCT.
2. Such a proof will not happen in my lifetime.
Ketan argued that any complexity theorist who really cares about P v. NP (such as myself) should spend a full month understanding this approach as it will give them a glimpse into how we will eventually separate P from NP. Given my limited knowledge of representation theory and algebraic geometry, I suspect it would take me much more than a month and doubtful that I could ever push the theory any further. Also while knowing the resolution of P v. NP is very important, knowing the details of the proof, especially if it requires deep and complex mathematics, is not nearly as important. I was excited to see Fermat's Last Theorem resolved in my lifetime but I have no desire to actually understand the proof.
Ketan is not even giving me that opportunity. Consider a huge mountain and you want to reach the mountaintop. Ketan comes along and says he'll teach you how to create the tools needed to climb the mountain. It will take a hard month of study and actually these tools aren't good enough to climb the mountain. They need to be improved and these improvements won't happen in your lifetime. But don't you want to learn how others will climb the mountain centuries from now?
If you want to spend the month, go here and start reading. Let me know when you've been enlightened.
## Thursday, April 10, 2008
### Applying Math to politics
In a prior post I tried to apply math to the problem of who to ask for a ride home. Todays post is about someone elses attempt to apply math to politics.
This is from This is from New York Magazine, Feb 4, 2008. (before McCain had clinched the Rep. Primary). The article was called Anatomy of a Freak Show by Kurt Andersen. Here is his formula and his justification.
(Romney + Huckabee)/3 + .01McCain + sqrt(Guilliani) = Bush
1. Romney and Bush were both Businessmen, though Bush was a pathetic one, while Romney was a good one.
2. Huckabee and Bush are both Evangelical Christians, though Bush is a pathetic one, while Huckabee is a good one.
3. McCain and Bush were both party boys in college who later became figher pilots, though Bush avoided real combat.
4. Guilliani and Bush both have a chip on their shoulder.
Is the formula true? Depends how you define true, but I'll say its not even wrong.
## Wednesday, April 09, 2008
### ICALP and EC
A busy conference day yesterday. In the morning I had two papers rejected by ICALP but later buffered by two papers accepted into Electronic Commerce. I'll take 2 for 4 any day.
The list of accepted papers for ICALP Track A has been posted. The EC list isn't out yet.
For ICALP, one my papers was rejected because the proof didn't seem hard enough and the other for having too many theorems.
Thus, while I do find that the results are interesting, reading the paper (including the appendix), I am not convinced that it is possible to present the results in a satisfying way within the page constraints. There simply seems to be too many results included for this to be feasible.
I need to listen to my own advice.
Update 4/10: EC accepted papers here.
## Tuesday, April 08, 2008
### Applying Math to getting rides
Here is an attempt, to apply math to the real real world and what the limits are.
I do not drive so I sometimes need a ride home (about once every two weeks). SO, who to ask? If giving me a ride home adds alot of time to their normal ride, then I would ask them less often. How to quanify this?
If person x has to go i minutes out of his way, then I will ask person x at most once every i weeks.
But here are problems with the formula:
1. Let say that person x normally takes 5 minutes to get home, but giving me a ride home will add 10 minutes, yielding a 15 minute ride. Let say that person y normally takes 30 minutes to get home, but giving me a ride home will add 10 minutes, yielding a 40 minute ride. Person x may view giving me a ride as tripling the time home, while Person y views it as adding just 10 minutes. On the other hand, Person y already has a 30 minute ride and may not want to add anything to it.
2. How much do they like my company? Is i minutes with Bill seem like log i minutes, &radic i , i/2, i, 2i, or i2, minutes (past i2 and I won't ever ask for a ride). (OFF TOPIC QUESTION- how do you do a good sqrt symbol in html? Whats above is the best I could find.)
3. Ditto for how much I like their company.
4. What if when giving me a ride home they pass by their own house and have to backtrack? Even if its not too many extra minutes it has a psycological effect.
5. How complicated is x's life? If x has to drop one of their kids at soccer practice, and one at Piano lessons then fitting a ride for me into it may be complicated even if it is not that many minutes out of the way.
6. Giving someone a ride TOO school is far worse then giving someone a ride FROM school, since FROM school both parties can be more flexible.
## Monday, April 07, 2008
### A Web 1.0 Guy in a Web 2.0 World
I consider myself reasonably Internet savvy. I've been using email since the early 80's, have been doing research via IM, I write a blog and have done some podcasts. But when it comes to social networking I find myself on the outside, in the wrong generation. It's not merely that I don't make much use of social networks, I just don't get it. I've tried out Facebook, Myspace and Linkedin, have loads of "friends" and haven't gotten much use out of them. I've asked younger people why they spend so much time on Facebook and what they get out of it. They tell me plenty and yet I'm still missing some crucial understanding of what makes these networks so popular. Perhaps like trying to understand why a roller coaster is fun without actually taking a ride.
Someone told me that if I started a Facebook page for my course, I could become the most popular professor in the University, but I don't know how to start or what to put up there.
In many of my classes, I start with the same question, "What is a computer?" The first response: "Something I can't live without." Computers have run the gamut from number crunchers, to word processing to a communications medium to an indispensable extension of oneself in a virtual world, a world I can enter but will never be more than a tourist.
## Friday, April 04, 2008
### If we didn't log on how much email would we get?
We all get lots of email. One reason is that we respond to it. When I go out of town I set in place a vacation program and do not log on (or I log on but do not respond to anything--- unless its REALLY important- a slippery slope). How much email do I get? What are the factors?
1. In 1998 I went to Italy for 6 weeks and did not log on at all. I came back to roughly 300 emails. (Very little Spam). This was far less than I thought I would get. The reason: Since I didn't respond and my vacation program said I was out of town, people did not re-email.
2. In 2007 I was on vacation for 11 days over Christmas/New Years and predicted I would get roughly 100 emails. Much to my surprise I got exactly 100 emails. Part of the reason it was so low was that it was most schools winter break. I predict that this will be less true over time- people seem to be working 24/7 and technology is allowing them to.
3. I was out of town and off of email from March 30 until April 3 (the last few days). I got exactly 150 emails. About 10% was from ORBITZ confirming my flights.
4. My spam filters are pretty good- most of the email that got through was not spam. Before I had good spam filters I would get lots of spam AND lots of bounced email when my spam was responded to by my vacation program which then got a bounce.
When I am at a conference I try to avoid logging on. I'm there to learn things, meet people, doing stuff I can't do normally. There is very little important email that needs to be dealt with now. CCC08 will be an exception as I expect there will be email telling me that there aren't enough bagels or whatnot.
## Thursday, April 03, 2008
### An Analog Guy in a Digital World
During my blogging hiatus last summer, some times I would see something that would make me want to write a post if I were still blogging. Such as the movie Live Free or Die Hard.
Bruce Willis reprises his role as policeman John McClane battling tech wizard Thomas Gabriel (Timothy Olyphant) who is creating havoc by taking over various computers controlling traffic, power and the like. For such a computer-oriented theme, the movie had a retro anti-tech feel. Though McClane is teamed up with a computer geek played by Justin Long (Mac from those Apple ads), he fights back mostly by crashing cars and blowing things up. McClane's aversion to technology is a running theme in the movie, where Gabriel at one point mocks him as an analog guy in a digital world. Without spoiling too much, you can guess who wins out in the end.
The movie itself has much less CGI than other recent action movies, relying on old fashioned stunts and lots of explosives. There is an interesting theme to this movie: Even in a technology dependent world, an old-fashioned hero can still save the day and have a lot of fun doing it.
## Wednesday, April 02, 2008
### Two Israels
My Israel trip started with a visit to the Technion with a short side trip to the University of Haifa. Visiting universities in Israel is not unlike visiting universities anywhere else in the world. I gave a seminar presentation, we talked research and gossiped about other computer scientists. We didn't talk politics much and then it was mostly American politics. Professors there have the usual problems balancing theorems and families.
After that, spurred by my daughter's upcoming Bat Mitzvah, I and the rest of my family did a tour with several other families from my congregation. This mission, as it was called, emphasized the Israel I grew up learning about, the Jewish state that we mention in many of our prayers. We examined the struggle of the Jews thousands of years ago, sixty years ago and today. I touched both the sacred Western wall of the old temple and the much newer wall that separates Israel proper from the territories.
Two very different experiences in the same country. But these worlds get very close. We drove by Tel Aviv University, visited a once-secret bullet factory near the Weizmann Institute and said our welcoming prayers to Jerusalem just downhill from Hebrew University. But more than that, one cannot help but notice the plaques on the walls at the Technion mentioning the various infrastructure donated from American Jews. These have been possibly the greatest gifts to the country as the strong Israeli University system has propelled an extremely successful high tech industry giving Israel an economic security that seemed unimaginable when I was a kid.
|
{}
|
# Article
Full entry | Fulltext not available (moving wall 24 months)
Keywords:
non-cyclic $p$-subgroup; $p$-nilpotent; self-normalizing subgroup; normal subgroup
Summary:
A theorem of Burnside asserts that a finite group $G$ is \mbox {$p$-nilpotent} if for some prime $p$ a Sylow \mbox {$p$-subgroup} of $G$ lies in the center of its normalizer. In this paper, let $G$ be a finite group and $p$ the smallest prime divisor of $|G|$, the order of $G$. Let $P\in {\rm Syl}_p(G)$. As a generalization of Burnside's theorem, it is shown that if every non-cyclic \mbox {$p$-subgroup} of $G$ is self-normalizing or normal in $G$ then $G$ is solvable. In particular, if $P\ncong \langle a,b\vert a^{p^{n-1}}=1,b^2=1, b^{-1}ab=a^{1+{p^{n-2}}}\rangle$, where $n\geq 3$ for $p>2$ and $n\geq 4$ for $p=2$, then $G$ is \mbox {$p$-nilpotent} or \mbox {$p$-closed}.
References:
[1] Huppert, B.: Endliche Gruppen I. Die Grundlehren der mathematischen Wissenschaften in Einzeldarstellungen 134 Springer, Berlin German (1967). MR 0224703 | Zbl 0217.07201
[2] Robinson, D. J. S.: A Course in the Theory of Groups. Graduate Texts in Mathematics 80 Springer, New York (1996). MR 1357169
Partner of
|
{}
|
SEARCH HOME
Math Central Quandaries & Queries
Question from Jerry, a parent: Aaron works part time as a salesperson for an electronics store. He earns $4.75 per hour plus a percent commission on all of his sales. Last week Aaron worked 20 hours and earned a gross income of$143.75. If his total sales for the week were $1,500, what percent commission does Aaron earn? Jerry, How much salary did Aaron earn at$4.75 per hour for 20 hours? The remainder of his income was from commissions. How much was that? This number is what percent of \$1,500?
Penny
Math Central is supported by the University of Regina and The Pacific Institute for the Mathematical Sciences.
|
{}
|
1. If you are a new user, please register to get an Indico account through https://login.ihep.ac.cn/registIndico.jsp. Any questions, please email us at helpdesk@ihep.ac.cn or call 88236855.
2. The name of any uploaded file should be in English or plus numbers, not containing any Chinese or special characters.
3. If you need to create a conference in the "Conferences, Workshops and Events" zone, please email us at helpdesk@ihep.ac.cn.
# International Conference on Technology and Instrumentation in Particle Physics 2017(TIPP2017)
21-26 May 2017
Beijing International Convention Center
Asia/Shanghai timezone
Home > Timetable > Session details > Contribution details
# Contribution oral
Beijing International Convention Center - Room 307
Photon detectors
# Results from Pilot Run for MEG II Positron Timing Counter
## Speakers
• Mitsutaka NAKAO
## Content
The MEG II experiment at Paul Scherrer Institut in Switzerland will search for the lepton flavour violating muon decay, $\mu^+\to e^+\gamma$, with a sensitivity ($4\times10^{-14}$) improving the existing limit of an order of magnitude. The positron Timing Counter (pTC) is the subdetector dedicated to the measurement of the positron emission time. It is designed on the basis of a new approach to improve a positron ($e^+$) timing resolution by a factor of two compared to MEG. pTC is composed of 512 ultra-fast plastic scintillator with SiPM readouts. The mean hit multiplicity for signal $e^+$ is evaluated to be $\sim$ 9 and a high timing resolution of $\sim$ 35 ps is expected by averaging the signal time of multiple hit counters. To achieve the target resolution, an internal time calibration with a precision of 10 ps or better is required. We have developed two new methods for the calibration, which meet the requirement: Track-based calibration and Laser-based calibration.
In 2016, We have finished construction of pTC and installed the first one-fourth of pTC into the MEG II experimental area to evaluate the performance using $\mu^+$ beam as a pilot run. We took data of $e^+$ from the dominant $\mu$ decay (Michel Decay, $\mu^+\to e^+\nu_e\overline{\nu}_{\mu}$) and applied both time calibration methods. The time offsets of each counter calculated independently from the two calibration methods were consistent and stable during the run within 6 ps. The systematic uncertainty between these methods was 39 ps, which is suppressed with $\frac{1}{\sqrt{N}}$ using multiple hits (N: number of hit counters). The overall timing resolution weighted with a distribution of the number of hit counters for signal $e^+$ of 38 ps was achieved in this pilot run. The prospects towards MEG II physics run are also discussed.
|
{}
|
# Monroe Eskew: Local saturation of the nonstationary ideals
Talk held by Monroe Eskew (KGRC) at the KGRC seminar on 2018-03-22.
Abstract: It is consistent relative to a huge cardinal that for all successor cardinals $\kappa$, there is a stationary $S \subseteq \kappa$ such that the nonstationary ideal on $\kappa$ restricted to $S$ is $\kappa^+$-saturated. We will describe the construction of the model, focusing how to get this property on all $\aleph_n$ simultaneously. Time permitting, we will also briefly discuss the Prikry-type forcing that extends this up to $\aleph_{\omega+1}$.
|
{}
|
Recognitions:
Gold Member
## Material Science Questions....
I finally got to exercise a bit of material science. I'd like to know whether what I'm saying correct or no. I also ran into two issues with Brinell's hardness calculations and SEA chart. I'd appreciate feedback! :)
1) Decreasing the size of the nucleus causes...
A) Increasing plasticity
B) Increasing strength
C) Thermic durability
D) A+B
I chose A. I do know for sure that decreasing the size of the nucelous makes the material stronger. It just makes more sense that when you have more little things to break
2) What's the correct sentence? (Based on the unit cell crsytal formation lattice)
A) Vanadium is more plastic than Magnesium
B) Vanadium is stronger than Magnesium
C) Vanadium is less hard than Magnesium in high temp'
Vanadium is a BCC. Mg is CPH.
CPH (and FCC) are denser than BCC. Therefor, I chooose B. Vanadium is stronger than Magnesium.
3) What thermal treatment is designed to unite steel?
A) Annealing
B) Homogenization
C) Tempering
I choose B. That's what I believe the manual is getting it.
4) What's the precentage of steel SAE10150?
A) 0.15%
B) 1.5%
C) 15%
I couldn't find a SEA chart in my book, so I looked at wiki, but wiki's list only starts from 201!
Does anyone have a good link for me for a chart?
5) In a normalization process the cooling is done:
A) In oil
B) In air
C) In an oven
"Air" form what I was able to google (also couln't find that in my book)
6) Calculate the hardness level by Brinell’s method
A) Ball diameter – 10mm
B) Dent size – 4mm
C) Pressing Force – 3000 [N
From some reason am getting syntax error!
Recognitions:
Gold Member
Homework Help
Quote by Femme_physics 1) Decreasing the size of the nucleus causes... A) Increasing plasticity B) Increasing strength C) Thermic durability D) A+B I chose A. I do know for sure that decreasing the size of the nucelous makes the material stronger. It just makes more sense that when you have more little things to break
Assuming "nucleus" means grain... Why choose A if you "know for sure" that B is correct?
Recognitions:
4) What's the precentage ?carbon? of steel SAE10150? A) 0.15% B) 1.5% C) 15%
SAE10XX are plain carbon steels. The percentage carbon is xx/100. For example SAE 1020 would be a low carbon steel (0.2%). SAE1060 would be fairly high carbon steel, 0.6% carbon. SAE10150 would have to be 1.5% carbon, a very high carbon steel.
Mentor
## Material Science Questions....
Quote by Femme_physics;33062996) Calculate the hardness level by Brinell’s method A) Ball diameter – 10mm B) Dent size – 4mm C) Pressing Force – 3000 [N From some reason am getting syntax error! [URL=http://imageshack.us/photo/my-images/839/synerror.jpg/ [/URL] Uploaded with ImageShack.us
What software is giving you the syntax error? I see two things that might be causing a syntax error:
1) The dot in the denominator. Maybe you need to use *, which is commonly used for multiplication.
2) The expression 10(10 - <other stuff>). Possibly whatever you're using requires an operator (e.g., *) for multiplication.
Recognitions:
Gold Member
Quote by uart SAE10XX are plain carbon steels. The percentage carbon is xx/100. For example SAE 1020 would be a low carbon steel (0.2%). SAE1060 would be fairly high carbon steel, 0.6% carbon. SAE10150 would have to be 1.5% carbon, a very high carbon steel. See : http://en.wikipedia.org/wiki/SAE_steel_grades
I see how it works, the "xx" represent the percentage then? So simple :)
Assuming "nucleus" means grain... Why choose A if you "know for sure" that B is correct?
Must've been a terrible mishap, was meant to be B :)
1) The dot in the denominator. Maybe you need to use *, which is commonly used for multiplication.
This is just a Microsoft Equation representation of what I plugged to my calculator.
2) The expression 10(10 - ). Possibly whatever you're using requires an operator (e.g., *) for multiplication.
I'll try it again tomorrow then and get back to you :) thanks.
Mentor Re the syntax error. It looks like you have the right parenthesis in the wrong place in the denominator, part of which is shown below. $$10(10 - \sqrt{10^2 - 4^2)}$$ This should be $$10(10 - \sqrt{10^2 - 4^2})$$ IOW, the right paren should be outside the radical. That would definitely cause a syntax error.
Blog Entries: 2
Recognitions:
Homework Help
|
{}
|
Question
Gwen and Cathy went shopping. After Gwen spent $52w, Cathy had 5 times as much as Gwen. If Gwen had$26w more than Cathy at first,
1. Express the amount of money that Gwen had at first in terms of w.
2. If w = 26, find the amount of money the two lady had left after shopping.
3 m
(a)
Click button first when a symbol is required. X
(b)
Click button first when a symbol is required. X
Gwen and Cathy went shopping. After Gwen spent $52w, Cathy had 5 times as much as Gwen. If Gwen had$26w more than Cathy at first,
1. Express the amount of money that Gwen had at first in terms of w.
2. If w = 26, find the amount of money the two lady had left after shopping.
|
{}
|
# HEP Experiments and Projects
The Experimental Particle Physics group is involved in the following projects:
The group was also involved in the CDF experiment at Fermilab, studying high-energy proton-antiproton collisions and in the ZEUS experiment at Hamburg, studying high-energy electron-proton collisions.
|
{}
|
# Measuring volume change from change in mass
I want to measure the expansion (by change in volume) of a material in liquid. However the shapes are so complex that I can only measure their mass. Is it possible to compute the volume of the expanded shape by measuring its mass alone?
Given mass $m_0$ at $t_0$ and mass $m_1$ at $t_1$ and the density $\rho_\mathrm m$ of the material and $\rho_\mathrm l$ of the liquid, can I find the volume change of the object from $t_0$ to $t_1$?
One could assume that $(m_0-m_1)/\rho_\mathrm l$ is the correct answer, but I am not sure if this is the right way to go about it?
• Can you measure the volume by displacement, as Archimedes did? [Do not run through the streets shouting 'Eureka!", though.] – DrMoishe Pippik Jul 11 '16 at 5:13
• I assume that "from t0 to t1" part of the question means that the temperature is changing. – MaxW Feb 14 '17 at 2:13
• Is there a way to know the starting volume? Is the material compact and easy to be handled? – Alchimista Jul 14 '17 at 20:30
|
{}
|
Students can access the CBSE Sample Papers for Class 11 Applied Mathematics with Solutions and marking scheme Term 2 Set 4 will help students in understanding the difficulty level of the exam.
## CBSE Sample Papers for Class 11 Applied Mathematics Term 2 Set 4 for Practice
Time : 2 Hours
Maximum Marks : 40
General Instructions:
• The question paper is divided into 3 sections -A, B and C.
• Section A comprises of 6 questions of 2 marks each. Internal choice has been provided in two questions.
• Section B comprises of 4 questions of 3 marks each. Internal choice has been provided in one question.
• Section C comprises of 4 questions. It contains one case study-based question. Internal choice has been provided in one question.
Section – A [2 Marks each]
Question 1.
The radius r of a right circular cone is decreasing at the rate of 3 cm/min and the height h is increasing at the rate of 2 cm/min. When r = 9 cm and h = 6 cm, find the rate of change of its volume.
OR
Find the value of k for which the function.
Question 2.
A dealer is in Jhansi buys some articles worth ? 8,000. If the rate of GST is 18%, find how much will the dealer pay for the articles bought.
Question 3.
₹ 1000 is invested every 3-months at 4.8% p.a. compounded quarterly. How much will the annuity be worth in 2 years ? [Given that (1.012)8 = 1.1001]
OR
What is the monthly equivalent interest rate to a quarterly interest 2.5% ? [Given that (1.025)1/3 = 1.008265]
Question 4.
If P(E) = $$\frac{7}{13}$$ P(F) = $$\frac{9}{13}$$ and P(E ∩ f)= $$\frac{4}{13}$$ then evaluate : (i) P$$\left(\frac{\bar{E}}{F}\right)$$ and (ii)P$$\left(\frac{\bar{E}}{\bar{F}}\right)$$
Question 5.
Determine ∠B of the triangle with vertices A(-2, 1), B(2, 3) and C(-2, -4).
Question 6.
Determine the number of 5 cards combinations out of a deck of 52 cards if there is exactly one ace in each combination.
Section – B (3 marks each)
Question 7.
An unbiased die is thrown twice, let the event A be ‘odd number on first throw’ and B the event ‘odd number on the second throw’. Check independence of the events.
Question 8.
Find the equation of the circle drawn on a diagonal of the rectangle as its diameter whose sides are the lines x = 4, x = – 5, y = 5 and y = – 1.
OR
An equilateral triangle is inscribed in the parabola y2 = 4ax, where one vertex is at the vertex of the parabola, Find the side of the triangle.
Question 9.
Two dice are thrown together. What is the probability that sum of the numbers on the two faces is neither divisible by 3 nor 4 ?
Question 10.
(a) In what time will ₹85000 amount to ₹157675 at 4.5% p.a. ?
(b) A sum of ₹46875 was lent out at simple interest and at the end of 1 year 8 months the total amount was ₹50,000. Find the rate of interest percent per annum.
Section – c [4 Marks each]
Question 11.
In how many ways three girls and nine boys can be seated in two vans, each having numbered seats, 3 in the front and 4 at the back ? How many seating arrangements are possible if 3 girls should sit together in a back row on adjacent seats ?
Question 12.
Compute the taxable value of the perquisite in respect of medical facilities availed of by X from his employer in the following situations:
(a) The employer reimburses the following medical expenses:
(i) Treatment of X by his family physician ₹ 8,400
(ii) Treatment of Mrs. X in a private nursing home ₹ 7,200
(iii) Treatment of X’s mother (dependent upon him) ₹ 2,400 by a private doctor
(iv) Treatment of X’s brother (not dependent upon him) ₹ 800
(v) Treatment of X’s grandfather (dependent upon him) ₹ 3,000
(b) The employer reimburses an insurance premium of ₹ 6,000 paid by X under a health insurance scheme on the life of X and his wife.
(c) The employer maintains a hospital for the employees where they and their family members are provided free treatment. The expenses on treatment of X and his family members during the previous year 2019-20 were as under:
Particulars Amount (₹) (i) Treatment of X’s major son (dependent upon him) 4,400 (ii) Treatment of X 10,400 (iii) Treatment of X’s uncle 9,200 (iv) Treatment of Mrs. X 16,000 (v) Treatment of X’s widowed sister (dependent upon him) 8,200 (vi) Treatment of X’s handicapped nephew 5,000
(d) Expenses on cancer treatment of married daughter of X at Tata Memorial Hospital, Mumbai paid by the employer ₹ 1,00,000 and reimbursement of expenses for medical treatment of himself amounting to ₹ 40,000.
(e) The following expenses on treatment of X’s major son outside India were paid by the employer:
Assume that the other income of X is (a) ₹ 1,20,000 (b) ₹ 1,80,000. (including income under the head salary excluding the above taxable perquisite)
OR
A manufacturer in a firm manufactures a machine and marks it at ₹ 80,000. He sells the machine to a wholesaler (in Gorakhpur) at a discount of 20%. The wholesaler sells the machine to a dealer (in Mathura) at a discount of 15% on the marked price. If the rate of GST 28%, find tax paid by the whole seller to Central Government.
Question 13.
Question 14.
Read the following text and answer the following questions on the basis of the same:
Water Bill: The amount one must pay to use water and sewage services each month. Normally, water and sewage is provided by a municipality, but this is not always the case. Water bills are usually based upon one’s usage, such that those who use more water are charged more.
The water bill invoice: is provided by a company that supplies water on a residential and/or commercial basis. A customer that receives their water supply from such a company will receive a water bill invoice complete with the charges for the company’s services and the amount owed for said services. From time to time, people forget to pay their utility bills, so the customer might see a summary of past due charges that must be paid on the next billing period. Payments not received on time could result in interest charges or additional fees. The water bill invoice will show the total amount due and the date upon which payment must be received.
There will be generally following three components of water / sewerage bill:
Fixed Water Charge: With a fixed water charge, the consumer pays a monthly water bill, which is the same independently of the volume consumed. In absence of a water metering system, a fixed water charge is the only possible tariff structure.
Sewerage maintenance charge: This charge is levied for the maintenance of sewerage system and is charged according to volumetric consumption of water.
Service charge: Service charge under the domestic category which is presently linked with the built up area of the property, that is, whether the covered area is more than or less than 200 sq. meters and this has been now delinked from the area concept. Instead, under the new tariff it will be linked with the consumption slab for all categories of consumers including the domestic category.
Based on the information given above, solve the following questions:
(a) What do you understand by is water tariff and fixed water charge? (2)
(b) For an industrial connection monthly consumption of water is 40 ki, calculate the Water bill. (2)
Tarrif rates can be considered as the table given below:
|
{}
|
# Linear regression
1. Oct 26, 2012
### iVenky
I read about "Linear regression" and I want to make sure that what I read is right
Just tell if these equations are right-
Slope of line of regression for y on x is given by
$m=\frac{E(XY)-E(X)E(Y)}{E(X^{2})-[E(X)]^{2}} \\ m=\frac{Cov(XY)}{Var(X)} \\ m=\frac{ρσ_{x}σ_{y}}{σ_{x}^{2}} \\ m=\frac{ρσ_{y}}{σ_{x}} \\and\ the\ equation\ is \\y-\bar{y}= m (x-\bar{x})$
Similarly the slope of line of regression of x on y is given by
$\\ \\ m=\frac{ρσ_{x}}{σ_{y}} \\and\ the\ equation\ is \\x-\bar{x}= m (y-\bar{y})$
Just tell me if the above equations are right.
Thanks a lot
Last edited: Oct 26, 2012
2. Oct 27, 2012
### chiro
Hey iVenky and welcome to the forums.
Those look correct if you swap the x's and x_bar's with the y's and y_bar's. So think about y - y_bar = m(x - x_bar) instead.
Also, we usually we write B0 = y_bar - B1_hat*x_bar (this is obtained by setting x = 0 and solving for y) and B1_hat = m (the gradient).
3. Oct 27, 2012
### iVenky
I mean, you should swap
$x\ and\ \bar{x}\ with\ y\ and\ \bar{y}$ for finding out the line of regression for x on y (not y on x) right?
4. Oct 27, 2012
### chiro
No you need to swap both.
Recall that the definition of a straight line in two dimensions has one form which is y - y0 = m(x - x0) and this is something from high school geometry. In this definition (x,y) is a point on the line and (x0,y0) is a specific point on the line with m being the gradient.
5. Oct 27, 2012
### iVenky
Please note that I have written the equation for two cases
i) Y is a function of X and the equation is given by the one that you have written
ii) X is a function of Y. By which I mean I have taken the values of Y along the X axis and values of X along the Y axis. If that is the case you have to swap them.
See my question. I have written the equation for both cases. :)
Thanks a lot
6. Oct 27, 2012
### chiro
If you changing the axis then recall that in two dimensions m1*m2 = -1 where m2 is the gradient of the line perpendicular to that involving the gradient m1.
7. Oct 27, 2012
### iVenky
If I change the axis the slope won't be perpendicular to the one before. For eg: Y increases as X increases (slope is positive). This means that X increases as Y increases. (once again slope is positive and not negative)
8. Oct 27, 2012
### chiro
Ohh yes, sorry you are spot on.
|
{}
|
# Elephant Specification Language Manual¶
Welcome to the Elephant Specification Language manual. This manual will introduce you to the reason behind and concepts within the Elephant Specification Language, or ESL for short. For a more in-depth explanation of ESL’s foundations check out the Elephant Specification Language Reference.
## Introduction¶
System Engineers play a vital role in the development of complex systems. Coordinating the design and production of thousands of components across modules, teams, departments and even companies is an immense challenge! Currently, the design documentation or “specs” of a system are often communicated in the form of PDF, Word, and Excel files. Several solutions exist that try to organize these files and tie them to the components they are supposed to describe.
Although this document organization is helpful, the documents themselves form a huge blind spot as they are often inconsistent, incomplete, and out-of-date. Enter ESL! A highly structured though human-readable specification language that introduces a fixed syntax while allowing the expression of any design requirement or constraint. Since its syntax is fixed, the ESL documents can be checked and analyzed automatically for their consistency and completeness.
Moreover, all relations, dependencies and links of requirements to components and variables are automatically derived from the specification instead of having to be maintained manually. Even for simple projects the number of dependencies quickly exceeds the 100 or even 1,000 mark! The automated dependency derivation reduces the risk of human error drastically, as dependencies cannot be forgotten to be added by definition.
Naturally, not every System Engineer you come across will work with ESL specifications, which is where the generated PDF output comes in. Nicely formatted documents, generated right from your ESL specification, which once more guarantees their consistency!
## Learning ESL¶
This tutorial touches the basic concepts of ESL and how to use them. We are going to specify a water pump, that consists of a centrifugal pump driven by an electromotor, with several requirements to the subcomponents, dimensions, and performance. ESL will provide a fixed syntax to describe the requirements, constraints and needs of your system. Even though the syntax is fixed, the specification documents themselves are designed to be human readable.
Note
Note that we use a certain spacing style when writing ESL for readability purposes, although the compiler itself ignores any indents.
### Hello world!¶
We start by specifying the modeling boundary. We do so by specifying the world. The absolute minimal ESL specification you can write is therefore:
world
empty
Thus introducing two new keywords:
• world: The modeling boundary, can only occur once in a ESL specification. We assume nothing exists outside of the world.
• empty: When we want to specify something without any further information, we write empty to signal this.
### Components¶
Components are at the core of ESL specifications. Typically, your (to be developed) systems will live as a component within the world. Components are introduced using a definition such that they can be instantiated in the world. In case of our water pump, this becomes:
world
components
pump is a centrifugal-pump
drive-mechanism is an electrical-drive-mechanism
define component centrifugal-pump
empty
define component electrical-drive-mechanism
empty
Where each component definition is defined using the define component keyword combination followed by the component [name], which cannot have any spaces. Dashes are fine.
Within the world definition, we instantiate defined components by given them a local name and referring to their definition with [name] is a [defined component].
Note
We can refer to the defined components before their written definitions. The ESL compiler is smart that way and resolves later define component [name] statements by default.
Note
Note the difference between definition and instantiation. Definitions are similar to classes or templates in other programming languages and instantiations are similar to the actual instances or objects.
### Nested components¶
Components can be defined with any number of subcomponents. This allows for a hierarchical layout of your specification. As such, you can start with your top-level components and work your way down as more details become available.
Let’s say that our electrical-drive-mechanism consists of both a brushless electromotor as well as a battery to power it. We then have to add those component definitions and can add them as subcomponents to the drive mechanism like so:
define component electrical-drive-mechanism
components
electric-motor is a brushless-motor
power-source is a battery
define component brushless-motor
empty
define component battery
empty
### Types, variables, and verbs¶
With components and subcomponents, the specification merely lays out the hierarchical component structure, but tells us little about what needs to go on inside the system. We would like to start expressing this using requirements and constraints, but we still miss a key element before we can do this: variables. They can be defined like so:
define type
mechanical-energy-flow is a real
define verb
provide to
world
variables
torque is a mechanical-energy-flow
Where we first define a mechanical-energy-flow type, and instantiate a variable on the global (world) level of that type named torque. Similar to components, we use a [name] is a [defined type] to describe this. There are several base types built in, being bool (Booleans), real (real valued numbers), integer(integer valued numbers) and string (for literals).
We also define a verb that we would like to use when we start expressing requirements. It consists of a verb (provide) and a preposition (to). By having to stick to a defined set of verbs and prepositions, the ambiguity in your specifications can be kept to an absolute minimum!
Note
Note how we define mechanical-energy-flow as a real, indicating that we would like to quantify this measure.
### Goal requirements¶
Great! With the groundwork set, we can start adding actual requirements! The goal-requirement defines the goal of a component with respect to another. In case of our pump, we would like to specify that it is the drive-mechanism that provides torque to the pump.
We express this using the sentence: drive-mechanism must provide torque to pump, thus following the format [active component] must [verb] [variable] [preposition] [passive component].
So a world definition illustrating this becomes:
world
variables
torque is a mechanical-energy-flow
components
drive-mechanism is an electrical-drive-mechanism
pump is a centrifugal-pump
goal-requirements
provide-torque: drive-mechanism must provide torque to pump
Note how the goal requirement has a label provide-torque. Requirements need to have a label for easy identification. This label only has to be unique in the definition of that single component (or the world) and can thus be re-used elsewhere.
### Transformation requirements¶
While goal-requirements are great to express the purpose of a component with respect to another, components often transform their inputs to their outputs internally, which we specify using transformation-requirements.
For instance, our brushless-motor is the component that converts power into torque. We describe this using the sentence must convert power into torque. Lets review its definition:
define verb
convert into
define component brushless-motor
variables
power is an electrical-energy-flow
torque is a mechanical-energy-flow
transformation-requirements
convert-power: must convert power into torque
And so, transformations are specified using [label]: must [verb] [variable in] [preposition] [variable out].
### Design requirements¶
To express requirements on the values of variables we use design-requirements. These are, for instance, useful if you want to specify a component’s size, or set the key performance indicator.
For instance, we can express a minimum required water flow of the pump:
define component centrifugal-pump
variables
min-water-flow is a liquid-material-flow
design-requirement
water-flow-min: water-flow must be at least min-water-flow
(In)equalities can be expressed using the design rule syntax [label]: [subject variable] must be [comparison] [bound variable] or [label]: [subject variable] must be [comparison] [value] [optional unit]. Where the comparison may be one of:
• equal to
• not equal to
• at least
• at most
• greater than
• smaller than
The logic may be extended using or.
### Constraints¶
We define requirements as that what is desired for the system. However, some parts may be already present. That what is given can be expressed using constraints. goal-, transformation-, and design-constraints exist, and are identical to their ...-requirements counterparts, apart from the change from must to does when going from requirements to constraints.
For instance, our battery power source will always convert the chemical power-potential to power regardless of our requirements:
define component battery
parameters
power-out is an electrical-energy-flow
variables
power-potential is a chemical-energy-flow
transformation-constraints
convert-potential: does convert power-potential into power-out
### Needs¶
Sometimes you cannot quantify a requirement and must state something qualitatively. For example, you may want to refer to an externally defined standard. This can be specified using needs, which should be used with caution. Needs serve a documentation purpose, but are mostly ignored by the compiler.
Let’s say our drive-mechanism needs to be waterproof up to IP68 compliance:
world
component
drive-mechanism is an electrical-drive-mechanism
need
compliance: drive-mechanism must be IP68 compliant
There are no further constraints on need lines than that it needs to start with a variable or a component-name as in [variable | component-name] [free text], which is why they are some form of ‘escape’ in the specifications. Use this wisely!
### Parameters & properties¶
Goals, transforms and needs give us the expressiveness to build a complete specification. However, sometimes, variables need to be available outside a component’s own definition. This way, multiple components can refer to the same torque, instead of each having their own unique variables.
To indicate the owner of a certain parameter, you can end the property keyword to the parameter declaration. Note that by definition flows cannot be properties of components as they flow through the system. As such, no single component owns a flow.
For instance, we can define the length of the drive mechanism like so:
define type
spatial-measure
world
variables
drive-length is a spatial-measure
component
drive-mechanism is an electrical-drive-mechanism with arguments
drive-length
define component electrical-drive-mechanism
parameters
length is a spatial-measure property
Note
Note that we do have to specify the spatial-measure type at every component definition. The typechecker will alert you if any type mismatches are found (or in case you forgot to add them).
### Relations¶
Variables are powerful in themselves, but they can be interdependent, too. They may be coupled via mathematical equations, models, laws of physics, or any other means of interdependency. We signal this by defining relations between them. Relations can have names and arguments, but contain no further (mathematical) detail in their specification. This is intentional, as the language is not designed to replace all kinds of complex computation environments and languages. Lets say we want to describe the battery’s efficiency as a relation between power potential (chemical) and the output power (electrical):
define relation
battery-efficiency-model
define component battery
parameters
power-out is an electrical-energy-flow
variables
power-potential is a chemical-energy-flow
relations
battery-efficiency: battery-efficiency-model with arguments
power-potential, power-out
Note
Note that the relation’s definition is merely a name, without any further (argument) specifications as of now. This is rather loose and may change in the future.
### Subclauses¶
The requirements are meant to be clear and concise, but can often do with a little hardening or additional measures. That is where subclauses come in. Essentially subclauses are design rules (similar to design-requirements), but attached to another goal-, transformation- or design-requirement.
You can add them like so:
goal-requirement
provide-power: power-source must provide power to electric-motor with subclauses
c0: power must be at least 300 Watt
end
Thus by simply adding with subclauses at the end of a requirement line, we can enumerate design rules until we end it using the end keyword. Design rules follow the design-requirement syntax.
### Nuancing requirements and constraints¶
Up until now, every requirement and subclause has been defined using must as the requirement specifier. One can add nuance by varying between any of the following options:
• shall
• must
• should
• could
• won't
There are two types of comments available in ESL specifications. Regular comments and attached comments. As a comment sign we use the pound sign (#). The attached comments are directly tied to their component and are included and available as additional information in any output document we generate. Attached comments can be added using the #< combination or in a special comment section, like so:
define component centrifugal-pump # I am a regular (ignored) comment.
parameter
torque is a mechanical-energy-flow #< Centrifugal pump requires
#< a lot of torque.
torque #< So much info! I'm attached to the torque, too, by the way.
Multi-comments are made by repeating the (attached) comment symbols on every line. The comment section works by naming the instance you want to attach a comment to and using the attached comment sign. This can help separating the comment documentation from the requirements. Valid targets are components, goal- and transformation-requirements, needs, and relations.
## Final specification¶
Finally, we can combine all the outlined concepts into a single specification:
define type
mechanical-energy-flow is a real
electrical-energy-flow is a real
chemical-energy-flow is a real
liquid-material-flow is a real
performance-measure is a real
spatial is a real
define verb
provide to
convert into
define relation
pump-efficiency-model
battery-efficiency-model
world
variables
reliability is a performance-measure
drive-length, pump-length is a spatial
components
drive-mechanism is an electrical-drive-mechanism with arguments
torque, drive-length
pump is a centrifugal-pump with arguments
torque, pump-length
goal-requirements
provide-torque: drive-mechanism must provide torque to pump with subclauses
c0: steady-state-torque must be equal to 40 Nm
c1: reliability must be at least 95 %
end
design-requirements
drive-length-eq: drive-length must be equal to pump-length
define component centrifugal-pump
parameters
torque is a mechanical-energy-flow
length is a spatial property
variables
water-flow, min-water-flow is a liquid-material-flow
efficiency is a performance-measure
transformation-requirements
convert-torque: must convert torque into water-flow with subclauses
c0: efficiency must be at least 75 %
end
design-requirements
water-flow-min: water-flow must be at least min-water-flow
relations
pump-efficiency: pump-efficiency-model with arguments
torque, water-flow, efficiency
define component electrical-drive-mechanism
parameters
torque is a mechanical-energy-flow
length is a spatial property
variables
power is an electrical-energy-flow
protection is a performance-measure
components
motor is a brushless-motor with arguments
power, torque
power-source is a battery with arguments
power
goal-requirements
provide-power: power-source must provide power to motor with subclauses
c0: power must be at least 300 Watt
end
needs
compliance: protection must be IP68 compliant
define component brushless-motor
parameters
power is an electrical-energy-flow
torque is a mechanical-energy-flow
transformation-requirements
convert-power: must convert power into torque
define component battery
parameters
power-out is an electrical-energy-flow #< Unit is in Watt.
variables
power-potential is a chemical-energy-flow
transformation-constraints
convert-potential: does convert power-potential into power-out with subclauses
c0: power-out is at most 300 W
end
relations
battery-efficiency: battery-efficiency-model with arguments
power-potential, power-out
|
{}
|
### send_nodes's blog
By send_nodes, history, 6 years ago,
Hi everyone, here are the solutions to the contest problems.
712A — Memory and Crow
Note that a[i] + a[i + 1] = b[i]. Use the initial condition b[n] = a[n] and we can figure out the entire array b.
Time Complexity: O(n)
Code
712B — Memory and Trident
First, if S has odd length, there is no possible string because letters must come in opposite pairs. Now, let's denote the ending coordinate after following S as (x, y). Since S has even length, |x| has the same parity as |y|. Suppose they are both even. Then clearly, we can make x = 0 in exactly moves, and same for y. If instead they are both odd, then we can change exactly one x-character into a y-character. With the correct choices of these characters, now our string has |x| and |y| with even parity, thus reducing to the problem above. Therefore, the answer is (|x| + |y|) / 2.
Time Complexity: O(|S|)
Code
712C — Memory and De-Evolution
Let's reverse the process: start with an equilateral triangle with side length y, and lets get to an equilateral triangle with side length x. In each step, we can act greedily while obeying the triangle inequality. This will give us our desired answer.
Time Complexity: O(log x)
Code
712D — Memory and Scores
One approach to this problem is by first implementing naive DP in O((kt)2). The state for this is (diff, turn), and transitions for (diff, turn) is the sum (diff - 2k, turn - 1) + 2(diff - 2k + 1, turn - 1) + 3(diff - 2k + 2, turn - 1) + ... + (2k + 1)(diff, turn - 1) + 2k(diff + 1, turn - 1) + ... + diff( + 2k, turn - 1).
Now, if we use prefix sums of all differences in (turn-1), along with a sliding window technique across the differences, we can cut a factor of k, to achieve desired complexity O(kt2).
However, there is a much nicer solution in O(kt log kt) using generating functions(thanks to minimario). We can compute the coefficients of , and the coefficient to xi corresponds to the number of ways we can form the difference i. To compute these coefficients, we can use the binomial theorem.
Time Complexity: O(kt2)
Code
Time Complexity: O(kt log kt)
Code
712E — Memory and Casinos
Lets think about two segments of casinos [i, j] and [j + 1, n]. Let L([a, b]) denote the probability we dominate on [a, b], and let R([a, b]) denote the probability we start on b and end by moving right of b. Let
l1 = L([i, j]),
l2 = L([j + 1, n]),
r1 = R([i, j]),
r2 = R([j + 1, n]).
You can use a geometric series to figure out both L([i, n]) and R([i, n]) using only l1,l2,r1, and r2. To derive these series, think about the probability we cross over from j to j + 1 once, twice, three times, and so on. The actual formulas are,
Now we can build a segment tree on the casinos, and use the above to merge segments.
Time Complexity: O(N + QlogN)
Code
• +46
» 6 years ago, # | +1 Can someone explain how to solve D using generating functions as mentioned in editorial ?
» 6 years ago, # | 0 Auto comment: topic has been updated by send_nodes (previous revision, new revision, compare).
» 6 years ago, # | ← Rev. 2 → +5 Could someone elaborate solution to D ? Didn't really understand it. How is the equation formed ?
• » » 6 years ago, # ^ | +2 The equation is divided by x^k, so actually the exponents of x are from -k to +k, and when raised to 2*t, coefficients of x^i will give the number of ways in which 2*t numbers between -k to +k will add up to i.And for Memory to win the final score should be greater than b — a, so we are interested in sum of all the coefficients whose exponents are greater than b — a.
• » » » 6 years ago, # ^ | 0 Can you explain the code , as far as I know we can solve the numerator as a geometric series , then the coefficients in denominator can be easily found but for the numerator I have not seen a method better than exponential in the number of brackets
• » » » 5 years ago, # ^ | 0 Can you please explain why the in the transformation (diff, turn) = ( diff — 1 , turn — 1 ) + 2 ( diff — 2*k + 1 , turn — 1 ) + ........ Why we are multiplying by 2, 3 here?
» 6 years ago, # | +28 For C, does anyone have a formal proof that greedy is okay? I had some intuition that's a very handwavy proof, but I don't have a formal proof of correctness. In general, I find that proving greedy algorithms will work is tricky.
• » » 6 years ago, # ^ | 0 Someone please prove this greedy algorithm!!!
• » » 6 years ago, # ^ | 0 The minimum side is a bound for velocity of increase.
• » » » 6 years ago, # ^ | ← Rev. 2 → 0 This thought was one of the two basis of my intuition, but it's not a formal proof at all.
• » » 6 years ago, # ^ | ← Rev. 2 → 0 Which part in particular did you failed to convince yourself that the greedy algorithm must be true?
• » » 6 years ago, # ^ | ← Rev. 2 → 0 It is interesting that, reversing the problem reduces the problem complexity to a lot than it sounds.If anyone with a greedy solution without "reversing approach" is most welcome.Proof regarding the approach mentioned here, Lets set (x,y,z) be length at any step. Initially x = b y = b z = b If you notice, the every side length should be increased to min(a, (y + z - 1)). This is the maximum length any side can take. Substituting maximum value obviously minimizes the steps taken. Array of lengths of triangle is always maintained in sorted order. When I read the problem statement, This idea didn't strike me. Reversing is too powerful tool here I think.
• » » » 6 years ago, # ^ | +5 This is more or less how my intuition went, but it's not a formal proof at all. Your statement "substituting maximum velocity obviously minimizes the steps taken", is extremely hand-wavy. You could say something similar for many kind of problems where greedy doesn't work. For instance, in the coin changing problem: https://en.wikipedia.org/wiki/Change-making_problemit's well known that greedy isn't optimal for certain coin denominations, yet you could say "maximizing the amount you take away obviously minimizes the steps taken" which is a completely wrong statement, but in my view, quite analogous to your argument here.
• » » » » 6 years ago, # ^ | -23 Intuition should be enough here because you can just reverse all your steps and it makes pretty good sense.Alternatively you could also write a brute force, and then just check it yourself.
• » » 6 years ago, # ^ | +24 Here is a more detailed write-up of the proof that the greedy algorithm works. As in the explanation, we start with an equilateral triangle with side length b and must get to one with side length a ≥ b.Proof. Call a triple (x, y, z) valid if x ≤ y ≤ z and x + y > z. In other words, a triple is valid if it is sorted and corresponds to the side lengths of a non-degenerate triangle. We will say (x1, y1, z1) ≤ (x2, y2, z2) (with ≤ read as "dominated by") if x1 ≤ x2, y1 ≤ y2, and z1 ≤ z2. Suppose we have (x1, y1, z1) ≤ (x2, y2, z2) ≤ (a, a, a) with all triples valid. We show that (x2, y2, z2) can reach (a, a, a) in fewer steps than (x1, y1, z1) when using an optimal algorithm. To see this note that any sequence of side increases valid on the smaller triple can also be applied to the larger triple. More specifically, suppose we wish to increase x1 → c. Then we can increase x2 → max(c, x2) with the modified triples (once sorted) still both valid and in the same domination order (requires a little checking). The same can be seen for modifying y1 or z1. Since the domination order remains respected, the increases can be repeated inductively until reaching (a, a, a).Starting from any valid triple (x, y, z), we finally note that the side increase to (y, z, y + z - 1) gives a valid triple that dominates all other possible valid triples achievable with 1 increase. To see this first consider increases in z → c. Then c ≤ x + y - 1 and we have (x, y, c) ≤ (y, z, y + z - 1). If instead we increase y → c then we either have (x, c, z) ≤ (y, z, y + z - 1) or (x, z, c) ≤ (y, z, y + z - 1). Lastly, if we increase x → c we either have (c, y, z), (y, c, z) or (y, z, c) all of which are dominated by (y, z, y + z - 1).
• » » » 6 years ago, # ^ | +5 Thanks. Aside from the minor issue that you didn't prove you should never decrease your edge, I think this is a complete proof.Nonetheless, I'm a bit perplexed that so many people solved this question during the contest. I did too, but only because I kinda gave up on proving its correctness and hoped my intuition was correct... I imagine most people did what I did which kinda makes this problem bizarre in that in punishes people who are more careful than me (I believe something they should be rewarded for). I think such questions aren't good for contests as it encourages just random stabs at a solution (which you would never do in real life).
» 6 years ago, # | +15 For Problem D, See my O(kt^2) implementation. HereMain idea : dp(t,i) can be calculated easily using dp(t,i-1) and dp(t-1,i+k) and dp(t-1,i-k-1).. See , dp(t,i) and dp(t,i-1) has much in common.here dp(t,i) means # of ways to achieve i, after t rounds.
• » » 6 years ago, # ^ | 0 Can u please clarify ur recurrence relation,especially the transition of dp(t,i) from dp(t,i-1). Thanks in advance.
• » » » 6 years ago, # ^ | ← Rev. 4 → +14 Let dp[i][j] be in how many ways you can achieve difference j in i turns.1.dp[0][0] = 1.2..You can easily check that dp[i][j] = dp[i - 1][j - 2k] + dp[i - 1][j - 2k + 1] × 2 + ... + dp[i - 1][j - 1] × 2k + dp[i - 1][j] × (2k + 1) + dp[i - 1][j + 1] × 2k + ... + dp[i - 1][j + 2k - 1] × 2 + dp[i - 1][j + 2k].Finally to get an O(kt2) solution you can see that: which can be found with partial sums.Here is my implementation.20514640
• » » » » 6 years ago, # ^ | ← Rev. 2 → 0 How did you simplify from 2nd last equation to last equation ??
• » » » » » 6 years ago, # ^ | ← Rev. 4 → 0 Try to find dp[i][j - 1] - dp[i][j] using the third relation.dp[i][j - 1] - dp[i][j] = - (dp[i - 1][j - 2k - 1]) × (1 - 0) - (dp[i - 1][j - 2k]) × (2 - 1) - ... - (dp[i - 1][j - 1]) × (2k - (2k - 1)) - (dp[i - 1][j]) × (2k + 1 - 2k) + (dp[i - 1][j]) × (2k + 1 - 2k) + ... + (dp[i - 1][j + 2k]) × (2 - 1) + (dp[i - 1][j + 2k + 1]) × (1 - 0)
• » » » » » » 6 years ago, # ^ | 0 Understood.
• » » » » 6 years ago, # ^ | ← Rev. 2 → 0 I tried to implement your formula but it gives me some negative intermediate result. With dp[i][j] = dp[i][j-1] - sigmaA + sigmaB, I found that there are cases where sigmaA > sigmaB so dp[i][j] goes negative. Is my implementation wrong 'cause sigmaA > sigmaB will never happen technically or do I need special treatment for this case?http://codeforces.com/contest/712/submission/20563138
• » » » » » 6 years ago, # ^ | 0 If sigmaA < sigmaB add sigmaA - sgmaB + MOD otherwise sigmaA - sigmaB.
• » » » » » » 6 years ago, # ^ | ← Rev. 2 → 0 I found the problem. it was modulo operation: for(int j = left; j <= right; j++){ dp[i][j] = j-1 >= 0 ? dp[i][j-1] : 0; dp[i][j] += (psum[min(MAXGAP, j + 2*k)] - (j - 1 >= 0 ? psum[j - 1] : 0)) % MOD; dp[i][j] -= (psum[max(0, j - 1)] - (j - 2*k - 2 >= 0 ? psum[j - 2*k - 2] : 0)) % MOD; dp[i][j] = dp[i][j] % MOD; } When I calculate dp[i][101], it uses not dp[i][100], but (dp[i][100] % MOD). How stupid I was :(I popped that final modulo calculation out to another loop and now it gives me correct answer.
• » » 6 years ago, # ^ | 0 ans += (dp[now][i + shift] * pr[max(0,o+shift)]) % MOD;Can you tell me what this line means? Why we need to multiply dp[][] with pr[]?
• » » 6 years ago, # ^ | ← Rev. 2 → 0 I can't understand this ans += (dp[now][i + shift] * pr[max(0,o+shift)]) % MOD; Can you tell me why?
» 6 years ago, # | ← Rev. 2 → 0 In questions C without reversing the question and decreasing the side instead of increasing, we can still solve the question greedily, right?What advantage is reversing the question giving us?
• » » 6 years ago, # ^ | ← Rev. 3 → 0 The advantage of reversing is that it let us know the max possible value to increase. You just do something like this: a = min(x, b + c — 1).And how can you solve the problem without reversing? My attempts have failed.Notice that something like a = max(y, c — b + 1) didnt' work — it can't take us the best answer, you can check it out even in example x = 22, y = 3.
» 6 years ago, # | +6 D can be solved with Inclusion–exclusion principle in O(Kt^2).we can calculate how many way to get X scores in O(t), and all possible scores in O(Kt^2), then we can easily get the answer.see Submisson
• » » 6 years ago, # ^ | 0 why this(http://beepaste.ir/view/e6b2016b) doesn't work?? I simply added k to everything ... I don't get why using Inclusion-Exclusion principle :(
• » » 6 years ago, # ^ | 0 Really beautiful way! If anyone stumbles upon, the inclusion-exclusion here is by number of steps in which we add >2*k points.
» 6 years ago, # | +2 Can anyone explain O(ktlogkt ) approach for problem D?? In,particular how to get expression for pref2[i]??
» 6 years ago, # | +3 i am getting runtime error on test case 15. but i am not able to see test case.can anybody help here is submission link:submission
» 6 years ago, # | ← Rev. 2 → 0 When will be the test cases be visible. I am getting runtime error on test 15 for Div2D. The solution is a simple 0(k^2 . t) top-down dp carrying differences of score with an offset. Need some help with this. My Code Thanks in advance.
• » » 6 years ago, # ^ | 0 Isn't your array a little bit to small? At most there are 2*k*t+100 points difference between the two players, I believe that is larger than 1e4.
• » » » 6 years ago, # ^ | 0 Well thanks for pointing that out , now after I corrected that mistake , and also reducing the time complexity to O(KT) , the solution is giving Time limit exceeded on test case 15. Can you point out the mistake. Where am I going wrong. 20525457
• » » » » 6 years ago, # ^ | 0 Your solution does not work in O(KT). There are O(K*T) states for each turn of the game, and a total of O(K*T*T) states for the whole game, and you're taking O(K) time to compute each state of it, so that's O((KT)^2) in total.The editorial have mentioned two ways of optimization. Both of them fits in the time limit, feel free to use the one that you are comfortable with. The first approach is about the sliding windows technique and prefix sum, and the second approach is about divide and conquer with an observation on the binomial coefficients.
» 6 years ago, # | ← Rev. 2 → 0 here is my submission for problem D use partial sum 20521519 . One loop I use about 4 * 2 * MAXSCORE. Total operation ~ 800000 * t (don't contain +mod or -mod) and runtime is 200ms. CF is very fast !! PS: My source code use tab 2 and when I submit it is hard to read. How can I fix it? sorry for my bad English
» 6 years ago, # | 0 Can someone explain me C in more detail, please.
» 6 years ago, # | +2 Can someone please elaborate on the explanation of E.
• » » 6 years ago, # ^ | ← Rev. 2 → +11 *It is strongly advised to pick up your pen to do a little math to understand the solution better.I will used the same terms that the editorial uses, L[a,b]= probability dominating from a to b while starting from casino a, R[a,b]= probability of starting from casino b and ending up at casino b+1 without losing at casino a. (The editorial missed out the never losing at casino a part)From the definitions we can find out the merging formulas of L&R[a,b] and L&R[b+1,c] (a,b,c, in range of n), c does NOT NECESSARILY needs to be n.For L[a,c], it's a GS sum of L[a,b]*L[b+1,c] (note that L[a,b] makes Memory ends at b+1) + L[a,b] * (1-L[b+1,c]) * R[a,b] + ... (If Memory fails to dominate on interval [b+1,c], then multiply R[a,b] to get him back to position b+1), from the GS sum formula you can end up with a similar formula as shown in the editorial.For R[a,c], it's a sum of two parts. The first part is R[b+1,c], this is pretty obvious. The second part is the GS sum when Memory fails at casino b+1 but never fails at casino a. The first term of the GS is (chance of failing at casino b+1) * (chance of getting back to b+1 w/o failing at casino a) * (dominating on [b+1,c]), that is (1-R[b+1,c]) * (R[a,b]) * (L[b+1,c]). The ratio of the sequence is (chance of failing to dominate on [b+1,c] = 1 — L[b+1,c]) * (chance of getting back to b+1 w/o failing at casino a = R[a,b]). That also brings us to the formula that the editorial gives.My implementation: http://codeforces.com/contest/712/submission/20521550*In case if you don't know understand segment tree, go to learn it before reading this code.
» 6 years ago, # | ← Rev. 5 → 0 Can someone explain in Problem D, How's total number of games played will be (2*k+1)^(2*t) ??
• » » 6 years ago, # ^ | +4 Size of interval [ - k, k] is 2k + 1, so for each turn player has 2k + 1 "choices". For t turns each player has (2k + 1)t "choices". For two players this becomes ((2k + 1)t)2 = (2k + 1)2t "choices".
» 6 years ago, # | +5 Can someone clearly explain how we can get the formulas in problem E?
• » » 6 years ago, # ^ | ← Rev. 4 → +27 Let's solve case, when (l, r) = (1, n), and there are no modifications: solve system of linear equationsConsider a walk on casinos 0, 1, 2, ..., n, n + 1 (0 corresponds to a state "lost in casino 1", n + 1 corresponds to "won in casino n").Denote fi = "the probability to get into casino n + 1, if we're in casino i", i = 0, 1, ..., n + 1, Δi = fi - fi - 1, i = 1, 2, ..., n + 1, , and .We know that f0 = 0, fn + 1 = 1, and fi = (1 - pi)·fi - 1 + pi·fi + 1 (i = 1, 2, ..., n). We need to calculate f1 = Δ1 (a.k.a. solve system of linear equations).Rewrite this equation as fi - fi - 1 = pi·(fi + 1 - fi - 1), rewrite using Δ to get Δi = pi·(Δi + 1 + Δi), rewrite again and this becomes .Remember that Δ1 + ... + Δn + 1 = (f1 - f0) + ... + (fn + 1 - fn) = fn + 1 - f0 = 1, so Δ1(1 + u1 + u1·u2 + ... + u1·u2·...·un) = 1, and finally (formula for f1 that we were looking for)Segment tree: segmentsGeneral case (note: in segments I store quantities different from those in authors solution)for segment [L, R] store values A[L, R] = (uL·uL + 1·uL + 2·...·uR) and B[L, R] = (uL + uL·uL + 1 + uL·uL + 1·uL + 2·... + uL·uL + 1·...·uR).Recalculate them like this: A[L, R] = A[L, M]·A[M + 1, R], B[L, R] = B[L, M] + B[M + 1, R]·A[L: M].The answer for [L, R] will be .I'm also not sure, why the constraint pi ≤ pi + 1 was given, I (probably) only used the fact that pi ≠ 0.
• » » » 6 years ago, # ^ | ← Rev. 2 → 0 It was given becauze consider a test case of 50000 casinos with probability (1 — 1e-9) followed by 50000 casinos with probability 1e-9. Answer over whole interval should be 0.5(I think), but precision will make it 0.
• » » » » 6 years ago, # ^ | +3 How would one prove that something "bad" (imprecision due to overflows) wouldn't happen with condidion pi ≤ pi + 1?
• » » » 6 years ago, # ^ | +1 Thanks for explanation. I couldn't get the editorial from the post :(
• » » » 6 years ago, # ^ | 0 nicely explained thanks
» 6 years ago, # | ← Rev. 2 → +5 For E I used sqrt decomposition with the following idea:Let D(l, r) denote probability to dominate on [l, r]if l = r, then D(l, r) = ar / brelse D(l, r) = pl * (D(l + 1, r) + (1 - D(l + 1, r)) * D(l, r))The idea is simple: if at beginning, we move to left at l (probabilty (1 - pl), we lose.So now we've moved from l to l + 1, and there are two cases: Either we dominate in (l + 1, r), or we move left (thus don't dominate) and return back to D(l, r).So we can compute D(l, r) from D(l + 1, r) like this transform:D(l, r) = D(l + 1, r) / (D(l + 1, r) - 1 + 1 / pl)Since D(l, r) are fractions, represent them as a 2x1 vector denoting And we see that if we apply the above transform, it's just multiplying matrix by vector and getting a new vector.Now just build a sqrt decomposition or a segment tree to quickly get matrix multiplication for segments and multiply it by pr.
• » » 6 years ago, # ^ | 0 Can you please elaborate more on how you join results from the various sqrt segments.i.e How to do D(l,r) given D(l,k) and D(k+1, r)? Given the definition of dominate, I am not sure how just multiplying these two would work.Also how did you come up with this: "Given D(l,r) are fractions represent them as 2x1 vector"..
• » » » 6 years ago, # ^ | ← Rev. 6 → +5 I treat fractions as vectors So we have the formulaD(l, r) = D(l + 1, r) / (D(l + 1, r) - 1 + 1 / pl)Let's shortcut it as following:D(l, r) = Fl(D(l + 1, r))Here Fl is a transform that takes D(l + 1, r) as input and outputs D(l, r)D(l, r) = Fl(D(l + 1, r)) = Fl(Fl + 1(D(l + 2, r)) = Fl(Fl + 1(Fl + 2(D(l + 3, r))) = ... = Fl(Fl + 1(Fl + 2(Fl + 3(... Fr - 1(D(r, r))))))Let's see how Fl acts on fractionsAfter simlplifying we getSo we see that Fl takes a fraction and transforms it into another fraction, whose coefficients are linear combination of source coefficients.This can be written like this:Let's call this Fl operator matrix Ml and let C(l, r) be D(l, r) represented as a 2x1 vector corresponding to its fraction. E.g. if D(l, r) is , C(l, r) will be C(l, r) = Ml * Ml + 1 * ... Mr - 1 * C(r, r)So all we need is to know how to quickly compute product of matrices on a segment. Use segment tree or sqrt decomposition for this.Also this solution doesn't seem to use the knowledge that pl is non-decreasing in any way. — Oops, this is false, this is actually needed since otherwise we would lose too much precision,as described here: http://codeforces.com/blog/entry/47050?#comment-314304
» 6 years ago, # | 0 Have a look at these two submissions for problem D: 20533661 20534239The first one fails on test 17, but the other gets AC. The only difference is changing the shift length and the limit. Why is this? I thought I checked whether or not the space was large enough already, and even now I'm not sure why it was a problem.
• » » 6 years ago, # ^ | ← Rev. 5 → +1 After t turn the minimal difference will be - 2 × t × k and the maximal will be 2 × t × k so you need 4 × t × k + 1 positions.Also if you want to have negative indices you can do it with this trick: # include using namespace std; int unused[1000005]; # define a (unused + 500000) int main(void) { a[-1] = -1; a[0] = 0; a[1] = 1; cout << a[-1] << ' ' << a[0] << ' ' << a[1] << '\n'; return 0; } I used this in my submission.20514640
• » » » 6 years ago, # ^ | 0 Nice trick, by the way. I think I understand where I went wrong now. Thanks.
» 6 years ago, # | 0 Probably there is solution for D which uses fact that after multiple convolution we get very close to normal distribution...
• » » 6 years ago, # ^ | 0 What?
» 6 years ago, # | 0 Hi, for problem C, can someone explain Why does counting down give a separate answer as compared to counting up? Greedy Approach Example 22 to 4 22,22,22 -> 22,22,4 -> 22,19,4 -> 16,19,4 -> 16,13,4 -> 10,13,4 -> 10,7,4 -> 4,7,4 -> 4,4,4 4 to 22 4,4,4 -> 4,4,7 -> 4,10,7 -> 16,10,7 -> 16,10,22 -> 16,22,22 -> 22,22,22 why counting up gives us the solution? (proof for correctness) I was thinking of a DP solution for it(counting down). Is it a special case, where the optimal solution of a subproblem same as the greedy solution? Counting down means moving from 22 to 4. Counting up means moving from 4 to 22.
• » » 6 years ago, # ^ | ← Rev. 3 → 0 I guess divergence of this process is faster than convergence, that's why.Let's look at divergence rate, (y,y,y) to (x,x,x):Take the smallest, and set it to sum of other two-1.1. Clearly, as long as at least one of the other two is larger than the smallest, the number is multiplies by a real factor of at least 2.0. The only case when the smallest is multiplied by a factor < 2.0 is when all three sides are equal.We can prove that only happens when all three sides are = y. This is because when we set the smallest to sum of other two-1, we are guaranteeing that this value becomes larger than both the remaining sides(with exception of side=1, which is outside our constraints). Now, from point (1.) we know that we are effectively taking at most log(x/y) steps, as y*(2^steps) >= x. Let's look at convergence rate, (x,x,x) to (y,y,y): In the first step, set one of the numbers = y. Then, Take the largest, and set it to absolute(difference of other two) + 1.Therefore, in second step, we have set the largest value = x-y+1.In third step, we have set the largest value = x-y+1-y+1 = x-2*y+2.In fourth step, we have set the largest value = x-2*y+2-y+1 = x-3*y+3. and so on.Thus, total number of steps = 1 + x/(y-1).Solving for 1+x/(y-1) <= log(x/y) keeping y constant (c = y-1) x/c <= logx — log((1+c)*2) where c>=2 (constraints)Using z = x/cz <= log(z) -log((1+c)*(2*c)). So, never. Another way of checking for 1+x/(y-1) <= log(x/y) y=3 : x/2 <= log(x/6) => nevery=4 : x/3 <= log(x/8) => never and so on.Therefore, rate of divergence is always more than rate of convergence.
• » » 6 years ago, # ^ | 0 I guess there has to be explanation of why this greedy work,i am really curious how did people come up with this solution during the contest,i was thinking of DP all the time,trying reverse counting would never come up to my mind and if it did i am not sure if i would write it as i am unable to prove the corectness of it.
• » » 6 years ago, # ^ | ← Rev. 3 → 0 in the name of allah, most mercifulwhen you counting up you follow this greedy:min side of three side = min(max amount of side that triangle has positive area, x)where x is side that we want have it by counting up.but when we count down we don't have main idea and we don't know what is next step.
» 6 years ago, # | ← Rev. 4 → 0 Blank
• » » 6 years ago, # ^ | ← Rev. 7 → 0 in the name of allah, most merciful(b1 = a1 + a2 and bn = an) is correct and (he or she) makes mistakei think (he or she) use a instate of b because at the end there is this statement we can figure out the entire array a.proof:if(x != n){a[x] == b[x] — b[x+1] + b[x+2] — b[x+3] + ... (-or+)* b[n];a[x+1] == b[x+1] — b[x+2] + b[x+3] — ... (+or -)*b[n];so a[x] + a[x+1] == b[x];}elsea[x] == b[x]
» 6 years ago, # | 0 I just solved Problem E with two segment trees, just like the official solution. But I still don't know why the problem needs the condition that the probabilities are in non-decreasing order. Even I solved without using that condition and got AC. Is it just a fake condition?
• » » 6 years ago, # ^ | 0 So that precision errors dont occur. Look at caze with 5e4 casinos probability (1-e-9) and 5e4 casinos probability (e-9). Precision will give you a very different answer than actual
• » » » 6 years ago, # ^ | 0 oh I see. thx
» 6 years ago, # | 0 Can someone explane me solution in complexity O(k t log kt) for fourth task in more details ? I solved task in the first complexity and that is not problem, but I can not understand how we generate polynom from the solution and why coefficient near xi is equal to DP [ t ][ i ] ?
• » » 6 years ago, # ^ | ← Rev. 2 → +5 The problem is actually more mathematical than DP-ish. Start from the value 'a'. There are 2k+1 edges going out, and so on for the further levels. This gives us binomial pyramid like structure. Simply apply formula for last level only.In case you are asking about the formula itself, I haven't worked it out yet. There's a comment about solving with inclusion exclusion. Maybe you can check it out as well. :)
» 6 years ago, # | ← Rev. 2 → 0 Hi every one , for problem D I didn't hear about sliding window technique before , so i read about it and i got a shallow idea about it, but still not unerstood how we can use it here , i didn't understand anything about that, reading the editorial and the implementation .can any one help me with details. thanks in advanse.
» 6 years ago, # | 0 Can you prove the greedy approach for Div2C , during the contest I first thought of doing greedily from larger sided triangle to smaller one , obeying triangle inequality , which was incorrect. But the doing the same thing from smaller to larger passes.
• » » 6 years ago, # ^ | 0 There are 2 comments above which try to prove it. here's my attempt to prove it. Have a nice day.
» 6 years ago, # | 0 Can someone explain to me the problem statement of E? What is a valid walk here? We can never go left of i but we can go right of j? Also, why is the statement given as Note that Memory can still walk left of the 1-st casino and right of the casino n and that always finishes the process. if we can never go left of i?Please help.
• » » 6 years ago, # ^ | 0 A walk is you can go wherever you want, including left of i and right of j. A valid walk, however, means you can't move left of i and you must walk right of j to finish the process.
• » » » 6 years ago, # ^ | 0 So, as soon as I walk right of j, is my walk over? Or can I go right of j as many times I want, but I have to return to j and then win in order to finish my walk?
• » » » » 6 years ago, # ^ | 0 Walk ends the first time you go right of j.
• » » » » » 6 years ago, # ^ | 0 Thank you :)
» 6 years ago, # | 0 problem D: for input 1 1 1 2how the ans can be 31 ?? please explain me someone :)
» 6 years ago, # | 0 I found myself trapped in some kind of (seemingly) paradox in problem E.For each position i, we can get a probability pr[i] for returning to position 1. Obviously, pr[] is non-decreasing. Now there is a problem. We have 1-pr[i] probability to go back to position 1, but pr[i] is definitely not the probability you can get to destination (it is just the probability to get to i+1). However as the procedure is infinitely repeated until getting to the destination(on either side), so outside 1-pr[i] we must reach the other end.Where is my fault in reasoning?
» 6 years ago, # | 0 Hey guys,Can anyone please take a look at my solution to D and suggest any improvements. It gets accepted and it seems to be running quite fast but I would like to improve it :)http://codeforces.com/contest/712/submission/20639410
» 6 years ago, # | 0 Why the complexity of problem C is O(log x). I didn't get it.
» 5 years ago, # | 0 Can you please explain why the in the transformation (diff, turn) = ( diff — 1 , turn — 1 ) + 2 ( diff — 2*k + 1 , turn — 1 ) + ........ Why we are multiplying by 2, 3 here?
• » » 5 years ago, # ^ | 0 To get diff-2*k, there's only one way since both players need to get -k. To get diff-2*k+1, there's two ways since players can get -k and -k+1 or -k+1 and -k, respectively, and so on. You can see it's like a dice roll probability distribution with the peak at 0.
• » » » 5 years ago, # ^ | 0 Got that. Thanks a lot.
• » » » » 5 years ago, # ^ | 0 can not understand the solution. if at current turn, both player get -k , then their new difference should not be diff+(-k-(-k))=diff+0 ??
• » » » » » 5 years ago, # ^ | 0 yes, it will be but the total number of ways will be ( 2k + 1 ) like ( — k + 1 — ( — k + 1 ) ) ( k — 1 — ( k — 1 ) ) and so on
» 3 years ago, # | 0 I believe D can be solved using an easier version of prefix sums.Just calculate number of ways to get a particular score after t turns for a person, then final answer is just the appropriate convolution of the score combinations which can again be done with the already calculated prefix sums.
» 2 years ago, # | 0 can anyone explain me in problem C why decreasing greedily wont work my solution is 84129102
|
{}
|
Question
# A line passes through (2,2) and has x-intercept and y-intercept as α units and β units respectively. It makes a triangle of area A with co-ordinate axes. Then the quadratic equation whose roots are α and β is : (α>0,β>0)x2−2Ax+2A=0x2−Ax+2A=0x2+2Ax+2A=0x2+Ax+2A=0
Solution
## The correct option is B x2−Ax+2A=0Let the equation of line be xα+yβ=1 it passes through (2,2) so 2α+2β=1 ⇒2(α+β)=αβ Here, α and β are intercepts So the area of triangle =12αβ=A ⇒αβ=2A ⇒ α+β=A So, Quadratic equation will be x2−(α+β) x+αβ=0 ⇒x2−Ax+2A=0
Suggest corrections
|
{}
|
Technical Support
Forums and Mailing Lists
Search Mackichan Web
SWP & SW
Version 5/5.5
Installation
Word Processing
Computations
Typesetting
File Issues
Graphics
Exam Builder
Style Editor
Other
Technical Articles
Scientific Notebook
Support Information
Troubleshooting
TeXnology
BibDB
Document 193
## How to create a multicolumn section in a document
Version: 2.5 - Scientific WorkPlace & Scientific Word
If you're using a single-column style, you can still have part of your document appear as multicolumn text. Insert a multicols TeX command according to the instructions below.
1. Modify the style of your document:
• If you're using a Standard LaTeX (non-Style Editor) style, add the multicol package to your document.
• If you're using a Style Editor style,
1. From the File menu, choose Document Info, then choose Preamble.
2. In the Preamble area, type \def\RestoreOutput{\relax}
2. Enter the two-column commands:
1. Place the insertion point on a new line before where you want the multicolumn text to begin.
2. From the Insert menu, choose Field and then choose TeX.
3. In the edit box, enter \begin{multicols}{x} where x is the number of columns you want.
4. Encapsulate the TeX field.
5. Choose OK.
6. Place the insertion point on a new line immediately following the text you want to appear in multiple columns.
7. From the Insert menu, choose Field and then choose TeX.
8. In the edit box, enter \end{multicols}.
9. Encapsulate the TeX field.
10. Choose OK.
Revised 06/19/96
This document was created with Scientific WorkPlace.
MacKichan Software, Inc. 19689 7th Avenue NE, Suite 183 #238 Poulsbo, WA 98370
Phone: 360-394-6033 Toll-free: 877-724-9673 Fax: 360-394-6039
|
{}
|
The extraction of Mn from pyrolusite with aluminum produces pure Mn metal and $\mathrm{Al}_{2} \mathrm{O}_{3}$. Calculate the heat of reaction
Question:
The extraction of Mn from pyrolusite with aluminum produces pure Mn metal and $\mathrm{Al}_{2} \mathrm{O}_{3}$. Calculate the heat of reaction $\left(\Delta H_{\mathrm{rxn}}^{\circ}\right)$
Similar Solved Questions
A CD has diameter of 12.0 cm. If the CD is rotating at a constant angular velocity of 25 radians per second_ then the tangential velocity of a point on the circumference10mls:0 1.5 m/s:20m/s:25m/s:3.0 m/s
A CD has diameter of 12.0 cm. If the CD is rotating at a constant angular velocity of 25 radians per second_ then the tangential velocity of a point on the circumference 10mls: 0 1.5 m/s: 20m/s: 25m/s: 3.0 m/s...
VII. Samples of 10 taken in 1985 and 1995 revealed the average time people Data set for those usi...
VII. Samples of 10 taken in 1985 and 1995 revealed the average time people Data set for those using statistics spend grocery shopping decreased from 18 minutes to 14 minutes. Respective standard deviations were 5 minutes and 4 minutes. Test at the .10 level of significance whether there has been a c...
W 2 1 3 1 1 1 8 1 L 3 Mi 3 3 1 0 1 1 1 1 j L 0 1 djd0
W 2 1 3 1 1 1 8 1 L 3 Mi 3 3 1 0 1 1 1 1 j L 0 1 djd 0...
Explain what it means to say that lim f(x) = 6 and lim flx) X-3- As X approaches 3 from the left; f(x) approaches As x approacnes from the right, f(x) approaches As X approaches 3, f(x) approaches 6, but f(3) = As X approaches 3 from the right, flx) approaches As x approaches from the left; f(x) approaches 9. As X approaches 3, f(x) approaches 9 but f(3)In this situation is it possible that lim f(x) exists? Explain_Yes_ f{x) could have hole at (3, 6) and be defined such that f(3) = 9Yes; f{x) co
Explain what it means to say that lim f(x) = 6 and lim flx) X-3- As X approaches 3 from the left; f(x) approaches As x approacnes from the right, f(x) approaches As X approaches 3, f(x) approaches 6, but f(3) = As X approaches 3 from the right, flx) approaches As x approaches from the left; f(x) app...
When an alpha particle, which has a charge of +2e and mass of 4u, is accelerated through a potential difference of 58 volts, its kinetic energy increases by how much? Answer in eV (electron-volts) to the nearest eV. One eV = 1.6x10-19J, e = 1.6x10-19C, u=1.67x10-27kg.
When an alpha particle, which has a charge of +2e and mass of 4u, is accelerated through a potential difference of 58 volts, its kinetic energy increases by how much? Answer in eV (electron-volts) to the nearest eV. One eV = 1.6x10-19J, e = 1.6x10-19C, u=1.67x10-27kg....
Suppose p, q, and r are simple statements. Use grouping symbols to write the following logical...
Suppose p, q, and r are simple statements. Use grouping symbols to write the following logical statement unambiguously: pVr 9 Vp Select the correct answer below: O IPVT) = [] Vp O [pV(r = q)] VP O pV[r = (qVp)] O (PVT) = (qVp)...
Finally; we consider the full second-order polynomial model Model I: " = Bo + 81T1 + B12 + Bs13 + Barz+ Bs11m2 +6 where iid N(O,03) . Fitting this model yields an SSE value of 0.0951_ (20 points) Test if all second-order polyuomial teFIS (, I3, Model III simultaneously: and T1f2) can be dropped [romn0^ At 5% signficance level, we fal to reject Hox and conclude that the ,_ At 59 signjicance polynomial terms should be drapped. evel we fan t0 reject Ho, and conckude tmt the 0C At 59 polyromil
Finally; we consider the full second-order polynomial model Model I: " = Bo + 81T1 + B12 + Bs13 + Barz+ Bs11m2 +6 where iid N(O,03) . Fitting this model yields an SSE value of 0.0951_ (20 points) Test if all second-order polyuomial teFIS (, I3, Model III simultaneously: and T1f2) can be droppe...
Discuss as a steps and do as a typing SCHOOL OF ENVIRONMENTAL ENGINEERING EAT 478-BUILDING INFORMATION MODELLING DR ZUL-ATFİ ISMAIL ASSIGNMENT 2 Question 1 Soalan 17 Discuss the ways in which BIM...
discuss as a steps and do as a typing SCHOOL OF ENVIRONMENTAL ENGINEERING EAT 478-BUILDING INFORMATION MODELLING DR ZUL-ATFİ ISMAIL ASSIGNMENT 2 Question 1 Soalan 17 Discuss the ways in which BIM can facilitate structural analysis. Bincangkan cara penggunaan BIM dalam memudahkan analisis str...
Restaurant Drive-Through Service Timesd.)Restaurant X Restaurant Y 85 119 119 143 266 101 127 157 115 172 178 123 152 165 214 137 108 127 125 130 334 313 178 113 152 131 136 227 217 294 150 96 234 240 189 130 140 242 140 158 202 171 122 142 204 148 139 141 197 176 118 145 169 146 161 138 170 136 186 195 233 194 349 238 235 256 236 238 302 213 196 179 183 172 108 168 103 146 172 159 169 149 138 128 161 170 124 139 310 143 133 181 151 129PrintDone
Restaurant Drive-Through Service Times d.) Restaurant X Restaurant Y 85 119 119 143 266 101 127 157 115 172 178 123 152 165 214 137 108 127 125 130 334 313 178 113 152 131 136 227 217 294 150 96 234 240 189 130 140 242 140 158 202 171 122 142 204 148 139 141 197 176 118 145 169 146 161 138 170 136 1...
0.0592 = EO logQ1of 2bC = 160 RTRQBGO = RTInk0.0592 logk450 ~nfeo6 = &0InQHoll-ReactlonDivi Hall-Reaction 0, - 30 - Eox ~Cu Hg Cl * ze FZH ? II Iad - e _ig - ( s0; . 4" - 2e CHSO, Ho (u" -_Cuiving"'369 CColH,O; 1 22 FZhO Cett ~0 Pbo. 4H' ~ Ze _PbSo. 2H,0 Mno= Cno ~2Ho ZH" D-0; HO Mno; 8H" _ & _ Mn'- 4ho Au"" + Be Pbo; 4H" + Ze" _Pb" -20p Cl; ~ 2e ~XI C5o, I4H" 6 _ICr - 740 SZHO Mno, % Mn' 2H00.2224" - 20.00
0.0592 = EO logQ 1of 2 bC = 160 RTRQ BGO = RTInk 0.0592 logk 450 ~nfeo 6 = &0 InQ Holl-Reactlon Divi Hall-Reaction 0, - 30 - Eox ~Cu Hg Cl * ze FZH ? II Iad - e _ig - ( s0; . 4" - 2e CHSO, Ho (u" -_Cu ivi ng"' 369 CCol H,O; 1 22 FZhO Cett ~0 Pbo. 4H' ~ Ze _PbSo. 2H,0 M...
5. An infant patient presents with stridor and profound inhalation. He has signs of a viral infection...
5. An infant patient presents with stridor and profound inhalation. He has signs of a viral infection (e.g. fever and general malaise). What upper respiratory obstructive disorder is most likely occurring in this patient? 1. A patient presents with signs of combative behavior, hypotension...
Draw the mechanism indicated. include all curved arrows and intermediates. draw all the products appropriate for...
draw the mechanism indicated. include all curved arrows and intermediates. draw all the products appropriate for this mechanism E1 OH sulfuric acid hot...
Chapter 3 Homework Exercise 3-3 Schedules of Cost of Goods Manufactured and Cost of Goods Sold...
Chapter 3 Homework Exercise 3-3 Schedules of Cost of Goods Manufactured and Cost of Goods Sold (LO3-3) Primare Corporation has provided the following data concerning last month's manufacturing operations. points eBook Purchases of raw materials Indirect materials included in manufacturing overhe...
Question 4. show all work please 3. A soil materi al was found to have the...
Question 4. show all work please 3. A soil materi al was found to have the a modulus of elasticity of 10 ksi and v 0.5. If the soil is loaded with a rigid rectangular slab 12' long and 9' wide and carrying a uniformly distributed load of 30 psi, estimate the immediate elastic settlement ...
What is the total solubility of promethazine HCl (MW=320.89) atpH 7.4? The intrinsic solubility of promethazine is 0.001 M.Promethazine, a weak base, has a Ka of8.3×10-10. What is the maximum pH that allows a clear solution of 5%promethazine HCl without precipitation? (Use the information givenon Problem 5) What is the maximum pH that allows a clear solution of 5%lidocaine HCl without precipitation? Molecular weight of lidocaineHCl is 270.79 g/mol and So of the drug is 0.00043M. pKa o
What is the total solubility of promethazine HCl (MW=320.89) at pH 7.4? The intrinsic solubility of promethazine is 0.001 M. Promethazine, a weak base, has a Ka of 8.3×10-10. What is the maximum pH that allows a clear solution of 5% promethazine HCl without precipitation? (Use the informatio...
Cyict Uunbialsclouic labicCalculating equilibrium concentrations when the net reaction proceeds forward Consider mixture B, which will cause the net reaction to proceed forward:net-_Concentration (M) [xYI initial: 0.500 change: equilibrium: 0.500 _ %[X] + [Y] 0.100 0.100 Tc +I 0.100 + x 0.100 + xThe change in concentration, I;, is negative for the reactants because they are consumed and positive for the products because they are produced.Part BBased on a Kc value of 0.240 and the given data tabl
cyict Uunbials clouic labic Calculating equilibrium concentrations when the net reaction proceeds forward Consider mixture B, which will cause the net reaction to proceed forward: net-_ Concentration (M) [xYI initial: 0.500 change: equilibrium: 0.500 _ % [X] + [Y] 0.100 0.100 Tc +I 0.100 + x 0.100 +...
Use generating functions to solve the following: In how many ways can Bob distribute 30 green hats and 26 red hats to Harold, Gertrude, Maude, Bert and Linus so that Harold gets at most green hats_
Use generating functions to solve the following: In how many ways can Bob distribute 30 green hats and 26 red hats to Harold, Gertrude, Maude, Bert and Linus so that Harold gets at most green hats_...
Exercise 6-21 (Algo) Long-term contract; revenue recognition over time; loss projected on entire project [LO6-9] On...
Exercise 6-21 (Algo) Long-term contract; revenue recognition over time; loss projected on entire project [LO6-9] On February 1, 2021, Arrow Construction Company entered into a three-year construction contract to build a bridge for a price of $8,300,000. During 2021, costs of$2,100,000 were incurred...
0/1 points Previous Answers SEssCalcET2 2.6.027 .Findby implicit differentiation_ 3x3 Sy3 = 5Need Help?Rced IlMaichllIdLteuuutetSubmit AnswerSave ProgressPractice Another Version
0/1 points Previous Answers SEssCalcET2 2.6.027 . Find by implicit differentiation_ 3x3 Sy3 = 5 Need Help? Rced Il Maichll IdLteuuutet Submit Answer Save Progress Practice Another Version...
Let F(z) = f(z*) and G(e) = (f(z))4 You also know that @ 13,f(o) = 3,f' (a) = 14,f' (a4) = 2Then F' (a)and G' (a)Submit Question
Let F(z) = f(z*) and G(e) = (f(z))4 You also know that @ 13,f(o) = 3,f' (a) = 14,f' (a4) = 2 Then F' (a) and G' (a) Submit Question...
(44) Let X and Y be two independent continuous random variables with identical distributions N(0. 1). Find fx+r(z) = (fx fy) (2).20
(44) Let X and Y be two independent continuous random variables with identical distributions N(0. 1). Find fx+r(z) = (fx fy) (2).20...
List a five year goal plan and give steps on your plan to reach them.
list a five year goal plan and give steps on your plan to reach them....
Find and simplify the derivative of the following functions 2x2+3X-4 f(x) = x2+1
Find and simplify the derivative of the following functions 2x2+3X-4 f(x) = x2+1...
Assume a slab of PMMA (0.12-m2) is burning with an average mass loss flux of 25 g.m-2.s-1. Comple...
Assume a slab of PMMA (0.12-m2) is burning with an average mass loss flux of 25 g.m-2.s-1. Complete heat of combustion of PMMA is 25-kJ.g-1, and assume a combustion efficiency of 65%. Calculate the heat release rate resulted from burning of this slab....
A projectile is launched with an initial speed of 51.0 m/s at an angle of 31.0°...
A projectile is launched with an initial speed of 51.0 m/s at an angle of 31.0° above the horizontal. The projectile lands on a hillside 3.65 s later. Neglect air friction. (Assume that the +x-axis is to the right and the +y-axis is up along the page.) (a) What is the projectile's velocity a...
.What are the financing techniques that the hospital can use to cover these costs, in risk management? Which technique w...
.What are the financing techniques that the hospital can use to cover these costs, in risk management? Which technique would be the best to use? Why?...
Problem 4.40 To borks conna ted by Eghi horzcntal rope $t Ji rest on honzontal fnctonlass surface Blck constnt horzonta Forc 0 Eoon applied DoctF Ue 6u5 C0$ alter Lne lorce Jppltd 6XAm 18 0 m Ine mphtmass 155 kR.and bock BhaamassPan A Wik me borts are moring xhal is te lension T i the rope thal connects te IwO bbcks? Pan 8Whali tte mass 0f block B7
Problem 4.40 To borks conna ted by Eghi horzcntal rope $t Ji rest on honzontal fnctonlass surface Blck constnt horzonta Forc 0 Eoon applied DoctF Ue 6u5 C0$ alter Lne lorce Jppltd 6XAm 18 0 m Ine mpht mass 155 kR.and bock Bhaamass Pan A Wik me borts are moring xhal is te lension T i the rope thal...
Find mod (2^1003+1003! ,5)
Find mod (2^1003+1003! ,5)...
Find the area of the region bounded by Ihe graph of f and ta x-axis on the given intervalf(x) = 2 30; [0,4]The area(Type integer 0l simplified fraction )
Find the area of the region bounded by Ihe graph of f and ta x-axis on the given interval f(x) = 2 30; [0,4] The area (Type integer 0l simplified fraction )...
Q26.4: The heating element of a hair dryer dissipates 1500 W when connected to a 150...
Q26.4: The heating element of a hair dryer dissipates 1500 W when connected to a 150 V outlet. You may want to review (Pages 841 - 843) . What is its resistance? Express your answer to two significant figures and include the appropriate units....
2. Why do you think that an MNC's strategy of diversifying projects internationally could achieve low...
2. Why do you think that an MNC's strategy of diversifying projects internationally could achieve low exposure to overall country risk? (2 points)...
Given the following molecular formulas. determine the empirical formula of each compound:NzOs, PCIz; HzOz CsHaClz:HTML EditoriI 020 * X E 2 12pt Paragraph
Given the following molecular formulas. determine the empirical formula of each compound: NzOs, PCIz; HzOz CsHaClz: HTML Editori I 020 * X E 2 12pt Paragraph...
Verify the statements in Exercises $19-24 .$ The matrices are square. If $A$ is invertible and similar to $B,$ then $B$ is invertible and $A^{-1}$ is similar to $B^{-1} .\left[\text { Hint } : P^{-1} A P=B \text { for some invertible }\right.$ $P .$ Explain why $B$ is invertible. Then find an invertible $Q$ such that $Q^{-1} A^{-1} Q=B^{-1} \cdot ]$
Verify the statements in Exercises $19-24 .$ The matrices are square. If $A$ is invertible and similar to $B,$ then $B$ is invertible and $A^{-1}$ is similar to $B^{-1} .\left[\text { Hint } : P^{-1} A P=B \text { for some invertible }\right.$ $P .$ Explain why $B$ is invertible. Then find an inve...
|
{}
|
# If the mean of observation x1, x2,
Question:
If the mean of observation $x_{1}, x_{2}, \ldots, x_{n}$ is $\bar{x}$, then the mean of $x_{1}+a, x_{2}+a_{1} \ldots \ldots, x_{n}+a$ is
(a) $a \bar{x}$
(b) $\bar{x}-a$
(c) $\bar{x}+a$
(d) $\frac{\bar{x}}{a}$
Solution:
The mean of $x_{1}, x_{2}, \ldots, x_{n}$ is $\bar{x}$.
$\therefore \frac{x_{1}+x_{2}+x_{3}+\ldots+x_{n}}{n}=\bar{x}$
$\Rightarrow x_{1}+x_{2}+x_{3}+\ldots+x_{n}=n \bar{x}$
Mean of $x_{1}+a_{1} x_{2}+a, \ldots, x_{n}+a$
$=\frac{\left(x_{1}+a\right)+\left(x_{2}+a\right)+\left(x_{3}+a\right)+\ldots+\left(x_{n}+a\right)}{n}$
$=\frac{\left(x_{1}+x_{2}+x_{3}+\ldots+x_{n}\right)+(a+a+a+\ldots+a)}{n}$
$=\frac{n \bar{x}+n a}{n}$
$=\bar{x}+a$
Hence, the correct option is (c).
|
{}
|
### hmehta's blog
By hmehta, history, 8 months ago, ,
Hey All!
Topcoder SRM 767 is scheduled to start at 07:00 UTC -4, Sept 18, 2019.
Problem Writers: Arpa and minimario
Registration is now open for the match in the Web Arena or Applet and will close 5 minutes before the match begins.
Good luck to everyone!
Match Results (To access match results, rating changes, challenges, failure test cases)
Problem Archive (Top access previous problems with their categories and success rates)
Problem Writing (To access everything related to problem writing at Topcoder)
Algorithm Rankings (To access Algorithm Rankings)
Editorials (To access recent Editorials)
• +15
» 8 months ago, # | 0 Gentle Reminder: Match begins in 45mins :)
» 8 months ago, # | 0 Auto comment: topic has been updated by hmehta (previous revision, new revision, compare).
» 8 months ago, # | ← Rev. 3 → 0 Ok, so you tell me that timestamps in easy were not sorted? Suuuuper -.-
• » » 8 months ago, # ^ | +43 No excuses. You're told to validate the input and it's also mentioned in one of the sample explanations.
• » » » 8 months ago, # ^ | -14 Who reads explanations of all samples if your code passes them? It seems I was not the only one to miss this since like half of participants was hacked on this problem and this is just plain stupid.
• » » » » 8 months ago, # ^ | 0 I don't think many people were hacked on this. It's much easier to miss that last <= a, last1 = last2 if a1 = a2 and last1 <= last2 if a1 <= a2 aren't sufficient conditions — for example, I missed it at first and I imagine it constitutes a large number of hacks. Then come the conditions mentioned in samples and various silly bugs.
» 8 months ago, # | +78 And, wtf was that med?
» 8 months ago, # | +74 Sorry to be a hater, but I think this is one of the worst topcoder SRMs ever in terms of problem quality
• » » 8 months ago, # ^ | +9 It could've been worse.
» 8 months ago, # | +53 The shit was this?To clarify: Dijkstra isn't a medium problem in \${CURRENT YEAR}. Hard wasn't all that bad (it required actually thinking, wow) but annoying to code. 250 is just difficult to understand, apparently for the author too. My result before resubmitting medium (350 -> 130 points just because I used long long once but forgot when rewriting stuff) is worse than my result after challenge phase and that's worse than after systests even though I maybe-failed 250.
• » » 8 months ago, # ^ | +9 Hard wasn't all that bad (it required actually thinking, wow) but annoying to code Emmm, and what is annoying to code here? I was amazed by how this problem was short and easy to code, I think it was one the most coding-friendly hard problems.(lacks obvious fenwick tree and dfs determining just preorders)
• » » » 8 months ago, # ^ | -10 My version is shorter :)
• » » » 8 months ago, # ^ | -13 Yes yes, if you cut off stuff and codegolf other stuff, it's very short. I was talking more about the parts that aren't the main point of the problem — generator, stuff I've written a million times before... it doesn't have to be long, but it's annoying standard stuff.
• » » » » 8 months ago, # ^ | -10 Is every problem where you need to code even the simplest possible segtree or simplest possible dfs annoying for you? I don't get you
• » » » » » 8 months ago, # ^ | 0 No.
• » » » » » » 8 months ago, # ^ | 0 Yes.
» 8 months ago, # | ← Rev. 2 → +10 Oh, and most stupid bruteforces you can imagine passed to hard as well :). Cause it is kinda impossible to hack it, especially since you made it so that f[i] = w[i+1] :).EDIT: Whoops, I have not read that solution I was talking about carefully. It is brute+opt that UnstoppableSolveMachine mentioned
» 8 months ago, # | +5 Dijkstra isn't a div1 medium problem. Div1 450 has a linear-time solution and the intention of the author was to let only O(n) solutions pass. Apparently, a few of the O(n log n) solutions managed to fit into the time limit anyway. This was not intended.
• » » 8 months ago, # ^ | +46 Surprise surprise
• » » 8 months ago, # ^ | +5 I figured it has a nicer solution, but decided to write Dijkstra because there's no way that can't be squeezed in. The first attempt, with vector< vector >(N), got predictable MLE. I removed that, computed next vertices directly and oh look, I can't make it fail. So I submitted. (Then I resubmitted at the end because I didn't pay attention to long long the second time around...) Seems good enough, just with basic bitch priority_queue of pairs, an array of distances and no other structures.
• » » 8 months ago, # ^ | 0 Ah, okay I just overlooked the extra 0 in the constraint to n, :(
» 8 months ago, # | ← Rev. 2 → 0 My div2 1000, failed with Dijsktra, why ?
» 8 months ago, # | ← Rev. 2 → +26 Got AC on 1000 with the following algorithm:Perform just what said in the statement.Optimization: if we are staying in the vertex v and the current amount of fuel is f and for some previous query we had less fuel and we were able to reach the root then break.
» 8 months ago, # | +43 Anyone else have issues with hacking? Is this just an issue with the web arena?Why is there both lowercase and uppercase n's in the pseudocode? (And why no sample explanations? At least provide the answers for each query :|)
• » » 8 months ago, # ^ | +13 Noboby could challenge in last few minutes in arena
• » » » 8 months ago, # ^ | ← Rev. 2 → 0 Oh, but in my case someone hacked (er I mean challenged) the solution I was looking at a minute later. :P
• » » » 8 months ago, # ^ | 0 I successfully challenged some 2 minutes before the end and got the OK reply immediately. Seems like it was a problem just for some people.
• » » » 8 months ago, # ^ | 0 I tried to hack in last minute and lost 25 points.
• » » 8 months ago, # ^ | +10 I used the old arena and got request timeout every single time I tried to hack a brute force on 1000 (tried at least 5 times before I gave up).
• » » » 8 months ago, # ^ | 0 Same. BTW what's the 'old' arena? Is there a new one?
• » » » » 8 months ago, # ^ | 0 The Java applet one. The web arena is more recent.
» 8 months ago, # | +64 After discussing the issues with the round we decided that Division 1 will not be rated. We deeply apologize for the issues.
• » » 8 months ago, # ^ | +28 So what's the issue? The jury doesn't have a correct solution for 250? So what's wrong in the original jury solution?
• » » » 8 months ago, # ^ | 0 Seems that the 250 pts problem has been rejudged and not everybody failed. Could you please announce something more accurate about why unrated for div. 1?
• » » 8 months ago, # ^ | 0 Is div2 rating updated? Because my rating not updated yet.
» 8 months ago, # | ← Rev. 2 → +65 Btw, I may rant at topcoder and its tasks for many reasons, but actually I would be very happy to see its improvement in the problem quality (both in the interestingness and in the quality of preparation), so I would suggest the following.If an input to the problem is given as a generator, provide this generator in the copy-pasteable format. It is a pain to always spend a minute adding semicolons, changing "modulo" to "%", adding parentheses and modifying loops. And then spend a minute thinking whether there is no overflow which may be tricky even for those who have define int long long in their codes since these computations are performed on function arguments which could be just ints. 95% of contestants use C++, but it shouldn't be much of an issue adding them in all available languages, right? That would be very helpful, especially if you are going to make bugs/typos (N instead of n) in your generators and if your generators could give segfaults (par[1]=0, w[1]=B, whereas n=1 is possible).
• » » 8 months ago, # ^ | +8 With the limited number of languages, I don't understand why they don't allow the problem setters to write a driver function. It can read the parameters, generate the input then call the class method.
» 8 months ago, # | +7 Hi all.I'm sorry for the problems occurred in the contest. Even now, I don't know what happened in 250 (Div. 1).Here is the editorial. I'm currently improving it. You can put comments, I'll answer.I promise I'll hold another round and remove this bad memory.
• » » 8 months ago, # ^ | +35 I promise I'll hold another round and remove this bad memory. Could you please not?
• » » » 8 months ago, # ^ | +59 That's rude
• » » » » 8 months ago, # ^ | +38 No, it's not, he said "please".
• » » » » 8 months ago, # ^ | +15 I'm ready to pay this price if it means cancelling another bad contest
• » » 8 months ago, # ^ | +2 You can hold it on Codechef — then if something fails, nobody will really care.
• » » » 8 months ago, # ^ | ← Rev. 2 → +1 DisclaimerIt's been ages since I last opened TopCoder. Don't know what exactly happened today. How serious it is.Let's reduce the space of people who will care. Why Codechef. Use hackerearth. Call for problem setting on HackerEarthNo of fks anyone gives to hackerearth = $-1$
» 8 months ago, # | +4 All these Red handles together, who is manning NASA headquarters
» 8 months ago, # | 0 Codeforces < Topcoder
• » » 8 months ago, # ^ | +24 Yeah, lexicographically.
|
{}
|
## andrewsurfs4995 Group Title What is the vertex for the graph of y = 2x2 + 12x + 10? one year ago one year ago
1. baldymcgee6 Group Title
|dw:1358913254412:dw|
2. satellite73 Group Title
first coordinate of the vertex of $$ax^2+bx+c$$ is always $$-\frac{b}{2a}$$
3. satellite73 Group Title
in your case it is $$-\frac{12}{2\times 2}=-3$$ second coordinate is what you get when you replace $$x$$ by $$-3$$
4. andrewsurfs4995 Group Title
you both gave good responses thank you ;)
|
{}
|
# 2014 USAJMO Problems/Problem 4
## Problem
Let $b\geq 2$ be an integer, and let $s_b(n)$ denote the sum of the digits of $n$ when it is written in base $b$. Show that there are infinitely many positive integers that cannot be represented in the form $n+s_b(n)$, where $n$ is a positive integer.
## Solution 1
Define $S(n) = n + s_b(n)$, and call a number unrepresentable if it cannot equal $S(n)$ for a positive integer $n$. We claim that in the interval $(b^p, b^{p+1}]$ there exists an unrepresentable number, for every positive integer $p$.
If $b^{p+1}$ is unrepresentable, we're done. Otherwise, time for our lemma:
Lemma: Define the function $f(p)$ to equal the number of integers x less than $b^p$ such that $S(x) \ge b^p$. If $b^{p+1} = S(y)$ for some y, then $f(p+1) > f(p)$.
Proof: Let $F(p)$ be the set of integers x less than $b^p$ such that $S(x) \ge b^p$. Then for every integer in $F(p)$, append the digit $(b-1)$ to the front of it to create a valid integer in $F(p+1)$. Also, notice that $(b-1) \cdot b^p \le y < b^{p+1}$. Removing the digit $(b-1)$ from the front of y creates a number that is not in $F(p)$. Hence, $F(p) \rightarrow F(p+1)$, but there exists an element of $F(p+1)$ not corresponding with $F(p)$, so $f(p+1) > f(p)$.
Note that our lemma combined with the Pigeonhole Principle essentially proves the claim. Therefore, because there are infinitely many intervals containing an unrepresentable number, there are infinitely many unrepresentable numbers.
## Operation Intuitive Number Theory (INT)- Solution 2
I hope this solution is quite intuitive, because it is without complicated notation. It didn't take me very long to discover. As with Solution 1, define $S(n)=n+s_b(n)$. We will use numbers from $b^k$ to $b^{k+1}-1$ for induction. Call this interval Class $k$.
Start with $b^0=1$ and go up to $b-1$. There are $b-1$ numbers covered. Easily $S(n)$ ranges from $1+1=2$ to $b-1+b-1=2b-2$, or a range of $2b-3$. Thus, there are at LEAST $b-2$ numbers lacking coverage. Now, in order to fill up these gaps, we consult Class $1$: $b^1=b$ to $b^2-1$. If we are to fill up all these gaps, then we need at least $b-2$ numbers in the next series with their $S$ values. Unfortunately, there are only at MOST $b-3$ numbers that can satisfy a value, from $b$ to $2b-3$, otherwise the $S$ value is too big (note that none of them are zero)! Thus, between Class 0 and 1, there is at least one value lacking.
After setting up the base case, consider the Class $k$ numbers from $b^k$ to $b^{k+1}-1$ there are clearly $b^{k+1}-b^k$ integers. Yet the $S(n)$ can range from $b^k+1=b^k+1$ to $(b^{k+1}-1)+(k+1)(b-1)=b^{k+1}+b(k+1)-(k+1)$. This gives a range of $b^{k+1}-b^k+b(k+1)-(k+1)$. This leaves an extra $b(k+1)-(k+1)$ numbers. Now, we invoke the next class: $k+1$. Again, to fill in the gap there are sadly only $b(k+1)-(k+2)$ numbers available: $b^{k+1}$ to $b^{k+1}+b(k+1)-(k+3)$, because $s_b(n)$ is at least 1.
By induction we are done. Because each Class $k$ for integral $k$ at least $0$ misses at least one $S$ value, we miss an infinite number of numbers. Game over!
Note: If you are unconvinced of the range of $S(n)$ values we go from the smallest value of both the number and the $s_b(n)$, or $b^k$ and $1$, to the greatest of both, i.e. $b^{k+1}-1$ and $k(b+1)$, because in that series the largest possible sum is a bunch of $b-1$s.
|
{}
|
# Begin by graphing $f(x)=2^{x} .$ Then use transformations of this graph to graph the given function. Be sure to graph
###### Question:
begin by graphing $f(x)=2^{x} .$ Then use transformations of this graph to graph the given function. Be sure to graph and give equations of the asymptotes Use the graphs to determine each function's domain and range. If applicable, use a graphing urility to confirm your hand-drawn graphs. $g(x)=2^{x}+2$
#### Similar Solved Questions
##### Of all the errors on computer failurescenaunivers ty, 209 of themhardware problem; and the remainlng 809 Is sofrware problem, sample of 13 errors on compurers examined:What Is the probability that exactly four of them are software problems?What [ the probability that fewer than five of them are hardware problems?What is the probability that none of them are hardware problems?d. Find the mean number of sofware problems.Find the standard deviation of the number of hardware problems
Of all the errors on computer failures cena univers ty, 209 of them hardware problem; and the remainlng 809 Is sofrware problem, sample of 13 errors on compurers examined: What Is the probability that exactly four of them are software problems? What [ the probability that fewer than five of them are...
##### A biophysics student measures her reaction time by having a friend drop a meterstick between her...
A biophysics student measures her reaction time by having a friend drop a meterstick between her fingers. The meterstick falls 0.32 m before she catches it. What is her reaction time? Reaction time = (s) Estimate the minimum average speed in cm s−1−1 of nerve impulses going from her e...
##### 1) Use the following descriptive statistics. The descriptive statistics are based on daily stock price returns...
1) Use the following descriptive statistics. The descriptive statistics are based on daily stock price returns over the last year for two companies: eBay and Disney. a) Does Disney appear to be more or less risky than eBay? b) Which company's stock performed better in the past year? Describe whi...
12. You have to pay $12.000 a year in school fees at the end of each of the next six years. If the interest is S how much do you need to set aside today to cover these bills?... 5 answers ##### Plot the point with the given polar coordinatesCnoose the graph bolow that represents tne point ( = 3,0A Plot the point with the given polar coordinates Cnoose the graph bolow that represents tne point ( = 3, 0A... 1 answer ##### Question 3 5 pts With a 85% learning curve, if item 25 took 48 minutes to... Question 3 5 pts With a 85% learning curve, if item 25 took 48 minutes to produce, how long would item 50 take to produce (in minutes)? 24.00 40.80 42.50 7.20 5p... 1 answer ##### Sierra Sierra is making a quilt and she has determined she needs 1012 square inches of green fabric and 809 square inches of burgundy. How many square yards of each material will she need? Round your answers up to the nearest quarter yard.... 1 answer ##### Consider the supersonic flow over a 5° half-angle wedge at zero angle of attack, as sketched in figure below. The freest... Consider the supersonic flow over a 5° half-angle wedge at zero angle of attack, as sketched in figure below. The freestream Mach number ahead of the wedge is 2.0, and the freestream pressure and density are 1.01 × 105 N/m2 and 1.23 kg/m3, respectively (this corresponds to standard sea lev... 1 answer ##### Your storage firm has been offered$96,900 in one year to store some goods for one...
Your storage firm has been offered $96,900 in one year to store some goods for one year. Assume your costs are$97,000, payable immediately, and the cost of capital is 8.4%. Should you take the contract? The NPV will be s. (Round to the nearest cent.) Should you take the contract? (Select from the d...
##### Tluc following tablcan Was arriverl at while solving StaIlard form Waximization LPP:2/37573 73 The eutrics O. 0.7 in te tablcau ukuown DArameters For cch of tc following stalements linc] SOme parameter values that will make the stalememt true.
Tluc following tablcan Was arriverl at while solving StaIlard form Waximization LPP: 2/3 7573 73 The eutrics O. 0.7 in te tablcau ukuown DArameters For cch of tc following stalements linc] SOme parameter values that will make the stalememt true....
##### Problem 10 [10 pts] Determine whether cach of the following series is convergent or divergent, You...
Problem 10 [10 pts] Determine whether cach of the following series is convergent or divergent, You must justify your conclusions by identifying the test used. (a) (b) + 13 - 12X-7 13 1 (lan...
##### The American Battery Company has two divisions, the Electrical Division and the Assembly Division. Both divisions...
The American Battery Company has two divisions, the Electrical Division and the Assembly Division. Both divisions have the full authority to make purchasing and selling decisions of their division output to both outsiders and the other division. (A highly decentralized structure) Each Division opera...
##### Suppose that busses arrive at station at constant rate of busses per minute: Let Tn denote the arival time o the nth bus measured in minutes: Find the couditional expectations EIT,IT; = t] and EJT;IT, = t] where Find the covariance Cov(T,, Tn) and the correlation coefficient 0r;,T where m.
Suppose that busses arrive at station at constant rate of busses per minute: Let Tn denote the arival time o the nth bus measured in minutes: Find the couditional expectations EIT,IT; = t] and EJT;IT, = t] where Find the covariance Cov(T,, Tn) and the correlation coefficient 0r;,T where m....
##### Question 155 PointsUse the given function value(s), and trigonometric identities (including the cofunction identities), to find the indicated trigonometric functions_sec 0 = 9cos 8cos 8 = 9cos 8 =cos @ =cos 0 = 0cos 6 =
Question 15 5 Points Use the given function value(s), and trigonometric identities (including the cofunction identities), to find the indicated trigonometric functions_ sec 0 = 9 cos 8 cos 8 = 9 cos 8 = cos @ = cos 0 = 0 cos 6 =...
##### Propose a complete mechanism for the following reaction: EtOzC 1, EtO Na t OEI (1eq )EtO,Csolvent: EtOH aq HCI
Propose a complete mechanism for the following reaction: EtOzC 1, EtO Na t OEI (1eq ) EtO,C solvent: EtOH aq HCI...
##### Find the general solution of *" + 2*'10c? + 4
Find the general solution of *" + 2*' 10c? + 4...
##### 11. What is the irony of landfills? Explain the procedure ANDresults of the experiment that gave credence to thisirony.
11. What is the irony of landfills? Explain the procedure AND results of the experiment that gave credence to this irony....
##### Only nitrogen and oxygen, hat is its formula? 12. Determine the number of € Hand 0 atoms in 50.0) g of sucrose. Cn2HzOn 13 The density o aluminum metal is 2.70 glem How many aluminum atOms an
only nitrogen and oxygen, hat is its formula? 12. Determine the number of € Hand 0 atoms in 50.0) g of sucrose. Cn2HzOn 13 The density o aluminum metal is 2.70 glem How many aluminum atOms an...
##### Soybean meal is 12% protein and corn meal is 6% protein how many pounds of each should be mixed together in order to get 240 pounds mixer that is 8% protein
soybean meal is 12% protein and corn meal is 6% protein how many pounds of each should be mixed together in order to get 240 pounds mixer that is 8% protein? help pleae??...
##### Assume you dissolve 0.235 g of the weak acid benzoic acid, C6H5CO2H, in enough water to...
Assume you dissolve 0.235 g of the weak acid benzoic acid, C6H5CO2H, in enough water to make 5.00 ✕ 102 mL of solution and then titrate the solution with 0.128 M NaOH. C6H5CO2H(aq) + OH-(aq) C6H5CO2-(aq) + H2O(ℓ) What are the concentrations of the following ions at the equivalence point...
##### NiCL; is & stong clectrolyie. concentration each of thc individual ions in 4 0.650 M NiCl; solution: Deamine the[Nil-] =I-] =
NiCL; is & stong clectrolyie. concentration each of thc individual ions in 4 0.650 M NiCl; solution: Deamine the [Nil-] = I-] =...
##### Which of the following pairs is a beta emitter and its decayelement? radium-226/radon-222B. cobalt-60m/cobalt-60C. carbon-14/nitrogen-14D. strontium-85m/strontium-85E. none of the above
Which of the following pairs is a beta emitter and its decay element? radium-226/radon-222 B. cobalt-60m/cobalt-60 C. carbon-14/nitrogen-14 D. strontium-85m/strontium-85 E. none of the above...
##### Suppose an individual is randomly selected from the population of all adult males in the US. Let A be the event that the selected individual is taller than 6ft in height; and B the event that the selected individual is professional Basketball player:Compare P(AIB) to P(BIA) in magnitude, which one do you think is larger? explain why you think so, list your assumptions; if you are making any:b) Also compare P(B) to P(BlA) in magnitude? List your assumptions, if you are making any:Assume we know t
Suppose an individual is randomly selected from the population of all adult males in the US. Let A be the event that the selected individual is taller than 6ft in height; and B the event that the selected individual is professional Basketball player: Compare P(AIB) to P(BIA) in magnitude, which one ...
##### Which of the following bonds will generally have the lowest default risk? a.) Bonds issued by...
Which of the following bonds will generally have the lowest default risk? a.) Bonds issued by the US Treasury b.) Municipal bonds c.) Corporate bonds of large companies d.) Corporate bonds of small companies e.) All of these bonds will have similar default risks...
##### Determine whether the matrix is in row-echelon form. If it is, determine whether it is in reduced row-echelon form. $\left[\begin{array}{cccc}1 & 0 & 0 & 5 \\0 & 1 & -2 & 3 \\0 & 0 & 1 & 0\end{array}\right]$
Determine whether the matrix is in row-echelon form. If it is, determine whether it is in reduced row-echelon form. $\left[\begin{array}{cccc}1 & 0 & 0 & 5 \\0 & 1 & -2 & 3 \\0 & 0 & 1 & 0\end{array}\right]$...
|
{}
|
| Home | E-Submission | Sitemap | Contact Us |
Environ Eng Res > Volume 20(4); 2015 > Article
Krueyai, Punyapalakul, and Wongrueng: Removal of haloacetonitrile by adsorption on thiol-functionalized mesoporous composites based on natural rubber and hexagonal mesoporous silica
### Abstract
Haloacetonitriles (HANs) are nitrogenous disinfection by-products (DBPs) that have been reported to have a higher toxicity than the other groups of DBPs. The adsorption process is mostly used to remove HANs in aqueous solutions. Functionalized composite materials tend to be effective adsorbents due to their hydrophobicity and specific adsorptive mechanism. In this study, the removal of dichloroacetonitrile (DCAN) from tap water by adsorption on thiol-functionalized mesoporous composites made from natural rubber (NR) and hexagonal mesoporous silica (HMS-SH) was investigated. Fourier-transform infrared spectroscopy (FTIR) results revealed that the thiol group of NR/HMS was covered with NR molecules. X-ray diffraction (XRD) analysis indicated an expansion of the hexagonal unit cell. Adsorption kinetic and isotherm models were used to determine the adsorption mechanisms and the experiments revealed that NR/HMS-SH had a higher DCAN adsorption capacity than powered activated carbon (PAC). NR/HMS-SH adsorption reached equilibrium after 12 hours and its adsorption kinetics fit well with a pseudo-second-order model. A linear model was found to fit well with the DCAN adsorption isotherm at a low concentration level.
### 1. Introduction
Haloacetonotriles (HANs) are nitrogenous species of disinfection by-products (DBPs), which can be formed by reactions between chlorine, chloramine or bromine disinfectants and natural organic matter (NOM). HANs are of concern because they cause problems in wastewater and tap water. The disinfection of drinking water reduces microbial risks but increases chemical exposure to man. Increasing of human health risks are due to the formation of disinfection by-products (DBPs) in the organic and inorganic precursors. Monochloroacetronitrile (MCAN), dichloroacetonitrile (DCAN), trichloroacetonitrile (TCAN), bromochloroacetonitrile (BCAN), and dibromoacetonitrile (DBAN) are common HANs.
Several techniques can remove DBPs from water such as adsorption [1], ozonation [2] and membrane filtration [3]. Among these three processes, however, the adsorption process has been most popular for water treatment because of its low cost and simple application. Although different materials have been used to adsorb DBPs, a suitable one for use with tap water has not been identified. Consequently, a study on the removal HANs from tap water using modified adsorbents can be deemed useful.
Among porous materials, mesoporous silicas provide a number of advantages as adsorbents because of their high surface area and narrow pore size, which improves their adsorption capacity and adsorption selectivity. Moreover, hexagonal mesoporous silica (HMS) is one of many kinds of adsorbents that have been used. In the past, a polymer/silica composite has been studied as a new material that has the advantages of silica and organic polymer. Natural rubber (NR) is an organic material that many have attempted to modify because of its thermal stability and mechanical properties. A NR/HMS composite has been synthesized and characterized as an attractive prospective adsorbent.
Consequently, the objectives of this study are to investigate the adsorption efficiency of a thiol-functionalized mesoporous com posite made from natural rubber and hexagonal mesoporous silica (NR/HMS-SH) in haloacetonitrile (HANs) removal. In this study, dichloroacetonitrile (DCAN) was selected as the representative HAN because it is a chloro-DBP that is highly toxic and has a considerable presence in water. The modification of hexagonal mesoporous silica (HMS) with the thiol group (R-SH) and natural rubber was employed to investigate the effect of surface functional groups on the DCAN adsorption. The adsorption experiments were carried out in batch experiments. The adsorption kinetics and the adsorption isotherm were studied to investigate the adsorption mechanisms. Moreover, the ionic strength and pH of synthesized water was controlled so that it could adequately represent tap water in this study.
### 2.1. Preparation of NR/HMS-SH
#### 2.1.1. Materials and reagents
TEOS (AR grade, >99%), MPTMS (AR grade, 95%), and DDA (AR grade, 98%) were purchased from Sigma-Aldrich (Germany), THF (AR grade, 99.5%), C2H5OH (AR grade, 99.9%), and H2SO4 (AR grade, 98%) were purchased from QRëC (New Zealand). The NR (commercial grade) was supplied by the Thai Hua Chumporn Natural Rubber Co., Ltd. (Thailand).
#### 2.1.2. Synthesis of NR/HMS-SH
The thiol-functionalized mesoporous composite was made of natural rubber and hexagonal mesoporous silicate (NR/HMS-SH) according to the condition of NR/HMS-SO3H synthesis [4]. Firstly, 1 g of a natural rubber (NR) sheet was mixed with 30 mL of tetrahydrofuran (THF) at room temperature overnight. The NR sheet was completely dissolved in the THF to obtain a homogeneous solution. Secondly, 7.51 g of dodecylamine (DDA) was added in the solution and stirred. After 0.5 h, 21 g of tetraethoxysilane (TEOS) was added and stirred for another 0.5 h. Next, 106 g of H2O was added and stirred at 40°C for 0.5 h and then 4.96 g of 3-mercaptopropyltrimethoxysilane (MPTMS) was added into the mixture by stirring and then it was left to stand at 40°C for 1 h. The gel was aged at 40°C for 3 days, after which it was precipitated in 100 mL of ethanol. The solid product was vacuum dried at 60°C for 2 h. Finally, the template in the composite was removed with 0.05 M of H2SO4/EtOH that from the concentrated sulfuric acid was dissolved in the 99.9% ethanol at 70°C for 8 h and filtered; it was then vacuum dried with ethanol. The pH was checked at around 7 and dried at 80°C overnight.
### 2.2. Characterization of NR/HMS-SH
Structural information on the NR/HMS-SH composite was obtained using X-ray powder diffraction (XRD) analysis on X-ray power of 40 kV and 40 mA. The repeating distance between the pore centers of the hexagonal structure was calculated from the XRD data [5]. A Fourier-transform infrared (FT-IR) spectrometer equipped with a mercury cadmium telluride detector was applied for the identification of NR and the functional groups in the composite. A self-supporting disk (20 mm, 10–20 mg) was placed in the quartz cell attached to a conventional closed circulation system. All the IR spectra were recorded under evacuation at 25°C. A total of 64 scans over 400–4000/cm at a resolution of 4/cm was averaged for each spectrum.
The particle size of the adsorbent was taken by using a scanning electron microscope (SEM). The sample was fixed on an aluminum stub before it was observed by the SEM. The samples on the copper grids were observed without any metal coating.
### 2.3. Adsorption Study
Stock solutions of HANs were prepared in deionized water, and a phosphate buffer was used to adjust the pH of the solution to 7 and ionic strength (IS) to 10 mM. The kinetic study was prepared by varying the contact time from 0 to 36 h. DCAN was selected as the model HAN. The initial DCAN concentration was 1 mg/L at pH 7. After that, 0.025 g of the adsorbent and 50 mL of the DCAN solution were mixed in a 125 mL Erlenmayer flask and covered with aluminium foil. The mixture was stirred at 25oC and the supernatant solution was filtrated through a nylon syringe filter (pore size 0.45 mm). The concentration of DCAN was analyzed by gas chromatography with an electron capture detector (GC/ECD) according to EPA Method 551.1 [6]. The concentrations in the isotherm studies were varied from 0.05 to 2 mg/L in single solutions at pH 7, while the ionic strength, adjusted by a phosphate buffer, remain fixed at 10 mM. The contact time was obtained from the kinetic study. After reaching equilibrium, the mixture was separated using a nylon filter and the solution was analyzed by gas chromatography with an electron capture detector (GC/ECD).
### 3.1. Physiochemical Characteristics of the Synthesized Adsorbents
#### 3.1.1. XRD analysis
Fig. 1 shows the XRD pattern of the NR/HMS-SH composite after the extraction of the template molecules. This material exhibited one diffraction peak at 2θ in the 2.2°, corresponding to the characteristics of the hexagonal porous structure.
#### 3.1.2. FTIR spectroscopy
FTIR analysis was used to confirm the presence of NR and the SH functional group in the HMS structure of the NR/HMS-SH composites (Fig. 2). The stretching vibration of the silica framework (Si-O-Si) appeared between 1000 and 1300/cm. The broad band at around 3500/cm can be assigned to the free silanol group. The band related to the thiol group was observed at around 2550–2620/cm (S-H), but it was a very weak band.
#### 3.1.3. Electron microscopy
The particle size of a NR/HMS-SH composite is shown in Fig. 3. The presence of NR/HMS-SH enhanced the agglomeration of HMS particles. It was observed that NR was homogenously dispersed throughout the particles of the NR/HMS composite.
### 3.2. Adsorption Kinetics
The kinetic curve for DCAN adsorption on the porous adsorbent is shown in Fig. 4. A large amount of DCAN was adsorbed over PAC and NR/HMS-SH, but with PAC equilibrium was reached after a short contact time (2 h). For NR/HMS-SH, the amount of adsorbed DCAN reached equilibrium at 12 h.
Kinetic modeling and pseudo-first-order and pseudo-second-order models were employed to investigate the adsorption mechanisms. Pseudo-first-order and pseudo-second-order equations can be defined as shown by Eqs. (1) and (2).
##### (1)
$qt=qe (1-exp(-k1t))$
##### (2)
$tqt=1k2qe2+tqe$
where qe is the amount of adsorbed contaminant at equilibrium (mg/g), qt is the amount of adsorbed contaminant at time t (mg/g), k1 is the rate constant of pseudo-first-order adsorption (min−1), and k2 is the rate constant of pseudo-second-order adsorption (g/mg·min).
Based on the pseudo-second-order model, the initial adsorption rate can be determined using Eq. (3).
##### (3)
$h=k2qe2$
In order to quantitatively compare the applicability of different kinetic models, a normalized standard deviation Δq (%) was used as shown in Eq. (4).
##### (4)
$Δq(%)=100×∑[qexp-qcal)/qexp]2N-1$
where N is the number of data points and qexp and qcal (mg/g) are the experimental and the calculated adsorption capacities. The best fit models should have the lowest Δq (%) values. The kinetic parameters were calculated and the results are shown in Table 1.
The data fitting curve for NR/HMS-SH is shown in Fig. 5. It displays three regimes; the first regime presents the external mass transfer in the boundary layer and the second presents the diffusion through the pores of the adsorbent. Moreover, the last step presents the DCAN adsorption on the internal site of the adsorbent. It occurred very quickly, so the rate limiting step of the adsorption must have happened during the first or the second step. Both steps together could thus be considered the controlling step.
### 3.3. Adsorption Isotherms
NR/HMS-SH is a functionalized silica-based material that had a comparable adsorption capacity to PAC. Moreover, the results indicate that increasing the specific surface area of the adsorbent cannot be used as an explanation for DCAN’s enhanced adsorption capacities when compared with the results of a previous study, which gave a similar specific surface area for HMS that functionalized with the thiol group without adding the natural rubber (SBET NR/HMS-SH of 880.4 m2/g). The results, however, can be used to explain the density of the functional group because NR/HMS-SH provided less density of sulfur than HMS in the previous study [7]. It can be argued that the natural rubber in the adsorbent may have affected the interaction that occurred between the functional group and HMS.
In order to model the adsorption mechanism, linear, Langmuir and Freundlich isotherm models were used to test the experimental adsorption process. The linear, Langmuir and Freundlich equations can be defined as shown in Eqs. (5), (6) and (7).
##### (5)
$qe=KpCe$
##### (6)
$1qe=1KLqmCe+1qm$
##### (7)
$qe=KFCe1/n$
where qe is the amount of adsorbate adsorbed at equilibrium (mg/g), Kp is the linear constant (L/mg), qm is the maximum adsorption capacity (mg/g), and KL is the Langmuir constant (L/mg). KF and n are constants, Ce is the equilibrium concentration (mg/L), and Ce is the concentration of the adsorbate at equilibrium (mg/L).
The isotherm parameters of DCAN adsorption on NR/HMS-SH are listed in Table 2. Meanwhile, the predicted and experimental data for the equilibrium adsorption of DCAN on NR/HMS-SH are shown in Fig. 6.
### 4. Conclusions
A NR/HMS-SH composite with a high structure order and mesoporosity was successfully prepared. The formation of the NR/HMS-SH composite induced the coalescence of HMS nanoparticles, resulting in enhanced textural porosity and hydrophobicity. DCAN adsorption on the NR/HMS-SH adsorbent followed a pseudo-second order rate kinetic model. A linear model fit well with the DCAN adsorption isotherm at a low concentration level.
### Acknowledgements
The authors gratefully acknowledge the financial support for this study from the International Postgraduate Program in Hazardous Substance and Environmental Management, under the Graduate School of Chulalongkorn University. This research was also supported by the National Research University Project, under the Office of the Higher Education Commission (WCU-014-FW-57). This work was carried out as part of a research cluster on the “Fate and Removal of Emerging Micropollutants in the Environment” under a grant by the Center of Excellence for Environmental and Hazardous Waste Management (EHWM) and Special Task Force for Activating Research (STAR), of Chulalongkorn University and conducted as part of a research program on the “Control of Residual Hormones and Antimicrobial Agents from Aquacultural and the Feedstock Industry” under a grant from the Center of Excellence on Hazardous Substance Management (HSM). Technical support was provided by the Department of Environmental Engineering, under the Faculty of Engineering, of Chulalongkorn University and the Department of Chemical Technology, under the Faculty of Science, of Chulalongkorn University. The authors finally wish to thank Asst. Prof. Dr. Chawalit Ngamcharussrivichai and Mr. Sakdinun Nuntang for their insightful suggestions.
### References
1. Uyak V, Yavuz S, Toroz I, Ozaydin S, Genceli EADisinfection by-products precursors removal by enhanced coagulation and PAC adsorption. Desalination. 2007;216:334–344.
2. Vaiopoulou E, Misiti TM, Pavlostathis SGRemoval and toxicity reduction of naphthenic acids by ozonation and combined ozonation-aerobic biodegradation. Bioresour Technol. 2015;179:339–347.
3. Uyak V, Koyuncu I, Oktem I, Cakmakci M, Toroz IRemoval of trihalomethanes from drinking water by nanofiltration membranes. J Hazard Mater. 2008;152:789–794.
4. Nuntang S, Poompradub S, Butnark S, Yokoi T, Tatsumi T, Ngamcharussrivichai COrganosulfonic acid-functionalized mesoporous composites based on natural rubber and hexagonal mesoporous silica. Mater Chem Phys. 2014;147:583–593.
5. Tanev PT, Pinnavaia TJMesoporous Silica Molecular Sieves Prepared by Ionic and neutral Surfactant Templating: A Comparison of Physical Properties. Chem Mater. 1996;8:2068–2079.
6. Hodgeson JW, Cohen ALDetermination of Chlorination Disinfection Byproducts, Chlorinated Solvents and Halogenated Pesticides/herbicides in Drinking Water by Liquid-Liquid Extraction and Gas Chromatography with Electron-capture Detection. Ohio: National Exposure Research Laboratory; 1990.
7. Prarat P, Ngamcharussrivichai C, Khaodhiar S, Punyapalakul PAdsorption characteristics of haloacetonitriles on functionalized silica-based porous materials in aqueous solution. J Hazard Mater. 2011;192:1210–1218.
##### Fig. 1
XRD pattern of the NR/HMS-SH that was extracted for removing the DDA template.
##### Fig. 2
FTIR spectra of NR/HMS-SH.
##### Fig. 3
SEM image of NR/HMS-SH at a magnification (a) ×15,000 and (b) ×25,000.
##### Fig. 4
DCAN adsorption kinetics of PAC and NR/HMS-SH at initial concentration 100 μg/L, pH 7, IS = 0.01 M.
##### Fig. 5
Plot of the intraparticle diffusion model for the adsorption of DCAN on NR/HMS-SH.
##### Fig. 6
Comparison of the predicted and experimental data for the equilibrium adsorption of DCAN on the NR/HMS-SH adsorbent.
##### Table 1
Kinetic Parameters of DCAN Adsorption on NR/HMS-SH using Pseudo-first-order and Pseudo-second-order Kinetic Models
Adsorbent qe, exp (μg/g) h (μg/g·min) qe, cal (μg/g) k1 (1/min) k2 (g/mg·min) R2 Δq (%) Kinetic model
NR/HMS-SH 274 6.13 260 0.0042 - 0.8173 65.23 pseudo-first-order
268 - 7.51×10-5 0.9969 27.22 pseudo-second-order
##### Table 2
Isotherm Parameters of the DCAN Adsorption on the NR/HMS-SH Adsorbent
Isotherms NR/HMS-SH
Langmuir
qm (mg/g) -
KL (L/mg) -
R2 0.6729
q (%) 235.39
Freundlich
1/n 1.6391
KF (mg/g) 872.3685
R2 0.7861
q(%) 35.70
Linear
Kp 102.48
R2 0.8961
TOOLS
PDF Links
PubReader
Full text via DOI
Download Citation
E-Mail
Print
Share:
METRICS
1 Crossref
1 Scopus
2,499 View
36 Download
Editorial Office
464 Cheongpa-ro, #726, Jung-gu, Seoul 04510, Republic of Korea
TEL : +82-2-383-9697 FAX : +82-2-383-9654 E-mail : eer@kosenv.or.kr
Copyright© Korean Society of Environmental Engineers. All rights reserved. Developed in M2community
About | Browse Articles | Current Issue | For Authors and Reviewers
|
{}
|
# Acceleration and Displacement from a Velocity vs. Time Graph
#### gnits
Homework Statement
Find the acceleration and displacement of a particle from its velocity versus time graph
Homework Equations
a=dv/dt
Part (a) is no problem. Acceleration is the gradient of the graph in regions OA and AB which gives 3 and 0.5
Part (b), I believe, requires me to calculate the greatest and least value of the gradient of the curve in region BC
Part (c), I believe, requires me to calculate the area under the whole graph.
My question is, can I solve parts (b) and (c) without working out the equation of the circle? Unless I have miscalculated, the equation of the circle is not particularly simple, the centre is not nicely located on integral values of x and y.
My working was that the length of the chord is sqrt(15^2+20^2) = 25 and so to get the centre of the circle in order to find it's equation I should have to solve the following system:
(x-15)^2+(y-20)^2 = 625
(x-30)^2+y^2=625
Once I had the equation of the circle I would differentiate it to find max min gradients and integrate it to find area under it.
is this the way to go?
The answers the book gives are:
(b) (1/3)*(24 + 13*sqrt(3)) and (1/3)*(24 - 13*sqrt(3))
(c) 258
Thanks,
Mitch.
Related Introductory Physics Homework Help News on Phys.org
#### Delta2
Homework Helper
Gold Member
I think for (c) you definitely need the equation of the circle. I don't see any other way cause you have to integrate to find the area under the curve BC.
But for (b) I got no clue what your book means by retardation. If it means deceleration then it wants you to find the maximum and minimum of v'(t), so i think you ll have to look for solutions to the equation v''(t)=0.
#### jbriggs444
Homework Helper
I think for (c) you definitely need the equation of the circle. I don't see any other way cause you have to integrate to find the area under the curve BC.
If you are looking to find the area under a curve, integration is not the only approach. Geometry works too.
#### Delta2
Homework Helper
Gold Member
If you are looking to find the area under a curve, integration is not the only approach. Geometry works too.
Ok I cant disagree with that but I still think he needs to find the center of the circle in order to do it with the geometric approach.
#### jbriggs444
Homework Helper
Ok I cant disagree with that but I still think he needs to find the center of the circle in order to do it with the geometric approach.
The area of an equilateral triangle and the area of a 60 degree sector of a circle can both be determined without bothering about the location of the center.
#### Delta2
Homework Helper
Gold Member
The area of an equilateral triangle and the area of a 60 degree sector of a circle can both be determined without bothering about the location of the center.
Ok I think I see your approach, quite clever I should say. It involves the area of an isosceles triangle as well right?
#### jbriggs444
Homework Helper
Ok I think I see your approach, quite clever I should say. It involves the area of an isosceles triangle as well right?
We are given that the chord is equal to the radius. To me, that says "equilateral". But yes, I think you have grasped the scheme.
#### Delta2
Homework Helper
Gold Member
We are given that the chord is equal to the radius. To me, that says "equilateral". But yes, I think you have grasped the scheme.
Besides the area of the equilateral triangle and the area of respective circular sector we also need the area of the triangle BCD into the calculations, where D the point on x axis below B, don't we?
#### jbriggs444
Homework Helper
Besides the area of the equilateral triangle and the area of respective circular sector we also need the area of the triangle BCD into the calculations, where D the point on x axis below B, don't we?
Oh, that. It is a right triangle (15, 20, 25 it looks like) not isosceles. I was just considering that as part of computing the area of a trapezoid quadrilateral -- too easy to worry about.
#### gnits
Thanks for all the replies. Looking at part (c) I want to calculate the area under the whole graph. Can anyone see an error in this calculation? Area of 60 degree sector = (1/2)*25^2*PI/3 = 327.25. Area of equilateral triangle = (1/2)*25^2*sin(60) = 625*sqrt(3)/4 = 270.63. So area of segment 327.25 - 270.63 = 56.62. So area under circular part of graph = Area of (15,20,25) triangle - 56.62 = (1/2)*15*20 - 56.62 = 93.38 and so area under whole graph = (1/2)*5*15+150+(1/2)*10*5+93.98 = 305.88 NOT 258 :( Thanks for any help, Mitch.
#### Delta2
Homework Helper
Gold Member
Thanks for all the replies. Looking at part (c) I want to calculate the area under the whole graph. Can anyone see an error in this calculation? Area of 60 degree sector = (1/2)*25^2*PI/3 = 327.25. Area of equilateral triangle = (1/2)*25^2*sin(60) = 625*sqrt(3)/4 = 270.63. So area of segment 327.25 - 270.63 = 56.62. So area under circular part of graph = Area of (15,20,25) triangle - 56.62 = (1/2)*15*20 - 56.62 = 93.38 and so area under whole graph = (1/2)*5*15+150+(1/2)*10*5+93.98 = 305.88 NOT 258 :( Thanks for any help, Mitch.
I cant spot any mistake in the above calculation...
#### gnits
Thanks for checking. It's really uncharacteristic for the book I am working from to have wrong answers. When it does it's usually just an obvious sign missing or something similar. I did part (b). Can anyone confirm whether my working looks correct. Firstly I derived the equation of the circle as (x - (45/2 + 10*sqrt(3)) )^2 + (y - ( 10+15*sqrt(3)/2) )^2 = 625. It can be easily seen that the points (15,20) and (30,0) are on this circle so I am confident that this is the correct equation. Now I differentiate this to get dy/dx = (-x-(45/2)+10*sqrt(3)) / (y -10-15*sqrt(3)/2) and then I put in x = 15 and y = 20 to get the maximum gradient. This gives dy/dx = -8.3 so my answer would be 8.3 but book says (1/3)*(24 + 13*sqrt(3)) which is 15.5. Thanks for any verification or otherwise. Mitch.
#### jbriggs444
Homework Helper
I ran the same numbers without looking at your calculations and got the same answer.
My notes:
Let us start with the area of a quadrilateral bounded below by the x axis and above by:
Line segment from (0,0) to (5,15). This triangle is 5 units wide by 15 units high. Area = 5*15/2 = 37.5
Line segment from (5,15) to (15,20). The average height is (15+20)/2. Multiply by width 10. Area = 35*10/2 = 175
Line segment from (15,20) to (30,0). This triangle is 15 units wide by 20 units high. Area = 15*20/2 = 150
Add them up and we get 325+37.5 = 362.5
But we have to subtract out that crescent-shaped segment where the circular arc eats into the quadrilateral.
I get $\frac{1}{6}\pi r^2$ = 327.25 for the area of the circular sector and $\frac{r^2\cos 30}{2}$ = 270.63 for the area of the equilateral triangle. (r=25 of course). The delta is 56.62 for the crescent.
Subtract 56.62 from 362.5 and get 305.88
#### neilparker62
Homework Helper
Thanks for checking. It's really uncharacteristic for the book I am working from to have wrong answers. When it does it's usually just an obvious sign missing or something similar. I did part (b). Can anyone confirm whether my working looks correct. Firstly I derived the equation of the circle as (x - (45/2 + 10*sqrt(3)) )^2 + (y - ( 10+15*sqrt(3)/2) )^2 = 625. It can be easily seen that the points (15,20) and (30,0) are on this circle so I am confident that this is the correct equation. Now I differentiate this to get dy/dx = (-x-(45/2)+10*sqrt(3)) / (y -10-15*sqrt(3)/2) and then I put in x = 15 and y = 20 to get the maximum gradient. This gives dy/dx = -8.3 so my answer would be 8.3 but book says (1/3)*(24 + 13*sqrt(3)) which is 15.5. Thanks for any verification or otherwise. Mitch.
Also getting 8.3 as maximum gradient by finding the centre D (22.5 + 10√3 ; 10 + 7.5√3) and hence gradients of DB and DC. Maximum gradient will be that of tangent line perpendicular to DB and minimum gradient that of tangent line perpendicular to DC.
#### Williambecker1
Great job you are doing. Keep it up. I really like it
#### gnits
Thanks averybody for all of your help. I'm going with our answers and in this case assuming that the book is incorrect. Thanks again, Mitch.
Last edited:
"Acceleration and Displacement from a Velocity vs. Time Graph"
### Physics Forums Values
We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving
|
{}
|
# What equal monthly investment is required over a period of 40 years to achieve a balance of...
What equal monthly investment is required over a period of 40 years to achieve a balance of $2,200,000 in an investment account that pays monthly interest of 0.25% if monthly interest is 2.5%? ## Savings accounts Savings account are used to save up money for a future time. They pay an interest which is compounded usually monthly. The key to accumulating a larger amount is to get the best interest and let it compound for a long time. ## Answer and Explanation: The answer is at .25% monthly interest for 40 years, the monthly investment required to achieve a balance of$2,200,000 is \$2,375.66.
At 2.5 % interest...
See full answer below.
Become a Study.com member to unlock this answer! Create your account
|
{}
|
New Titles | FAQ | Keep Informed | Review Cart | Contact Us Quick Search (Advanced Search ) Browse by Subject General Interest Logic & Foundations Number Theory Algebra & Algebraic Geometry Discrete Math & Combinatorics Analysis Differential Equations Geometry & Topology Probability & Statistics Applications Mathematical Physics Math Education
Complexity of Proofs and Their Transformations in Axiomatic Theories
SEARCH THIS BOOK:
Translations of Mathematical Monographs
1993; 153 pp; hardcover
Volume: 128
ISBN-10: 0-8218-4576-4
ISBN-13: 978-0-8218-4576-9
List Price: US$75 Member Price: US$60
Order Code: MMONO/128
The aim of this work is to develop the tool of logical deduction schemata and use it to establish upper and lower bounds on the complexity of proofs and their transformations in axiomatized theories. The main results are establishment of upper bounds on the elongation of deductions in cut eliminations; a proof that the length of a direct deduction of an existence theorem in the predicate calculus cannot be bounded above by an elementary function of the length of an indirect deduction of the same theorem; a complexity version of the existence property of the constructive predicate calculus; and, for certain formal systems of arithmetic, restrictions on the complexity of deductions that guarantee that the deducibility of a formula for all natural numbers in some finite set implies the deducibility of the same formula with a universal quantifier over all sufficiently large numbers.
|
{}
|
# What is the simplest chiral spherical harmonic?
What is the simplest (lowest order) real spherical harmonic that cannot be rotated into its own mirror image? I would like to make an illustration of a simple smooth chiral object, but cannot find a good example.
Edit: I should have clarified that I need a linear combination of spherical harmonics, not just one of them.
-
I'm not sure if I understand chirality correctly, but maybe you could take a linear combination of basic harmonics composed with incommensurate rotations? – anon Aug 26 '11 at 9:58
Apologies for my incorrect answer; you're quite right that linear combinations of $l=1$ harmonics can be rotated into their mirror images. Adding an $l=0$ function won't help, so that shows you'll need at least $l=2$. An arbitrary linear combination of the nine functions up to $l=2$ is chiral (even three $l=1$ and one $l=2$ should suffice), so that raises the question how exactly you define "simplest", since "lowest order" doesn't distinguish between the various possibilities including $l=2$ functions. – joriki Aug 26 '11 at 14:10
Thanks for your comment. I don't think arbirary combinations are chiral, for exampel you can make the functions independent of z and then they are not chiral. But maybe most functions are indeed chiral. For example we can take the harmonic polynomial $x+y+z + xy$, can you prove that this is chiral? I am also interested in how to prove these things. – uekstrom Aug 26 '11 at 14:41
I've updated and undeleted my answer. About "arbitrary combinations": By "arbitrary" I meant something like "in general position". As it happens, your example is in the set of measure zero of achiral combinations :-) – joriki Aug 29 '11 at 9:20
Here's a way to determine rigorously whether a given linear combination of spherical harmonics is chiral. The functions with different values of $l$ don't get mixed under rotations. Thus, a combination is achiral if and only if there's a single rotation that rotates the mirror images of all $l$ components into their originals. So you can look at the $l$ values individually and see whether there's any intersection among the sets of rotations.
For your example, $x+y+z+xy$, the $l=1$ component $x+y+z$ is a rotated version of $Y_{10}$ oriented towards $(1,1,1)$. The rotations that rotate the mirror image of this into the original are rotations through $\pi$ about an axis perpendicular to $(1,1,1)$. (Subsequently rotating about $(1,1,1)$ doesn't add anything; it just corresponds to a different choice of axis for the first rotation.)
The $l=2$ component $xy$ is invariant under inversion, so we need to find the rotations that leave it invariant. Its angular dependence is $\sin^2\theta\sin\phi\cos\phi$, and the only rotations that leave the $\theta$ dependence invariant are ones with axes in the $x$-$y$ plane or orthogonal to it (since the poles have to end up at the poles). There's one axis in the $x$-$y$ plane that's orthogonal to $(1,1,1)$, namely $(1,-1,0)$. Coincidentally, rotating through $\pi$ about that axis corresponds to $x\to-y$, $y\to-x$, $z\to-z$, and thus works for $x+y+z+xy$. However, this is only one of two axes in the $x$-$y$ plane for which rotating through $\pi$ leaves $xy$ invariant, since rotating the rotation axis about the $z$ axis by $\alpha$ amounts to rotating the result about the $z$ axis by $2\alpha$, and that only leaves $xy$ invariant if $2\alpha$ is a multiple of $\pi$. So you can pick almost any other combination of the $l=1$ functions, as long as the associated vector isn't orthogonal to either the $z$ axis or the axes $(1,-1,0)$ and $(1,1,0)$, and add $xy$ to get a chiral combination.
|
{}
|
# Evaluating an integral using Cauchy's Integral Formula
I am having a little bit of trouble with the following:
$$\int_{\gamma}\frac{z^2-1}{z^2+1}dz$$ where $\gamma$ is a circle of radius $2$ centered at 0. I am trying to separate this or simplify it into the form in which we can maybe apply Cauchy's differentiation formula but this isn't working. I did get this:
$$\frac{z^2-1}{z^2+1} = 1-\frac{2}{z^2+1}$$ but this doesn't seem to take me anywhere. Any hints would be helpful.
Edit:
Would it be possible to do this:
$$\frac{z^2-1}{z^2+1} = 1-\frac{2}{z^2+1}= 1-\frac{\frac{2}{z+i}}{z-i}$$ and then just apply cauchy's integral formula with $z_0=i$? and $f(z)=\frac{2}{z+i}$?
Using this, we would have
$$\frac{2}{i+i}2\pi i + \frac{2}{-i-i}2\pi i = 0$$ as the answer?
• You have another pole, at $-i$, that is inside the circle. ${}\qquad{}$ – Michael Hardy Mar 9 '15 at 7:10
• @MichaelHardy My trouble is I don't particularly understand what you mean by pole. I'm assuming you mean a point in which we have a problem as in $i$ and $-i$ because this gives us the two trouble points inside the curve. I don't know where to go from here. I just learned about Cauchy's Integral Formula. – H5159 Mar 9 '15 at 7:11
• @MichaelHardy Would we do two different integrals one for $-i$ and one for $i$ and add them together? Should the answer be 0? – H5159 Mar 9 '15 at 7:15
• I posted a method using partial fraction expansion, which leads to a fairly quick result that might be surprising. – Mark Viola Mar 9 '15 at 16:41
• I guess the problem is that Frumpy knows a "Cauchy differentiation formula" but doesn't know the "Cauchy integral formula". And doesn't know what is a pole, so certainly doesn't know what is a residue. – GEdgar Mar 9 '15 at 16:50
$$\int_\gamma 1\,dz=0$$ because $\gamma$ returns to its starting point.
Next we have $$z^2+1 = (z-i)(z+i).$$
So the integral should involve the sum of residues at $\pm i$, since $\gamma$ winds once around each of those two points.
PS: Your proposal to apply Cauchy's formula at $i$ to the function $$1-\frac{\frac{2}{z+i}}{z-i}$$ would work if not for the fact that the numerator $\dfrac2{z+i}$ also has a pole inside the curve $\gamma$. You need to take the residue at that point into account as well.
Partial fraction expansion yields $$\frac{z^2-1}{z^2+1}=\frac{z^2-1}{2i}\left(\frac{1}{z-i}-\frac{1}{z+i}\right)$$ and application of the Residue Theorem reveals that the integral is $$=2\pi i\left(\frac{(i)^2-1}{2i}-\frac{(-i)^2-1}{2i}\right)=0$$
|
{}
|
Next: , Previous: Laguerre Functions, Up: Special Functions
### 7.23 Lambert W Functions
Lambert's W functions, W(x), are defined to be solutions of the equation W(x) \exp(W(x)) = x. This function has multiple branches for x < 0; however, it has only two real-valued branches. We define W_0(x) to be the principal branch, where W > -1 for x < 0, and W_{-1}(x) to be the other real branch, where W < -1 for x < 0. The Lambert functions are declared in the header file gsl_sf_lambert.h.
— Function: double gsl_sf_lambert_W0 (double x)
— Function: int gsl_sf_lambert_W0_e (double x, gsl_sf_result * result)
These compute the principal branch of the Lambert W function, W_0(x).
— Function: double gsl_sf_lambert_Wm1 (double x)
— Function: int gsl_sf_lambert_Wm1_e (double x, gsl_sf_result * result)
These compute the secondary real-valued branch of the Lambert W function, W_{-1}(x).
The GNU Scientific Library - a free numerical library licensed under the GNU GPL
Back to the GNU Scientific Library Homepage
|
{}
|
• 论文 •
### 中国桂樱属植物的分类研究
1. 北京, 中国科学院植物研究所
• 出版日期:1984-12-15 发布日期:2016-06-13
### TAXONOMIC STUDY ON THE GENUS LAUROCERASUS TOURN EX DUH.OF CHINA
Lu Ling-ti
1. Institute of Botany, Academia Sinica, Beijing
• Online:1984-12-15 Published:2016-06-13
Abstract: The present paper is a preliminary study on Chinese specics of the genus Laurocerasus Tourn. ex Duh., in which Tournefort(Institutiones rei Herbariae 627. t. 403. 1700) first put two species known as Prunus laurceerasus and P. lusitanica, but Linn-aeus transferrd them to the genus Padus Mill. and in Sp Plant. 1753 those species were placed in the rather inclusive genus Prunus L. The systematic position of this genus has been much disputed during more than two centuries after its delimitation. Roemer, Schneider, Komarov and Hutchinson recognized Lcu-rocerasus as an independent genus. In the early part of the 19th century it was reduced by some authors, e. g. Seringe, Rei-chenbach and others to a subgenus or section of the genus Cerasus Mill. Up to the late 19th century to early 20th centuryBentham, Hooker J. D. and Koehne coined an inclusive genus Prunus, therefore the genus Laurocerasus was brought to Prunus as a section or subsection. In later times, Rehder, Kalkman and Vidal considered it to be a subgenus of Prunus. During the course of our writing the Flora of China all the species of Laurocerasus and its allied genera such as Pygeum Gaertn. and Padus Mill. collected from the whole China for the past more than fifty years have been examined. After a detailed investigation of the morphological characters and analysis of geographical distribution of Laurocerasus we accepted the generic concept of Roemer, Schneider and so on, regarded it as an independent genus. The genus Laurocerasus differs from the closely related genus Padus by its leaves usually evergreen, coriaceous, margin entire or sparsely serrate and inflorescence leafless at base, from the genus Pygeum, which was referred by Kalkman to a section of Prunus subgen. Laurocerasus in having flowers with a biseri-ate, usaally regularly 5-merous perianth, petals in shape and demensions distinct from sepals, 2 or more times as long as the latter, leaves sparsely serrate or rarely entire, while in the genus Pygeum the perianth segments 5-10 (-15) and in most of its species petals not or hardly differt from sepals, or, if sometimes petals and sepals distinguishable in shape, but petalsat most 1½ (very rarely up to 2) times as long as the latter. leaves always entire. The 13 species and 9 forms of Laurocerasus recognized inChina, among them 4 new species and 5 new forms have been discri-bed, are divided into two groups which are regarded inthis treat-ment as sections, i. e. Sect. Phaeostictae Yü et Lu and Sect. Laurocerasus. The Sect. Phaeostictae Yü et Lu, containing 2 speciesand 5 forms, is characterized by leaves with densely dark glan-dular points beneath, while Sect. Laurocerasus, comprising 11 species and 4 forms, is grouped into 4 series which distingtuish from Sect. Phaeostictae by leaves epunctate on lower surface. In respect to the geographical distribution the genus Laurocerasus occurs mainly in tropical regions, i. e. from Africa, southern Asia, southeastern Asia, New Guinea to South America, some species spread to subtropical and cooltemperate areas, i. e. from southwestern and southeastern Europe to Near East, China and Japan. There are about 20 species distributed in the Old world, of which 13 species have been recorded in southern China and other regions (Vietnam, Upper Burma to northeastern India) which are adjacent to it, amount to 65% of the total number of the Old world. It has shown clearly that the distribution centre of Laurocerasus lies in southern China and its neighbouring regions. There are only few species extending their areas outside this distributional centre. In China the genus is restricted in southern parts of southern slopes of the mountain Qin Ling, the southwestern part of Shanxi and southeastern part of Gansu are the northern dist-ributional limits of this genus. By analysis of the distribution of species in China the great majority of them inhabit the southern parts of Yangtze River where exist the greatest number of species of Laurocerasus, particularly abundant in southern and southwestern parts of China. There are 13 species occupied in China, of which the species distributed in southern (8) and southwestern China (7) amount to 61. 5% and 54% of the total number of the whole China respeetively. From the above faets we may now concluded that the greatest density of species in the distributional centre of Laurocerasus is now found in southern and southwestern Parts of Cilina.
|
{}
|
$$\displaystyle{\left({3}{x}^{{2}}+{2}{x}+{4}\right)}+{\left({8}{x}^{{2}}-{x}+{1}\right)}=$$
• Questions are typically answered in as fast as 30 minutes
### Solve your problem for the price of one coffee
• Math expert for every subject
• Pay only if we can solve it
Anonym
Example:
$$\displaystyle{\left({x}+{2}\right)}+{\left({x}^{{2}}+{3}{x}+{4}\right)}={x}+{2}+{x}^{{2}}+{3}{x}+{4}$$
$$\displaystyle={x}^{{2}}+{3}{x}+{x}+{2}+{4}$$
$$\displaystyle={x}^{{2}}+{4}{x}+{6}$$
Here we have combined the terms x with x and constants with constants.
Applying addition property of the polynomial for the given expression, we get
$$\displaystyle{\left({3}{x}^{{2}}+{2}{x}+{4}\right)}+{\left({8}{x}^{{2}}-{x}+{1}\right)}={3}{x}^{{2}}+{2}{x}+{4}+{8}{x}^{{2}}-{x}+{1}$$
$$\displaystyle={3}{x}^{{2}}+{8}{x}^{{2}}+{2}{x}-{x}+{4}+{1}$$
$$\displaystyle={11}{x}^{{2}}+{x}+{5}$$
Hence, we can fill in the blanks:
$$\displaystyle{\left({3}{x}^{{2}}+{2}{x}+{4}\right)}+{\left({8}{x}^{{2}}-{x}+{1}\right)}={11}{x}^{{2}}+{x}+{5}$$
|
{}
|
Canadian Mathematical Society www.cms.math.ca
location: Publications → journals → CJM
Abstract view
# Group Actions and Codes
Published:2001-02-01
Printed: Feb 2001
• V. Puppe
Format: HTML LaTeX MathJax PDF PostScript
## Abstract
A $\mathbb{Z}_2$-action with maximal number of isolated fixed points'' ({\it i.e.}, with only isolated fixed points such that $\dim_k (\oplus_i H^i(M;k)) =|M^{\mathbb{Z}_2}|, k = \mathbb{F}_2)$ on a $3$-dimensional, closed manifold determines a binary self-dual code of length $=|M^{\mathbb{Z}_2}|$. In turn this code determines the cohomology algebra $H^*(M;k)$ and the equivariant cohomology $H^*_{\mathbb{Z}_2}(M;k)$. Hence, from results on binary self-dual codes one gets information about the cohomology type of $3$-manifolds which admit involutions with maximal number of isolated fixed points. In particular, most'' cohomology types of closed $3$-manifolds do not admit such involutions. Generalizations of the above result are possible in several directions, {\it e.g.}, one gets that most'' cohomology types (over $\mathbb{F}_2)$ of closed $3$-manifolds do not admit a non-trivial involution.
Keywords: Involutions, $3$-manifolds, codes
MSC Classifications: 55M35 - Finite groups of transformations (including Smith theory) [See also 57S17] 57M60 - Group actions in low dimensions 94B05 - Linear codes, general 05E20 - Group actions on designs, geometries and codes
top of page | contact us | privacy | site map |
© Canadian Mathematical Society, 2015 : https://cms.math.ca/
|
{}
|
### Home > CCA2 > Chapter 3 > Lesson 3.2.2 > Problem3-79
3-79.
How many solutions does each equation below have?
What is the highest power of $x$? That is the number of solutions you should be trying to find. But, remember that some equations have no solution and others have infinitely many solutions. Try solving each equation.
1. $4x + 3 = 3x + 3$
1 solution
1. $3(x − 4) − x = 5 + 2x$
1. $(5x − 2)(x + 4) = 0$
In this problem you have to multiply the two factors to see the highest power of $x$ if you don't recognize the number of solutions it will have.
1. $x^2 − 4x + 4 = 0$
|
{}
|
# Center for Analysis and Design of Intelligent Agents
### Site Tools
public:t-701-rem4:scales_display_of_data
# Differences
This shows you the differences between two versions of the page.
public:t-701-rem4:scales_display_of_data [2008/10/29 14:58]
thorisson
public:t-701-rem4:scales_display_of_data [2008/10/29 15:00] (current)
thorisson
Line 76: Line 76:
===Edward Tufte's Six Grand Principles of Information Display=== ===Edward Tufte's Six Grand Principles of Information Display===
-| The First Grand Principle: Enforce Wise Visual Comparisons., i.e., force answers to the question "Compared with What?" | +| The First Grand Principle: **Enforce Wise Visual Comparisons**, i.e., force answers to the question "Compared with What?" |
-| The Second Grand Principle: Show Causality. We are looking at information to understand mechanisms. Policy reasoning is about examining causality. Napoleon was defeated by the winter, not the opposing army, as shown by the temperature scale o n the bottom of Minard's graph. | +| The Second Grand Principle: **Show Causality**. We are looking at information to understand mechanisms. Policy reasoning is about examining causality. Napoleon was defeated by the winter, not the opposing army, as shown by the temperature scale o n the bottom of Minard's graph. |
-| The Third Grand Principle: The World We Seek to Understand is Multivariate, as Our Displays Should Be. The Minard graph has six dimensions: size of the army, the two dimensional route of the march, the direction of the march, the temperatures and the dates. | +| The Third Grand Principle: **The World We Seek to Understand is Multivariate, as Our Displays Should Be**. The Minard graph has six dimensions: size of the army, the two dimensional route of the march, the direction of the march, the temperatures and the dates. |
-| The Fourth Grand Principle: Completely Integrate Words, Numbers and Images. Don't let the accidents of the modes of production break up the text, images and data. Just because the artists, technical writers and database people work in differen t buildings doesn't mean reports should be disjoint with text, graphs and images in different boxes or on different pages. | +| The Fourth Grand Principle: **Completely Integrate Words, Numbers and Images**. Don't let the accidents of the modes of production break up the text, images and data. Just because the artists, technical writers and database people work in differen t buildings doesn't mean reports should be disjoint with text, graphs and images in different boxes or on different pages. |
-| The Fifth (most important) Grand Principle: Most of What Happens in Design Depends upon the Quality, Relevance and Integrity of the Content. Minard's graphic was made as an anti-war poster. To improve a presentation, get better content. If yo ur numbers are boring you have the wrong numbers. Design won't help, it is too late. \\ Page 18 of Envisioning Information shows a book by Galileo published in 1613 which reported the discovery of sunspots and the rings of Saturn for the first time. He wrote in Italian, not Latin, because he wanted to reach a wider audience than the scie ntific elite. His tone of writing is wide eyed, straight-forward, undiplomatic, sardonic and sounds a lot like the modern voice of Richard Feynman. The report of the discovery of sunspots has a simple drawing of the sun on each page to show daily obser vations. From these observations he learned that the sun was rotating as the spots moved across the page and changed apparent shape at the edges due to foreshortening. It is easy to make comparisons between the left hand and right hand pages because the y are within the eye span. | +| The Fifth (most important) Grand Principle: **Most of What Happens in Design Depends upon the Quality, Relevance and Integrity of the Content**. To improve a presentation, get better content. If your numbers are boring you have the wrong numbers. Design won't help, it is too late. \\ Page 18 of Envisioning Information by Edward Tufte shows a book by Galileo published in 1613 which reported the discovery of sunspots and the rings of Saturn for the first time. He wrote in Italian, not Latin, because he wanted to reach a wider audience than the scie ntific elite. His tone of writing is wide eyed, straight-forward, undiplomatic, sardonic and sounds a lot like the modern voice of Richard Feynman. The report of the discovery of sunspots has a simple drawing of the sun on each page to show daily obser vations. From these observations he learned that the sun was rotating as the spots moved across the page and changed apparent shape at the edges due to foreshortening. It is easy to make comparisons between the left hand and right hand pages because the y are within the eye span. |
-| The Sixth Grand Principle: Information for Comparison Should be Put Side by Side., i.e., within the eye span, not stacked in time on subsequent pages, which is known as 'one damn thing after another', and also known as the computer interface. T he computer interface is a low-resolution display device compared to paper, so we have a relentless sequentiality. The most common user question after a sequence of computer operations is "Where am I?" The lesson: get the biggest monitor of the highest resolution that you possibly can. \\ \\ One of Tufte's students scanned Galileo's images and animated them so the sun of 1612 could be seen to rotate. At a couple points in the annimation the images skip forward because there was missing data due to clouds, or Galileo taking a day off. \\ \\ A Jesuit rival of Galileo republished the sunspot data (see p17 of Envisioning Information). He used the single most effective tool of information design, the small multiple, which puts all 38 images within the eye span. |+| The Sixth Grand Principle: **Information for Comparison Should be Put Side by Side**, i.e., within the eye span, not stacked in time on subsequent pages, which is known as 'one damn thing after another', and also known as the computer interface. The computer interface is a low-resolution display device compared to paper, so we have a relentless sequentiality. The most common user question after a sequence of computer operations is "Where am I?" The lesson: get the biggest monitor of the highest resolution that you possibly can. \\ \\ One of Tufte's students scanned Galileo's images and animated them so the sun of 1612 could be seen to rotate. At a couple points in the annimation the images skip forward because there was missing data due to clouds, or Galileo taking a day off. \\ \\ A Jesuit rival of Galileo republished the sunspot data (see p17 of Envisioning Information). He used the single most effective tool of information design, the small multiple, which puts all 38 images within the eye span. |
\\ \\
|
{}
|
Calculate current flowing from one cylinder to another
1. Jul 23, 2017
1. The problem statement, all variables and given/known data
Two long concentric cylinders of radii .04 m and .08 m are separated by aluminum. The inner cylinder has a charge per unit length of $\Lambda$ at any time. When the two cylinders are maintained at a constant potential difference of 2 V via an external source, calculate the current from one cylinder to the other if the cylinders are 1 m long
2. Relevant equations
potential difference = IR
R = pl / A
I = dq/dt
J = I/A
J = neV
E = pJ
3. The attempt at a solution
So my first attempt I used:
potential difference = IR
(2 V ) / (R) = I
(R) = pl / A
length is 1 meter
I took area by pi(r)^2 outside - pi(r)^2 inside, pi(.08)^2 - pi(.04)^2 , .020 - .005 = .015 m^2
p is given in the book, p for aluminum is 2.655x10^-8
plugging in, R = 1.77 x 10^-6
so (2 V) / R = I
(2 V ) / (1.77 x 10^-6) = 1.1 x 10^6
But my answer is way off from my books, with the answer being: 6.8 x 10^8 A
What did I do wrong? I think the lambda at the beginning is a hint and that I might have to use Guass law??
My next thought was:
E according to gauss law = (lambda)/(2pi\epsilon0r)
E/p = J
J * A = I
but how am I suppose to get a value for lambda?? sorry that part gets me mixed up :/
Can anyone comment on which method they think is leading me to the right direction? Also something I might be doing wrong??
Last edited: Jul 23, 2017
2. Jul 23, 2017
Staff: Mentor
Your method for the calculation of the resistance of the aluminum tube is not correct. The current is flowing laterally, from the inside surface of the tube to its outside surface, not lengthwise down the tube. You'll want to find that resistance using integration -- think of the tube as consisting of nested cylinders of differential thickness.
3. Jul 23, 2017
Ok this is starting to make sense to me... am I right when I say:
I am using (dR) = d(pl / A)
I am going to be integrating with respect to dr? my integral is going to go from inside radius to big radius?
Since im taking the Area of the tube part, the inside tubes area is circumfrence * length right?
so A = 2pi(r_inside) * length of the cylinders (L)
but I'm confused on the top part.
Since l is not the length of the cylinders, but instead its coming from the surface of the inside cylinder to the surface of the outside one.
So if I replaced this with r, considering that I have A = 2pi(r_inside) * length of the cylinders (L) on the bottom, the r's would cancel and I would be left with dr?
so $R = (p/2piL) \int_{r_in}^{r_out} dr$
is that my integral?
Last edited: Jul 23, 2017
4. Jul 23, 2017
Or maybe because I know that .08 m - .04 m = .04 m, I can substitute that for l, pull p and l out, a = 2 pi r l, so I pull out pl/2pi(l)
that way I'm just integrating dr/r, which is ln(x) = ln(outside/inside radius)
5. Jul 23, 2017
Ok, ignore that top post.
I have R = pL / A
so I must integrate
p dr / 2pirL
p/2piL integral dr/r from .04 to .08
p/2piL ln(.08/.04) = R
wow guys.
After subbing that value of R into I = RV, i finally got the same answer as my text book.
I can't believe I made the integration right thanks as usual Gniel your a gangsta bro
|
{}
|
If $$f$$ and $$g$$ are respectively a differentiable function and a convex, lower semi-continuous function, then the algorithm defined by:
$$x^{k+1} = \text{prox}_{\gamma{g}}[x^{k} - \gamma\nabla{f(x^{k}})]$$
converges to $$\text{argmin}[f+g]$$.
This is justified by the fact that if $$x^{*}$$ is a minimizer of $$f+g$$, then: $$x^{*} = \text{prox}_{\gamma{g}}[x^{*} - \gamma\nabla{f(x^{*}})]$$
But I do not understand this relation. Why is it true?
That is, why $$x^{*} = \text{argmin}[f+g] \Leftrightarrow x^{*} =\text{prox}_{\gamma{g}}[x^{*} - \gamma \nabla{f(x^{*}})]$$ ?
I will show below that if $$x^* = \text{prox}_{\gamma{g}}[x^{*} - \gamma \nabla{f(x^{*}})]$$ then $$x^* \in \text{argmin } f(x)+g(x)$$.
Plugging in the definition of proximal operator, we have $$\text{prox}_{\gamma{g}}[x^{*} - \gamma \nabla{f(x^{*}})] = \text{argmin}_{x} \left\{\gamma g(x) + \frac{1}{2}\|x- (x^* - \gamma \nabla f(x^*))\|^2\right\}$$ Now since $$x^* = \text{prox}_{\gamma{g}}[x^{*} - \gamma \nabla{f(x^{*}})]$$, we have by Fermat's rule at $$x=x^*$$, the following $$0 \in \partial (\gamma g(x)) + (x-(x^*-\gamma\nabla f(x^*)))$$ Now just substitute $$x=x^*$$, we get $$0\in \gamma \partial g(x^*) + \gamma \nabla f(x^*)$$ This is equivalent to saying that $$0 \in \partial F(x^*)$$, where $$F(x) = g(x)+f(x)$$, so $$x^*$$ is the minimizer. The other direction of the proof is very similar.
Note: The last step $$0 \in \partial F(x^*)$$ need not always hold (check subdifferential properties).
Here's an explanation which assumes that we already understand the idea that the prox-operator of $$g$$ with parameter $$t > 0$$ is the operator $$(I + t \partial g)^{-1}$$, where $$\partial g$$ is the subdifferential of $$g$$.
I'll assume that $$f$$ is convex as well as differentiable, and that $$g$$ is convex and closed. Let $$t >0$$. A point $$x$$ is a minimizer of $$f + g$$ if and only if \begin{align} &0 \in \nabla f(x) + \partial g(x) \\ \iff &x \in x + t \nabla f(x) + t\partial g(x) \\ \iff &x - t \nabla f(x) \in (I + t \partial g)(x) \\ \iff &x = (I + t \partial g)^{-1}(x - t \nabla f(x)). \end{align} The final equation is another way of saying that $$x = \text{prox}_{tg}(x - t \nabla f(x)).$$
We can then solve this equation using fixed point iteration, which yields the proximal gradient method.
By the way, if this derivation of the proximal gradient method doesn't seem intuitive, there are other ways to discover the proximal gradient method that are more obvious. The viewpoint given here has the advantage that it shows that the proximal gradient method is a fixed point iteration, which helps us with convergence proofs.
|
{}
|
58KiB, 165x236, i30814.png
No.12008161
Why is Ten related to Akagi? i know Ten is an Akagi sequel, but it seems totally unrelated to me.
|
{}
|
## Book Store
Download books and chapters from book store.
Currently only available for.
CBSE Gujarat Board Haryana Board
## Previous Year Papers
Download the PDF Question Papers Free for off line practice and view the Solutions online.
Currently only available for.
Class 10 Class 12
Two trains, each of length 100 m, are running on parallel tracks. One overtakes the other in 20 seconds and one crosses the other in 10 second. Calculate the velocities of each train?
Let u and v be the velocity of two trains A and B. While overtaking the relative velocity of train A w.r.t B = u -v.
While crossing, the relative velocity of train A w.r.t B = u+ v.
Total distance to be travelled by car A while crossing = 100 + 100 = 200
So,
1518 Views
What is a vector quantity?
A physical quantity that requires direction along with magnitude, for its complete specification is called a vector quantity.
835 Views
Give three examples of vector quantities.
Force, impulse and momentum.
865 Views
What is a scalar quantity?
A physical quantity that requires only magnitude for its complete specification is called a scalar quantity.
1212 Views
Give three examples of scalar quantities.
Mass, temperature and energy
769 Views
What are the basic characteristics that a quantity must possess so that it may be a vector quantity?
A quantity must possess the direction and must follow the vector axioms. Any quantity that follows the vector axioms are classified as vectors.
814 Views
|
{}
|
Select Page
Directions: Given below are the forecasts of the World and SA energy demand for the years 1980, 1990 and 2000 AD and the forecast made for these years follows for the subsequent decade as well. For example, the forecast for 1990 means the forecast for years 1990-1999. The demand is given in million barrels per day, crude oil equivalent.
Question 1. Over 1990 – 2010, which two fuels meet more than 60 percent of the total energy demand of both World and SA?
(a) A&B
(b) A&C
(c) B&C
(d) None of the above
Question 2. Which fuel’s proportion in the total energy demand increases over the decade 1980–1990 and decreases over the decade 1990 – 2000 for both the World and SA?
(a) A
(b) B
(c) C
(d) D
Question 3. Which is the fuel whose proportion to the total energy demand of the world will remain constant over the period 1980 – 2000 but whose proportion will increase in the total energy demand in SA?
(a) C
(b) D
(c) E
(d) B
Question 4. Which is the fuel whose proportion in the total energy demand will decrease continuously over the period 1980 – 2000, in SA?
(a) B
(b) C
(c) D
(d) E
### Answers and Explanations
Answer
Answer 1. (B)
Thus we can see that C and A combined constitute more than 60% of total energy in both World & SA for the given period.
Answer 2. (A)
As seen from the above table, A is the fuel whose proportion in the total energy demand increases during 1980-1990 and decreases during 1990-2000 for both world & SA.
Answer 3. (D)
For the answer choices given and for world we can make the following table.
Hence we can see that the proportion of B remains constant over the given period.
Answer 4 (D)
For the answer choices given and for SA we can observe that proportion of E is 10%, 7% and 6% in 1980, 1990 and 2000 respectively. Hence we can see that the proportion of E goes on decreasing over the period.
|
{}
|
Formats
Format
BibTeX
MARC
MARCXML
DublinCore
EndNote
NLM
RefWorks
RIS
### Abstract
Linear-response functions and second-order transition amplitudes can be calculated exactly provided one effective state, the solution of an inhomogeneous Schrodinger equation, is known. We show how variational principles can be applied to the calculation of this effective state; an important advantage is the possibility of using variational techniques, which are very efficient when large basis sets (such as plane waves) are used, or when the solution is required as a function of a free parameter (for instance, the frequency in dynamical response functions). The variational principle applies to solid-state as well as to atomic problems. We illustrate the use of one variational technique in the simple case of two-photon transitions in the hydrogen atom.
|
{}
|
## Mixed-effects modeling — four hour workshop — part II: Data Exploration
We got our data together in Part I of this workshop, eventually arriving at the collation of a single dataframe in ‘long format’ holding all our data on subject attributes, item attributes and behavioural responses to words in a lexical decision experiment. That collated database can be downloaded here.
The code we will be using in this post can be downloaded from here.
We should begin by examining the features of our collated dataset. We shall be familiarising ourselves with the distributions of the variables, looking for evidence of problems in the dataset, and doing transformation or othe operations to handle those problems.
Some of the problems are straightforward. We should look at the distributions of scores to check if any errors crept in to the dataset, miscoded or misrecorded observations. Other problems are a little more demanding both to understand and to diagnose and treat. The running example in this workshop requires a regression analysis in which we will be required to enter a number of variables as predictors, including variables characterizing word attributes like the frequency, length and imageability of words. The problem is that all these variables share or overlap in information because words that appear often in the language also tend to be short, easy to imagine, learnt earlier in life and so on. This means that we need to examine the extent to which our predictors cluster, representing overlapping information, and presenting potential problems for our analysis.
Much of what follows has been discussed in previous posts here on getting summary statistics and plotting histograms to examine variable distributions, here on plotting multiple histograms in a grid, see also here on histograms and density plots, here on plotting scatterplots to examine bivariate relationships, and here on checking variable clustering. This post constitutes an edited, shortened, version of those previous discussions.
Examine the variable distributions — summary statistics
I am going to assume you have already downloaded and read in the data files concerning subject and item or word attributes as well as the collated database merging the subject and item attribute data with the behavioural data (see Part I). What we are going to do now is look at the distributions of the key variables in those databases.
We can start simply by examining the sumaries for our variables. We want to focus on the variables that will go into our models, so variables like age or TOWRE word or non-word reading score, item length, word frequency, and lexical decision reaction time (RT) to words.
If you have successfully read the relevant data files into the R workspace then you should see them listed by the dataframe names (assigned when you performed the read.csv function calls) in the workspace window in RStudio (top right).
R provides a handy set of functions to allow you to examine your data (see e.g. Dalgaard, 2008):
```
summary(subjects)
describe(subjects)
str(subjects)
psych::describe(subjects)
length(subjects)
length(subjects\$Age)
```
Here we are focusing on the data in the subject scores dataframe.
Copy-paste these commands into the script window and run them, below is what you will see in the console:
Notice:
1. the head() function gives you the top rows, showing column names and data in cells; I asked for 2 rows of data with n = 2
2. the summary() function gives you: minimum, maximum, median and quartile values for numeric variables, numbers of observations with one or other levels for factors, and will tell you if there are any missing values, NAs
3. psych::describe() (from the psych package) will give you means, SDs, the kind of thing you report in tables in articles
4. str() will tell you if variables are factors, integers etc.
If I were writing a report on this subjects data I would copy the output from describe() into excel, format it for APA, and stick it in a word document.
As an exercise, you might apply the same functions to the items or the full databases.
Notice the values and evaluate if they make sense to you. Examine, in particular, the treatment of factors and numeric variables.
Examine the variable distributions — plots
Summary statistics are handy, and required for report writing, but plotting distributions will raise awareness of problems in your dataset much more efficiently.
We will use ggplot to plot our variables, especially geom_histogram, see the documentation online.
We are going to examine the distribution of the Age variable, age in years for ML’s participants, using the ggplot() function as follows:
```
pAge <- ggplot(subjects, aes(x=Age))
pAge + geom_histogram()
```
Notice:
— We are asking ggplot() to map the Age variable to the x-axis i.e. we are asking for a presentation of one variable.
If you copy-paste the code into the script window and run it three things will happen:
1. You’ll see the commands appear in the lower console window
2. You’ll see the pAge object listed in the upper right workspace window
Notice:
— When you ask for geom_histogram(), you are asking for a bar plot plus a statistical transformation (stat_bin) which assigns observations to bins (ranges of values in the variable)
— calculates the count (number) of observations in each bin
— or – if you prefer – the density (percentage of total/bar width) of observations in each bin
— the height of the bar gives you the count, the width of the bin spans values in the variable for which you want the count
— by default, geom_histogram will bin data to the range in values/30 but you can ask for more or less detail by specifying binwidth, this is what R means when it says:
```>stat_bin: binwidth defaulted to range/30. Use 'binwidth = x' to
3. We see the resulting plot in the plot window, which we can export as a pdf:
Notice:
— ML tested many people in their 20s – no surprise, a student mostly tests students – plus a smallish number of people in middle age and older (parents, grandparents)
— nothing seems wrong with these data – given what we know about the sample.
We can move on to examine the distribution of multiple variables simultaneously by producing a grid of histograms, as follows. In Exploratory Data Analysis, we often wish to split a larger dataset into smaller subsets, showing a grid or trellis of small multiples – the same kind of plot, one plot per subset – arranged systematically to allow the reader (maybe you) to discern any patterns, or stand-out features, in the data. Much of what follows is cribbed from the ggplot2 book (Wickham, 2009; pp.115 – ; see also Chang, 2o13). Wickham (2009) advises that arranging multiple plots on a single page involves using grid, an R graphics system, that underlies ggplot2.
Copy-paste the following code in the script window to create the plots:
```
pAge <- ggplot(subjects, aes(x=Age))
pAge <- pAge + geom_histogram()
pEd <- ggplot(subjects, aes(x=Years_in_education))
pEd <- pEd + geom_histogram()
pw <- ggplot(subjects, aes(x=TOWRE_wordacc))
pw <- pw + geom_histogram()
pnw <- ggplot(subjects, aes(x=TOWRE_nonwordacc))
pnw <- pnw + geom_histogram()
pART <- ggplot(subjects, aes(x=ART_HRminusFR))
pART <- pART + geom_histogram()
```
Notice:
— I have given each plot a different name. This will matter when we assign plots to places in a grid.
— I have changed the syntax a little bit: create a plot then add the layers to the plot; not doing this, ie sticking with the syntax as in the preceding examples, results in an error when we get to print the plots out to the grid; try it out – and wait for the warning to appear in the console window.
We have five plots we want to arrange, so let’s ask for a grid of 2 x 3 plots ie two rows of three.
```
grid.newpage()
pushViewport(viewport(layout = grid.layout(2,3)))
vplayout <- function(x,y)
viewport(layout.pos.row = x, layout.pos.col = y)
```
Having asked for the grid, we then print the plots out to the desired places, as follows:
```
print(pAge, vp = vplayout(1,1))
print(pEd, vp = vplayout(1,2))
print(pw, vp = vplayout(1,3))
print(pnw, vp = vplayout(2,1))
print(pART, vp = vplayout(2,2))
```
As R works through the print() function calls, you will see the Plots window in R-studio fill with plots getting added to the grid you specified.
You can get R to generate an output file holding the plot as a .pdf or as a variety of alternatively formatted files.
Wickham (2009; p. 151) recommends differing graphic formats for different tasks, e.g. png (72 dpi for the web); dpi = dots per inch i.e. resolution.
Most of the time, I output pdfs. You call the pdf() function – a disk-based graphics device -specifying the name of the file to be output, the width and height of the output in inches (don’t know why, but OK), print the plots, and close the device with dev.off(). You can use this code to do that.
I can wrap everything up in a single sequence of code bracketed by commands to generate a pdf:
```
pdf("ML-data-subjects-histograms-061113.pdf", width = 15, height = 10)
pAge <- ggplot(subjects, aes(x=Age))
pAge <- pAge + geom_histogram()
pEd <- ggplot(subjects, aes(x=Years_in_education))
pEd <- pEd + geom_histogram()
pw <- ggplot(subjects, aes(x=TOWRE_wordacc))
pw <- pw + geom_histogram()
pnw <- ggplot(subjects, aes(x=TOWRE_nonwordacc))
pnw <- pnw + geom_histogram()
pART <- ggplot(subjects, aes(x=ART_HRminusFR))
pART <- pART + geom_histogram()
grid.newpage()
pushViewport(viewport(layout = grid.layout(2,3)))
vplayout <- function(x,y)
viewport(layout.pos.row = x, layout.pos.col = y)
print(pAge, vp = vplayout(1,1))
print(pEd, vp = vplayout(1,2))
print(pw, vp = vplayout(1,3))
print(pnw, vp = vplayout(2,1))
print(pART, vp = vplayout(2,2))
dev.off()
```
Which results in a plot that looks like this:
There is nothing especially concerning about these distributions though the bimodal nature of the age distribution is something worth remembering if age turns out to be a significant predictor of reading reaction times.
Further moves
Histograms are not always the most useful approach to visualizing data distributions. What if you wanted to compare two distributions overlaid on top of each other e.g. distributions of variables that should be similar? It is worth considering density plots, see here.
Exploring data and acting on what we find
Let’s consider an aspect of the data that, when revealed by a distribution plot (but was also evident when we looked at the summary statistics), will require action.
I am talking about what we see if we plot a histogram of the RT distribution using the following code. If we copy this into RStudio and run it:
```
pRT <- ggplot(RTs.norms.subjects, aes(x=RT))
pRT + geom_histogram()
```
We get this:
Notice that there are a number of negative RTs. That is. of course, impossible but makes sense if you know that DMDX, the application used to collect lexical decision responses, encodes RTs for errors as negative RTs.
Notice also that even if we ignore the -RT values, the RT distribution is, as usual, highly skewed.
Data transformations I/II
We can remove the error -RT observations by subsetting the full collated database, filtering out all observations correspondong to errors to create a new ‘correct responses only’ database. We can also log transform the RTs for correct responses in that same new database. We can do this in a few lines of code in the script as follows:
```
# the summary shows us that there are negative RTs, which we can remove by subsetting the dataframe
# by setting a condition on rows
nomissing.full <- RTs.norms.subjects[RTs.norms.subjects\$RT > 0,]
# inspect the results
summary(nomissing.full)
# how many errors were logged?
# the number of observations in the full database i.e. errors and correct responses is:
length(RTs.norms.subjects\$RT)
# > length(RTs.norms.subjects\$RT)
# [1] 5440
# the number of observations in the no-errors database is:
length(nomissing.full\$RT)
# so the number of errors is:
length(RTs.norms.subjects\$RT) - length(nomissing.full\$RT)
# > length(RTs.norms.subjects\$RT) - length(nomissing.full\$RT)
# [1] 183
# is this enough to conduct an analysis of response accuracy? we can have a look but first we will need to create
# an accuracy coding variable i.e. where error = 0 and correct = 1
RTs.norms.subjects\$accuracy <- as.integer(ifelse(RTs.norms.subjects\$RT > 0, "1", "0"))
# we can create the accuracy variable, and run the glmm but, as we will see, we will get an error message, false
# convergence likely due to the low number of errors
# I will also transform the RT variable to the log base 10 (RT), following convention in the field, though
# other transformations are used by researchers and may have superior properties
nomissing.full\$logrt <- log10(nomissing.full\$RT)
```
Notice:
— You can see that there were a very small number of errors recorded in the experiment, less than 200 errors out of 5000+ trials. This is a not unusual error rate, I think, for a dataset collected with typically developing adult readers.
— I am starting to use the script file not just to hold lines of code to perform the analysis, not just to hold the results of analyses that I might redo, but also to comment on what I am seeing as I work.
— The ifelse() function is pretty handy, and I have started using it more and more.
— Here, I can break down what is happening step by step:-
1. RTs.norms.subjects\$accuracy <- — creates a vector that gets added to the existing dataframe columns, that vector has elements corresponding to the results of a test on each observation in the RT column, a test applied using the ifelse() function, a test that results in a new vector with elements in the correct order for each observation.
2. as.integer() — wraps the ifelse() function in a function call that renders the output of ifelse() as a numeric integer vector; you will remember from a previous post how one can use is.[datatype]() and as.[datatype]() function calls to test or coerce, respectively, the datatype of a vector.
3. ifelse(RTs.norms.subjects\$RT > 0, “1”, “0”) — uses the ifelse() function to perform a test on the RT vector such that if it is true that the RT value is greater than 0 then a 1 is returned, if it is not then a 0 is returned, these 1s and 0s encode accuracy and fill up the new vector being created, one element at a time, in an order corresponding to the order in which the RTs are listed.
I used to write notes on my analyses on paper, then in a word document (using CTRL-F that will be more searchable), to ensure I had a record of what I did.
Check predictor collinearity
The next thing we want to do is to examine the extent to which our predictors cluster, representing overlapping information, and presenting potential problems for our analysis. We can do this stage of our analysis by examining scatterplot matrices and calculating the condition number for our predictors. Note that I am going to be fairly lax here, by deciding in advance that I will use some predictors but not others in the analysis to come. (If I were being strict, I would have to find and explain theory-led and data-led reasons for including the predictors that I do include.)
Note that we will first run some code to define a function that will allow us to create scatterplot matrices, furnished by William Revelle on his web-site (see URL below):
```
#code taken from part of a short guide to R
#Version of November 26, 2004
#William Revelle
# see: http://www.personality-project.org/r/r.graphics.html
panel.cor <- function(x, y, digits=2, prefix="", cex.cor)
{
usr <- par("usr"); on.exit(par(usr))
par(usr = c(0, 1, 0, 1))
r = (cor(x, y))
txt <- format(c(r, 0.123456789), digits=digits)[1]
txt <- paste(prefix, txt, sep="")
if(missing(cex.cor)) cex <- 0.8/strwidth(txt)
text(0.5, 0.5, txt, cex = cex * abs(r))
}
```
We can then run the following code:
```
# we can consider the way in which potential predictor variables cluster together
summary(nomissing.full)
# create a subset of the dataframe holding just the potential numeric predictors
# NB I am not bothering, here, to include all variables, e.g. multiple measures of the same dimension
nomissing.full.pairs <- subset(nomissing.full, select = c(
"Length", "OLD", "BG_Mean", "LgSUBTLCD", "brookesIMG", "AoA_Kup_lem", "Age",
"TOWRE_wordacc", "TOWRE_nonwordacc", "ART_HRminusFR"
))
# rename those variables that need it for improved legibility
nomissing.full.pairs <- rename(nomissing.full.pairs, c(
brookesIMG = "IMG", AoA_Kup_lem = "AOA",
TOWRE_wordacc = "TOWREW", TOWRE_nonwordacc = "TOWRENW", ART_HRminusFR = "ART"
))
# subset the predictors again by items and person level attributes to make the plotting more sensible
nomissing.full.pairs.items <- subset(nomissing.full.pairs, select = c(
"Length", "OLD", "BG_Mean", "LgSUBTLCD", "IMG", "AOA"
))
nomissing.full.pairs.subjects <- subset(nomissing.full.pairs, select = c(
"Age", "TOWREW", "TOWRENW", "ART"
))
# draw a pairs scatterplot matrix
pdf("ML-nomissing-full-pairs-items-splom-061113.pdf", width = 8, height = 8)
pairs(nomissing.full.pairs.items, lower.panel=panel.smooth, upper.panel=panel.cor)
dev.off()
# draw a pairs scatterplot matrix
pdf("ML-nomissing-full-pairs-subjects-splom-061113.pdf", width = 8, height = 8)
pairs(nomissing.full.pairs.subjects, lower.panel=panel.smooth, upper.panel=panel.cor)
dev.off()
# assess collinearity using the condition number, kappa, metric
collin.fnc(nomissing.full.pairs.items)\$cnumber
collin.fnc(nomissing.full.pairs.subjects)\$cnumber
# > collin.fnc(nomissing.full.pairs.items)\$cnumber
# [1] 46.0404
# > collin.fnc(nomissing.full.pairs.subjects)\$cnumber
# [1] 36.69636
```
You will see when you run this code that the set of subject attribute predictors (see below) …
… and the set of item attribute predictors (see below) …
… include some quite strongly overlapping variables, distinguished by high correlation coefficients.
I was advised, when I started doing regression analyses, that a rule-of-thumb one can happily use when examining bivariate i.e. pairwise correlations between predictor variables is that there is no multicollinearity to worry about if there are no coefficients greater than $r = .8$ however, I can think of no statistical reason for this number, while Baayen (2008) recommends using the condition number, kappa, as a diagnostic measure, based on Belsley et al.’s (1980) arguments that pairwise correlations do not reveal the real, underlying, structure of latent factors (i.e. principal components of which collinear variables are transformations or expressions).
If we calculate the condition number for, separately, the subject predictors and the item predictors, we get values of 30+ in each case. Baayen (2008) comments that (if memory serves) a condition number higher than 12 indicates a dangerous level of multicollinearity. However, Cohen et al. (2003) comment that while 12 or so might be a risky kappa there are no strong statistical reasons for setting the thresholds for a ‘good’ or a ‘bad’ condition number; one cannot apply the test thoughtlessly. That said, 30+ does not look good.
The condition number is based around Principal Components Analysis (PCA) and I think it is worth considering the outcome of such an analysis over predictor variables when examining the multicollinearity of those variables. PCA will often show that set of several item attribute predictors will often relate quite strongly to a set of maybe four orthogonal components: something to do with frequency/experience; something relating to age or meaning; something relating to word length and orthographic similarity; and maybe some thing relating to sublexical unit frequency. However, as noted previously, Cohen et al. (2003) provide good arguments for being cautious about performing a principal components regression in response to a state of high collinearity among your predictors, including problems relating to the interpretability of findings.
Data transformations II/II: Center variables on their means
One perhaps partial means to alleviate the collinearity indicated may be to center predictor variables on their means.
Centering variables is helpful, also, for the interpretation of effects where 0 on the raw variables is not meaningful e.g. noone has an age of 0 years, no word has a length of 0 letters. With centered variables, if there is an effect then one can say that for unit increase in the predictor variable (where 0 for the variable is now equal to the mean for the sample, because of centering), there is a change in the outcome, on average, equal to the coefficient estimated for the effect.
Centering variables is helpful, especially, when we are dealing with interactions. Cohen et al. (2003) note that the collinearity between predictors and interaction terms is reduced (this is reduction of nonessential collinearity, due to scaling) if predictors are centered. Note that interactions are represented by the multiplicative product of their component variables. You will often see an interaction represented as “age x word length effects”, the “x” reflects the fact that for the interaction of age by length effects the interaction term is entered in the analysis as the product of the age and length variables. In addition, centering again helps with the interpretation of effects because if an interaction is significant then one must interpret the effect of one variable in the interaction while taking into account the other variable (or variables) involved in the interaction (Cohen et al., 2003; see, also, Gelman & Hill, 2007). As that is the case, if we center our variables and the interaction as well as the main effects are significant, say, in our example, the age, length and age x length effects, then we might look at the age effect and interpret it by reporting that for unit change in age the effect is equal to the coefficient while the effect of length is held constant (i.e. at 0, its mean).
As you would expect, you can center variables in R very easily, we will use one method, which is to employ the scale() function, specifying that we want variables to be centered on their means, and that we want the results of that operation to be added to the dataframe as centered variables. Note that by centering on their means, I mean we take the mean for a variable and subtract it from each value observed for that variable. We could center variables by standardizing them, where we both subtract the mean and divide the variable by its standard deviation (see Gelman & Hill’s 2007, recommendations) but we will look at that another time.
So, I center variables on their means, creating new predictor variables that get added to the dataframe with the following code:
```
# we might center the variables - looking ahead to the use of interaction terms - as the centering will
# both reduce nonessential collinearity due to scaling (Cohen et al., 2003) and help in the interpretation
# of effects (Cohen et al., 2003; Gelman & Hill, 2007 p .55) ie if an interaction is sig one term will be
# interpreted in terms of the difference due to unit change in the corresponding variable while the other is 0
nomissing.full\$cLength <- scale(nomissing.full\$Length, center = TRUE, scale = FALSE)
nomissing.full\$cBG_Mean <- scale(nomissing.full\$BG_Mean, center = TRUE, scale = FALSE)
nomissing.full\$cOLD <- scale(nomissing.full\$OLD, center = TRUE, scale = FALSE)
nomissing.full\$cIMG <- scale(nomissing.full\$brookesIMG, center = TRUE, scale = FALSE)
nomissing.full\$cAOA <- scale(nomissing.full\$AoA_Kup_lem, center = TRUE, scale = FALSE)
nomissing.full\$cLgSUBTLCD <- scale(nomissing.full\$LgSUBTLCD, center = TRUE, scale = FALSE)
nomissing.full\$cAge <- scale(nomissing.full\$Age, center = TRUE, scale = FALSE)
nomissing.full\$cTOWREW <- scale(nomissing.full\$TOWRE_wordacc, center = TRUE, scale = FALSE)
nomissing.full\$cTOWRENW <- scale(nomissing.full\$TOWRE_nonwordacc, center = TRUE, scale = FALSE)
nomissing.full\$cART <- scale(nomissing.full\$ART_HRminusFR, center = TRUE, scale = FALSE)
# subset the predictors again by items and person level attributes to make the plotting more sensible
nomissing.full.c.items <- subset(nomissing.full, select = c(
"cLength", "cOLD", "cBG_Mean", "cLgSUBTLCD", "cIMG", "cAOA"
))
nomissing.full.c.subjects <- subset(nomissing.full, select = c(
"cAge", "cTOWREW", "cTOWRENW", "cART"
))
# assess collinearity using the condition number, kappa, metric
collin.fnc(nomissing.full.c.items)\$cnumber
collin.fnc(nomissing.full.c.subjects)\$cnumber
# > collin.fnc(nomissing.full.c.items)\$cnumber
# [1] 3.180112
# > collin.fnc(nomissing.full.c.subjects)\$cnumber
# [1] 2.933563
```
Notice:
— I check the condition numbers for the predictors after centering and it appears that they are significantly reduced.
— I am not sure that that reduction can possibly be the end of the collinearity issue, but I shall take that result and run with it, for now.
— I also think that it would be appropriate to examine collinearity with both the centered variables and the interactions involving those predictors but that, too, can be left for another time.
Write the file out as a csv so that we do not have to repeat ourselves
```
write.csv(nomissing.full, file = "RTs.norms.subjects nomissing.full 061113.csv", row.names = FALSE)
```
What have we learnt?
In this post, we learnt about:
— How to inspect the dataframe object using functions like summary() or describe().
— How to plot histograms to examine the distributions of variables.
— How to use grid() to plot multiple variables at once.
— How to output our plots as a pdf for later use.
— How to subset dataframes conditionally.
— How to log transform variables.
— How to create a variable conditionally dependent on values in another variable.
Key vocabulary
working directory
workspace
object
dataframe
NA, missing value
csv
histogram
graphics device
ggplot()
geom_histogram()
geom_density()
facet_wrap()
pdf()
grid()
print()
log10()
ifelse()
Baayen, R. H. (2008). Analyzing linguistic data. Cambridge University Press.
Belsley, D. A., Kuh, E., & Welsch, R. E. (1980). Regression diagnostics: Identifying influential data and sources of collinearity. New York: Wiley.
Chang, W. (2013). R Graphics Cookbook. Sebastopol, CA: o’Reilly Media.
Cohen, J., Cohen, P., West, S. G., & Aiken, L. S. (2003). Applied multiple regression/correlation analysis for the behavioural sciences (3rd. edition). Mahwah, NJ: Lawrence Erlbaum Associates.
Dalgaard, P. (2008). Introductory statistics with R (2nd. Edition). Springer.
Gelman, A., & Hill, J. (2007). Data analysis using regression and multilevel/hierarchical models. New York,NY: Cambridge University Press.
Wickham, H. (2009). ggplot2: Elegant graphics for data analysis. Springer.
This entry was posted in 2.2 LME workshop II/IV, Uncategorized and tagged , , , , . Bookmark the permalink.
|
{}
|
# Difference: PhysicsResultsSUS14016 (10 vs. 11)
#### Revision 112015-09-03 - JohannesSchulz
Line: 1 to 1
META TOPICPARENT name="SUSYPhotonWorkingGroup"
Line: 43 to 43
Figure 8: Measurement of the trigger efficiency of the photon part as a function of the transverse momentum of the photon with the highest transverse momentum. The plateau region is reached for $p_{\text{T}}$ = 40 GeV and the efficiency for $p_{\text{T}}$ $>$ 40 GeV is given by $\varepsilon_{\text{$\gamma$-req}}$ = (88.0 $\pm$ 0.7({stat.)}) $\%$. Figure 9: Control region where the $V\gamma$ and $\gamma$jets background simulations are fitted to the data using the template variable $E_{\mathrm{T}}^{\text{miss}}$/$\sqrt{\text{$H_{\text{T}}$}}$. Table 3: Number of signal events and the selection efficiency after each selection step for three signal points, each corresponding to a different signal scenario. The TChiNg_500 point corresponds to a NLSP mass of 500 GeV, The TChiWg_650 point corresponds to a NLSP mass of 650 GeV and the GGM_640_630 point corresponds to a wino mass of 640 GeV and a bino mass of 630 GeV.
Changed:
<
<
Figure 10: Efficiencies for the GGM signal points in the plane spanned by of the wino and bino mass. Figure 11: Effciencies for the TChiNg signal scenario depending on the mass of the NLSP after the full selection. Figure 12: Effciencies for the TChiWg signal scenario depending on the mass of the NLSP after the full selection. The TChiWg simulation has a higher granularity compared to the TChiNg simulation.
>
>
Figure 10: Efficiencies for the GGM signal points in the plane spanned by of the wino and bino mass. Electronic version: WinoBino_Acceptance.root Figure 11: Effciencies for the TChiNg signal scenario depending on the mass of the NLSP after the full selection. Eelectronic version: TChiNg_acc.root Figure 12: Effciencies for the TChiWg signal scenario depending on the mass of the NLSP after the full selection. The TChiWg simulation has a higher granularity compared to the TChiNg simulation. Electronic version: TChiwg_acc.root
-- JohannesSchulz - 2015-07-22
Deleted:
<
<
META FILEATTACHMENT attachment="Full_signal.pdf" attr="" comment="" date="1437648549" name="Full_signal.pdf" path="Full_signal.pdf" size="65325" user="jschulz" version="1" attachment="TChiNg_xs.pdf" attr="" comment="" date="1437648550" name="TChiNg_xs.pdf" path="TChiNg_xs.pdf" size="62844" user="jschulz" version="1" attachment="TChiNg_xs_CN.pdf" attr="" comment="" date="1437648550" name="TChiNg_xs_CN.pdf" path="TChiNg_xs_CN.pdf" size="59217" user="jschulz" version="1"
Copyright &© 2008-2019 by the contributing authors. All material on this collaboration platform is the property of the contributing authors.
Ideas, requests, problems regarding TWiki? Send feedback
|
{}
|
# Isogonal trajectory
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)
Jump to: navigation, search
A plane curve intersecting the curves of a given one-parameter family in the plane at one and the same angle. If
$$F(x,y,y')=0\label{1}\tag{1}$$
is the differential equation of the given family of curves, then an isogonal trajectory of this family intersecting it at an angle $\alpha$, where $0<\alpha<\pi$, $\alpha\neq\pi/2$, satisfies one of the following two equations:
$$F\left(x,z,\frac{z'-\tan\alpha}{1+z'\tan\alpha}\right)=0,\quad F\left(x,z\frac{z'+\tan\alpha}{1-z'\tan\alpha}\right)=0.$$
In particular, the equation
$$F\left(x,z,-\frac{1}{z'}\right)=0\label{2}\tag{2}$$
is satisfied by an orthogonal trajectory, that is, a plane curve that forms a right angle at each of its points with any curve of the family \eqref{1} passing through it. The orthogonal trajectories for the given system \eqref{1} form a one-parameter family of plane curves — the general integral of equation \eqref{1}. For example, if the family of lines of force of a plane electrostatic field is considered, then the family of orthogonal trajectories are the equipotential lines.
#### References
[1] W.W. [V.V. Stepanov] Stepanow, "Lehrbuch der Differentialgleichungen" , Deutsch. Verlag Wissenschaft. (1956) (Translated from Russian)
#### References
[a1] H.S.M. Coxeter, "Introduction to geometry" , Wiley (1961) [a2] D. Hilbert, S.E. Cohn-Vossen, "Geometry and the imagination" , Chelsea (1952) (Translated from German)
How to Cite This Entry:
Isogonal trajectory. Encyclopedia of Mathematics. URL: http://encyclopediaofmath.org/index.php?title=Isogonal_trajectory&oldid=44769
This article was adapted from an original article by r equation','../w/w097310.htm','Whittaker equation','../w/w097840.htm','Wronskian','../w/w098180.htm')" style="background-color:yellow;">N.Kh. Rozov (originator), which appeared in Encyclopedia of Mathematics - ISBN 1402006098. See original article
|
{}
|
# Printing documents to tiff
I have a faxing program that only accepts TIFF files to send. It don't have any utility to convert documents to TIFF and i have to convert them myself. Even worse, it only accepts monochrome TIFF Files with 200x200 dpi and Width of 1728 pixels!
I found Peernet TIFF Image Printer that is really perfect. It sets up a Virtual Printer and i can print anything and anywhere to TIFF format! But it is not free and i cannot pay any more extra \$100 !
Is there any software that could do it for me?
Is there any way to install a tiff printer? I found something about ghostscript and how to setup a Virtual Tiff Printer, But i was not simple and i don't know where i should set 200x200dpi and Width=1728pixel
-
You could try BullZip PDF Printer. It has an option for printing to TIFF and allows you to set the resolution. I'm not sure about how you would guarantee a width of 1728 pixels (probably by changing the size of the original document), but it might be worth having a look.
-
You can use PDFCreator. From the website:
Key Features:
• safe-to-install awardCreate PDFs from any program that is able to print
• Security: Encrypt PDFs and protect them from being opened, printed etc
• New: Digitally sign your PDFs to ensure that you are the author and the file has not been modified
• New: Create PDF/A files for long term archives
• Send generated files via eMail
• Create more than just PDFs: PNG, JPG, TIFF, BMP, PCX, PS, EPS
• AutoSave files to folders and filenames based on Tags like Username, Computername, Date, Time etc.
• Merge multiple files into one PDF
• Easy Install: Just say what you want and everything is installed
• Terminal Server: PDFCreator also runs on Terminal Servers without problems
• And the best: PDFCreator is free, even for commercial use! It is Open Source and released under the Terms of the GNU General Public License.
-
I have had good experience with both
Bullzip
and Polestar
Polestar is slightly better, as it allows you to specify resolution etc before you print
-
Polestar doesn't have any option for setting resolution of TIFF images. It only have this option for PDFs. – Isaac Sep 28 '09 at 15:01
It does right next to the PDF settings under "Image" section – admintech Sep 28 '09 at 15:32
A lot of PDF printers have an option to print to tiff. I haven't got one that prints to tiff installed myself, but that may be an option.
I also found some instructions for setting up ghostscript with tiff printing. You need to add a line to the tif.rsp file to set the resolution.
I think this is -r200. e.g.
-Ic:\gs\gs8.00\lib;c:\gs\fonts
-sDEVICE=tiffg4
-sOutputFile=c:\print.tif
-dNOPAUSE
-dSAFER
-dBATCH
-sPAPERSIZE=letter
-r200
-
1728 pixels width at 200dpi gives a page width of 8.64in, which translates to 622pt (PostScript points). This is roughly the width of Letter (612pt)... So what is the page height you are allowed to use? If you use Ghostscript's "tiffg4" output and specify the width in pixels, you also need to specify the height in pixels.
Note, the above suggestion to run Ghostscript with sPAPERSIZE=letter will not create a TIFF with a width of 1728 pixels as required by you. It will use 612 PostScript points, which translate into 1720 pixels only.
You can use Ghostscript on Windows to create the TIFF with exactly the resolution and width/height you need. Here is an example command (using a height of 2200 pixels and a width of 1728 pixels at 200dpi):
gswin32c.exe ^
-dBATCH ^
-dNOPAUSE ^
-sDEVICE=tiffg4 ^
-dPDFFitPage ^
-sOutputFile=c:\temp\page-%04d.tif ^
-r200x200 ^
-g1728x2200 ^
c:\path\to\input\pdf
-
|
{}
|
# Minimum distance of nodes from a set of two nodes
In an unweighted tree, suppose that we want to delete (or mark) any node which is closer to node $$v$$ than node $$w$$ ($$dist(x,v) < dist(x,w)$$). The solution that comes to my mind is running two BFS, which gives us $$\mathcal{O}(V + E)$$ running time.
Is there any better way to do this? possibly with one BFS?
EDIT: We can start two BFS from $$w$$ and $$v$$ simultaneously and one step at a time. First we find the nodes in distance 1 of the $$w$$ then the nodes in distance 1 of $$v$$ and then nodes at distance 2 of $$w$$ and so on. by the time there is no node in the queue of BFS of node $$v$$, we can end the search. any node which is first visited with $$v$$ BFS is closer to $$v$$ than $$w$$.
But again I suppose there is a more efficient way to do this.
• Please clarify that whether "any node which is closer to node $v$ than node $w$" means "any node $x$ such that $dist(x,v)<dist(w,v)$ " or "any node $x$ such that $dist(x,v)<dist(x,w)$". – Apass.Jack May 23 at 2:49
• I edited the post. – mahdi gh May 23 at 9:04
I will assume that you have as input a graph $$G = (V, E)$$. Let $$dist$$ be a vector which stores the distance of all nodes of your graph to node $$v \in V$$.
You must run a BFS algorithm starting on node $$v$$ and store the distance from each node to $$v$$ in $$dist$$. This value can be simply obtained during the BFS execution. Then, you can iterate through this vector marking (or deleting) all nodes $$i$$ such that $$dist[i]$$ < $$dist[w]$$.
The overall complexity of this algorithm is $$\mathcal{O}(V + E + V)$$. In a tree, we have that $$|E| = |V|-1$$. Thus, the complexity is $$\mathcal{O}(3 V) = \mathcal{O}(V)$$.
• I am afraid this answer is wrong. Consider the graph with edges $xu$, $uv$ and $vw$. $dist_v[x]=2 > dist_v[w]=1$. However, $x$ is nearer to $v$ than $w$. – Apass.Jack May 23 at 0:30
• Why $x$ is nearer to $v$ than $w$? In your example, $dist_v[w] = 1$ (it follows the path $<w, v>$) and $dist_v[x] = 2$ (it follows the path $<x, u, v>$). Thus, you will not delete or mark node $x$. – Iago Carvalho May 23 at 1:21
• I see now... There are two possible interpretations of this question. The first one (which I answered) is to find the nodes whose distance to $v$ are smaller than the distance from $v$ to $w$. The second interpretation (which I now answered) is @Appas Jack interpretation. – Iago Carvalho May 23 at 1:28
|
{}
|
64 667
Assignments Done
99,2%
Successfully Done
In September 2018
# Answer to Question #65485 in Microeconomics for pepper
Question #65485
Suppose that Billy's preferences over baskets containing milk (good x), and coffee (good
y ), are described by the utility function U(x; y ) = xy +2x. Billy's corresponding marginal
utilities are, MUx = y + 2 and MUy = x:
Use Px to represent the price of milk, Py to represent the price of coffee, and I to represent
Billy's income.
Suppose that Px = $1 and I =$40. Find the equivalent variation for an
increase in the price of coffee from Py1 = $4 to Py2 =$5
Need a fast expert's response?
Submit order
and get a quick answer at the best price
for any assignment or question with DETAILED EXPLANATIONS!
|
{}
|
Calculating amount of HCL required
1. Aug 16, 2016
JonnyG
1. The problem statement, all variables and given/known data
Suppose Iron(II) Sulfide is reacted with hydrogen chloride. How many grams of HCl is required is react with 75.0 grams of Iron (II) Sulfide ore of which 30% is inactive.
2. Relevant equations
3. The attempt at a solution
EDIT: Nevermind I solved it. Obviously 43.6 g of $2 \mathrm{HCl}$ is the same as 43.6 g of $\mathrm{HCl}$.
Only 70% of the Iron (II) Sulfide is active, so we can just pretend we are dealing with 52.5 g of Iron (II) Sulfide. Now, the equation is $\mathrm{FeS} + 2\mathrm{HCl} \rightarrow \mathrm{FeCl}_2 + \mathrm{H}_2 \mathrm{S}$.
1 mole of $\mathrm{FeS} = 87.95$ g and 1 mole of $2\mathrm{HCl} = 73$ g. Because 52.5/87.95 = 0.597 then we only need (0.597)(73) = 43.6 g of $2 \mathrm{HCl}$ for the reaction, which is equivalent to (43.6)(2) = 87.2 g of $\mathrm{HCl}$. Is this correct? A couple answers that I read online says that the answer is 43.6 g of $\mathrm{HCl}$ rather than 43.6 g of $2 \mathrm{HCl}$.
2. Aug 16, 2016
BvU
Well, on a scale it comes down to exactly the same quantity of $HCl$
|
{}
|
# What's this called? A small Gaussian at every data point to estimate probability
Say I have a dataset $X \subset{\mathbb R}^d$ (assumed to be iid samples) and I want to estimate the probability density of some unseen point y under that (unknown distribution). One way of doing that is by placing a Gaussian with small variance $\sigma$ at every datapoint and averaging their densities: $$p(y) = \frac{1}{|X|}\sum_{x \in X} N(y \mid x, \sigma I)$$ Where $N(y|x, \sigma I)$ is the probability density of y under a multivariate Gaussian distribution with mean $y$ and covariance matrix $\sigma I$ (ie. the identity matrix scaled by \sigma).
This seems like such a basic method that there should be a name for it. Is there? I thought it might be convolution, but that's not getting me anywhere.
$p_0(x)={1 \over |X|}\sum\limits_{x_i \in X} \delta_{x_i}(x)$
$p(x) = p_0(x) * \mathcal{N}(x| 0,\sigma)$
|
{}
|
Convergence of seriesApplication of the ratio, root or integral test on a series.Calculation of the Taylor seriesCalculates and draw the graph of the first n elements of the Taylor series for an entered function.Calculation of the Fourier seriesCalculates and draw the graph of the first n elements of the Fourier series for an entered function.Direction fieldDraws the direction field for a differential equation $y'=f(x,y)$.Direction field for a system of equationsDraws the direction field for a system of two differential equations $y_1'=f(y_1,y_2)$$y_2'=g(y_1,y_2)$Solution in the direction fieldDraws the direction field and several different solutions for a differential equation $y'=f(x,y)$.Separation of variablesGeneral solution of a differential equation which can be written in the form $y'=f(x)g(y)$.Linear differential equation of the first orderGeneral solution of a differential equation which can be written in the form $y'+py=q$.Homogeneous linear ODEs with constant coefficientsCalculation of the basis of the space of solutions of a homogeneous linear ODE with constant coefficients.Finite difference method for the Poisson equationBoundary value problem.$-\frac{\partial^2 u}{\partial x^2}-\frac{\partial^2 u}{\partial y^2}=f(x,y)$ inside the square $\Omega = [0,1]x[0,1]$, $u=0$ on the boundary $\partial \Omega = \Gamma$,is being solved by the finite difference method. $n$ inner nodes on the side of the square generates a system of $n^2$ linear equations.
|
{}
|
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.