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INDUCTORS

# Inductors

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## Overview

Inductors are passive electronic components which store energy as a magnetic field. They are made by coiling wire around a material (be it air, or something else). They are well used but not as common as resistors and capacitors in electronic circuits.

The mechanical equivalent of an inductor is mass. The larger the inductance, the larger the mass. This is when using the force-voltage equivalence.

Two inductances which are in close proximity they can couple with each, this is called mutual inductance. This is the basic principle behind a transformer.

Inductors are used in many things, including:

## How They Work

When current is flowing through an inductor, some electrical energy is converted and stored as magnetic energy in the inductors magnetic field. Inductors like staying at a constant current, and they “actively resist” current changes. They do this by generating a voltage across their terminals which resists the change in current.

The defining equation relating the voltage, inductance and change in current is:

$$v = L \frac{di}{dt}$$

where:
$$v$$ is the voltage across the inductor
$$L$$ is the inductance of the inductor
$$\frac{di}{dt}$$ is the instantaneous change in current through the inductor

For example: If you connected 12V across an inductor, initially there would be no current, as current cannot change instantaneously through an inductor. Then, because the voltage across the inductor is a constant 12V, the current through the inductor would rise linearly. The bigger the inductance, the slower the current would increase, as per the equation above. In a perfect world this would continue forever, but in reality either your power supply will current limit or your inductor will start to heat up and smoke.

## Schematic Symbols and Designator

There are a few common schematic symbols for an inductor. L is almost always used as the designator prefix on schematics, e.g. inductors will be named L1, L2, e.t.c. The figure below shows the symbols used for common sub-families of inductors. Not all circuit designers will be specific and use a special symbol to indicate iron/ferrite cored inductors (vs. air-cored inductors), and the generic symbol may be used for all types.

## Important Parameters

### Inductance

• Symbol: $$L$$
• Units: Henries (or Henrys)1 ($$H$$)

The main thing you care about! The inductance determines the relationship between the rate of change in current through the inductor and the inductors voltage. It determines the voltage the inductor will generate in response to a specific rate of current change. This parameter is used to calculate it’s impedance, for a given frequency.

### K-factor

The K-factor (a.k.a.) coupling coefficient is a fractional value of the amount of flux linkage between two coils. It is not applicable for a single inductor, but is a useful property when describing transformers.

### DC resistance

• Symbol: $$DCR$$
• Units: Ohms ($$\Omega$$)

The DC resistance is the resistance of the inductor’s coil of wire, measured at DC. You can use this to calculate resistive losses through the inductor. An ideal inductor has no DCR. The DC resistance of an inductor can be easily measured by a good quality multimeter or benchtop Ohm meter. For a fixed volume, a larger valued inductor usually has a higher DCR as it has a larger number of thinner guage turns.

Typical DC resistances range from $$10m\Omega$$ to $$10\Omega$$.

### Saturation Current

• Symbol: $$I_{sat}$$
• Units: Amps ($$A$$)

The saturation current is the maximum current the inductor can take before it stops working like an inductor (it’s not a sudden point, but rather defined as when the inductance drops to a fixed percentage of it’s nominal value). Above this point, the magnetic field does not increase proportionally with the current (the core is “saturated”), and the apparent inductance of the inductor is decreasing. This also means that above the saturation current the current will rise/fall much faster, which can cause it to shoot up to dangerous levels. It is not the same thing as the rated current.

For example, Vishay Dale specified the saturation current for it’s IHLP-1212BZ-11 family (datasheet) of inductors as the current at which the inductance drops 20% down from the value at 0A (it’s rated inductance)2. The following graph shows the inductance vs. DC current plot for the IHLP1212BZER1R0M11 inductor in this family. This inductor has a saturation current of 4.5A.

Typical saturation currents range from 10-100mA’s for small inductors, up to 1-10A for larger PCB mount inductors.

### Rated current

• Symbol: $$I_{rated}$$
• Units: Amps ($$A$$)

The rated current is typically given as the amount of average current required to created a fixed temperature rise above ambient due to resistive losses in the copper winding. The temperature rise is usually 40°C. Inductors can usually handle current peaks above the rated current as long as the average is still less.

### Self-resonant Frequency

• Symbol: $$SRF$$
• Units: Megahertz ($$MHz$$)

The self-resonant frequency of an inductor is the frequency at which the parasitic capacitance of the inductor resonates with the inductance, resulting in a very high impedance (e.g. open circuit)3. The inductor stops behaving like a inductor above the self-resonant frequency.

The below plot shows a plot of inductance vs. frequency for the Vishay Dale IHLP1212BZER1R0M11 inductor. The self-resonant frequency occurs where the inductance $$L$$ spikes at about 75MHz:

The equation for the SRF is4:

\begin{align} SRF = \frac{1}{2\pi \sqrt{LC}} \end{align}

where:
$$L$$ is the inductance in Henries
$$C$$ is the parasitic capacitance, in Farads
$$SRF$$ is the self-resonant frequency, in Hertz

The main contributor to the parasitic capacitance is the distributed capacitance between each coil windings.

## Types Of Inductors

### Ferrite Drum Core Inductors

Ferrite drum core inductors can be optionally shielded. They allow for high saturation currents and high inductances (due to having a large number of turns).

### Flat Coil Inductors

Flat coil inductors have a very low profile (height from the surface of the PCB) allowing use in space-constrained designs. Flat coil inductors usually have a low number of turns, resulting in less inductance than ferrite drum core or toroidal inductors. The core is typically made up of powdered iron.

### Metal Composite Inductors

Metal composite inductors (also called moulded power inductors) are made with metal powders moulded around copper windings at high pressure.

### Air-cored Inductors

Air-cored inductors represent any inductor which contains no ferromagnetic material. This means that “air-cored” inductors also covers a range of inductors which use an insulator instead of air for the core, which includes materials such as glass or PTFE.

• Unlike a ferro-magnetic cored inductor, an air-cored inductors inductance does not vary with the amount current that is flowing with it. Ferromagnetic cores work fine up until a point at which they saturate.
• Air-cores do not have any iron losses. This advantage becomes more significant with increasing frequency.

• Air-cored inductors have to have more and/or larger turns to achieve the same inductance value as a ferro-magnetic core. This is because ferro-magnetic cores have a higher permeability than air.
• Air-cored inductors radiate more electromagnetic fields over a wider area, and also pick up more radiation. This is because ferro-magnetic-cores constrain the magnetic field lines better.

## Common Core Materials

### Air Core

Air core is used to describe any inductor which does not use a magnetic material for it’s core (so they may in fact, have something other than air for their core). Used in low inductance, RF applications because they lack any magnetic core to boost their inductance, but at the same time benefit from smaller core losses which can be significant in the RF spectrum. Go to the Air-Cored Inductors section for more info.

### Powdered Ferrite Core

The powder is compressed into blocks before using (but don’t confuse these with solid ferrite core inductors). Very lossy (resistive) at high frequencies. Used to make ferrite beads.

### Solid Iron (or Steel) Core

Used in low resistance, high inductance applications. Has a higher saturation current than air-cored inductors, but lower one than powdered iron cores.

### Powdered Iron Core

Powered iron core inductors have a distributed air gap, which gives them a higher saturation current then solid iron cored inductors.

### Metal Composite

While ferrite cored inductors seem to have similar current ratings (based of thermal performance) as their saturation current, metal composite inductors typically have a saturation current between 1-2x their current rating (e.g. a current rating of 3A, and a saturation current between 3-6A).

## Inductors In Series And In Parallel

The behaviour of inductors when connected together in series and in parallel is exactly the same behaviour resistors exhibit, and exactly the opposite behaviour of what capacitors exhibit.

### Inductors In Parallel

When two inductors are connected in parallel, the equivalent total inductance follows the inverse law, as long as there is no mutual coupling of their magnetic fields:

$$L_{total} = \dfrac{1}{\dfrac{1}{L1} + \dfrac{1}{L2}}$$

It is usually easier to remember this equation as:

$$\dfrac{1}{L_{total}} = \dfrac{1}{L1} + \dfrac{1}{L2}$$

The following diagram shows this:

### Inductors In Series

The equivalent inductance of two inductors connected in series is the sum of the individual inductances, as long as there is no mutual coupling of their magnetic fields.

$$L_{total} = L1 + L2$$

This is shown in the diagram below:

## The Energy Stored In An Inductor

The energy stored in a inductor is given by:

$$E = \frac{1}{2}LI^2$$

where:
$$E$$ is the energy stored in the inductor, in Joules $$J$$
$$L$$ is the inductance of the inductor, in Henries $$H$$
$$I$$ is the current flowing through the inductor, in Amps $$A$$

This equation is only valid when the inductor is operating in it’s linear region, that is, before the current reaches the point where the magnetic field begins to saturate. Notice that it is similar to the equation for energy in a capacitor.

## The Impedance Of An Inductor

### Simple Model

An ideal inductor (no parasitic elements) has an impedance $$Z_L$$ given by:

$$Z_L = 2\pi fL$$

where:
$$Z_L$$ = inductor's impedance, in Ohms ($$\Omega$$)
$$f$$ = frequency of analysis, in Hertz ($$Hz$$)
$$L$$ = inductance of inductor, in Henries ($$H$$)

Sometimes $$X_L$$ is used instead of $$Z_L$$. $$X$$ represents reactance (purely imaginary), whilst $$Z$$ is for impedance (which may contain real and imaginary parts). For a simple inductor with no parasitic elements the reactance and impedance is the same, i.e. $$X_L = Z_L$$.

This shows that the impedance goes up as the frequency goes up. At DC levels, the inductor has no impedance and acts like a short-circuit, while at high frequencies the inductor approaches an open-circuit.

You may also see this written as:

$$Z_L = \omega L$$

where:
$$\omega$$ = radian frequency (rad/s) (which is equal to $$2\pi f$$)

Impedance, although measured in Ohms, is not the same as resistance. A resistance converts the energy into heat, the impedance of the inductor however cause the energy to be stored in it’s magnetic field, which is returned to the circuit when the current falls.

For a more accurate but complex model, see the Parasitic Model section.

### Parasitic Model

An inductor with parasitic components may be modelled by:

$$Z_L = \frac{1}{ \frac{1}{R_P} + \frac{1}{j \omega L + R_S} + j \omega C_P }$$

where:
$$Z_L$$ = inductor's impedance ($$\Omega$$)
$$R_P$$ = parasitic parallel resistance (which is negligible for most simulations) ($$\Omega$$)
$$\omega$$ = frequency of analysis ($$rad/s$$)
$$L$$ = inductance of inductor ($$H$$)
$$R_S$$ = parasitic series resistance ($$\Omega$$)
$$C_P$$ = parasitic parallel capacitance ($$F$$)

The main parasitic components are the series resistance ($$R_S$$) and the parallel capacitance ($$C_P$$). The series resistance arises from the resistance of the coil of wire which makes up the inductor. This thinner and longer the coil, the larger this resistance. The parallel capacitance is due to coil windings being very close to one another, each coil forming small capacitors to surrounding coils. This small capacitors can be lumped together and form the parallel capacitance.

This model is a pretty good representation for most simulation purposes, and gives that characteristic decrease in performance (non-inductor-like behaviour) at high-frequencies.

Typical values might be:

• $$R_P = 0\Omega$$
• $$L = 1uH$$
• $$R_S = 10m\Omega$$
• $$C_P = 10pF$$.

## Inductor Networks

### Inductors In Parallel

Inductors in parallel behave just like resistors in parallel.

$$\frac{1}{L_{eq}} = \frac{1}{L_1} + \frac{1}{L_2} + \ldots + \frac{1}{L_N}$$

### Inductors In Series

Inductors in series behave just like resistors in series.

$$L_{eq} = L_1 + L_2 + \ldots + L_n$$

## Inductor Kickback

Inductor kickback is when an inductor the current through an inductor is quickly stopped, usually by making an open circuit. Because inductors “want to keep the current flowing”, they generate a very large voltage spike in attempt to keep the current going. This large voltage spike can jump switch gaps, fry MOSFETs, destroy other sensitive circuitry, and give people dangerous shocks.

A more technical way of looking a kickback is by analysing the voltage-inductance equation. From this is should be obvious that the inductors voltage is proportional to the change in current. Making an inductor open-circuit is creating a very large change in current, and hence it produces a very large voltage spike.

Inductor kickback is a significant design issue when using relays or motors.

## Saturation Current

Inductors only behave like inductors over a limited current range. As the current continues to increase, the apparent inductance of the inductor decreased. Above a certain current, they become what is called saturated.

Saturation is the state reached when an increase in applied external magnetizing field H cannot increase the magnetization of the material any further.

The below image shows a typical inductance vs. current curve for a ferrite core, SMD inductor that could be used on a 3A SMPS design. It’s rated saturation current is 4.2A. Notice how after the current increases beyond this point, the inductance of the inductor drops of rapidly.

As governed by the basic induction equation, applying a constant voltage across an inductor will lead to a linear increase in current. This will be the case until the saturation point, at which the current will begin to increase more rapidly, until is reaches a limit determined only by the DC resistance of the windings.

Reaching saturation is not always a bad thing. Saturation is desirable in self-switching power supplies, timing circuits and blocked oscillators, and saturable-core transformers (e.g. arc welding). It is not desirable in typical SMPS, or power/signal transformers.

## Rated Current

The rated current of an inductor is the maximum DC current allowed before the inductor gets"too hot". “Too hot” is usually defined as a certain temperature rise above ambient (e.g. $$40^{\circ}C$$) when the inductor is mounted using a standard footprint on a standard PCB; again, both defined by the manufacturer.

## Saturation Current vs. Rated Current

There are usually two maximum current ratings on an inductor’s datasheet, the saturation current and the rated current.

Confusing? The most simple Simon design approach is to pick the lower maximum current of the two and treat that as the maximum current allowed through the inductor. A seasoned inductor sensei will realise that the rated current is not applicable to short current peaks, and so it is safe to exceed this for brief periods of time, as long as the saturation current is not exceeded (disclaimer: in some cases, even exceeding the saturation current limit is o.k. or even desirable, but see the Saturation Current section for more info on this.

## How To Measure Inductance In An Electronics Lab

Inductor markings can be confusing (or completely absent!), making it hard to determine the part number and inductance of a particular inductor. Additionally, basic multimeters have no ability to measure inductance. Luckily, there are many ways to measure inductance in a laboratory setting.

### Measuring Inductance with a Signal Generator and Oscilloscope

One way to measure the inductance of an inductor is with a signal generator and oscilloscope5.

1. Set the signal generator to:

• Waveform: Sine wave
• Voltage: 1V peak-to-peak
• Frequency: 10kHz
• Output type: High-Z
2. Connect the oscilloscope to the signal generator and measure the open-circuit voltage. It should be about 1V peak-to-peak. Record down the exact peak-to-peak value, as signal generators are not usually that accurate with their voltages!

If you read 2V peak-to-peak, you probably have the signal generators output type set to 50Ω, which means it doubles the voltage. Switch to high-Z or change the voltage to 500mV peak-to-peak to compensate.
3. Now connect the inductor to the signal generator, whilst still measuring the voltage with the oscilloscope (this will be the output voltage of the signal generator, as well as the voltage across the inductor). The circuit should look like this:

4. Adjust the frequency of the signal generator until the peak-to-peak voltage across the inductor is half the measured open-circuit peak-to-peak voltage.

5. Calculate the inductance! It is equal to:

\begin{align} L = \sqrt{\frac{1}{3}} \cdot \frac{50\Omega}{2\pi f} \end{align}

where:
$$f$$ is the frequency in Hertz at which the inductors peak-to-peak voltage was half the signal generators open-circuit peak-to-peak voltage
$$50\Omega$$ is the output impedance of the signal generator (measure and tweak this value for a more accurate calculation)
$$L$$ is the calculated inductance, in Henries

The above method only works well for inductors which have a low series resistance[^bib-dos4ever-meas-unknown-inductors].

How It Works

The internal resistance of the signal generator $$50\Omega$$ and the inductance form a voltage divider. The impedance of the resistor stays constant at $$50\Omega$$ no matter the frequency, but the impedance of the inductor does change with frequency ($$Z_L = X_L = \omega L$$). You vary the frequency until the voltage at the mid-point of the voltage divider is half the total voltage, which means that the impedances of the resistor and inductor are the same.

We arrive at the equation for the inductance by using the voltage-divider rule for a complex circuit:

\begin{align} \left| \frac{V_{SCOPE}}{V_{SIG}} \right| = \left| \frac{j\omega L}{R + j\omega L} \right| \end{align}

Simplifying:

\begin{align} \left| \frac{V_{SCOPE}}{V_{SIG}} \right| &= \left| \frac{j\omega L}{R + j\omega L} \cdot \frac{R - j\omega L}{R - j\omega L} \right| \nonumber \\ &= \left| \frac{j\omega RL + \omega^2 L^2}{R^2 + \omega^2 L^2} \right| \end{align}

Now let’s separate the equation into a real and imaginary part:

\begin{align} \left| \frac{V_{SCOPE}}{V_{SIG}} \right| &= \left| \frac{\omega^2 L^2}{R^2 + \omega^2 L^2} + j\frac{\omega RL}{R^2 + \omega^2 L^2} \right| \\ \end{align}

Now we can take the magnitude of this and simplify:

\begin{align} \left| \frac{V_{SCOPE}}{V_{SIG}} \right| &= \sqrt{ \frac{\omega^4 L^4}{(R^2 + \omega^2 L^2)^2} + \frac{\omega^2 R^2L^2}{(R^2 + \omega^2 L^2)^2} } \nonumber \\ &= \frac{\sqrt{\omega^4 L^4 + \omega^2R^2L^2}}{R^2 + \omega^2L^2} \nonumber \\ &= \frac{\sqrt{\omega^2 L^2(R^2 + \omega^2L^2)}}{R^2 + \omega^2L^2} \nonumber \\ &= \frac{\omega L}{\sqrt{R^2 + \omega^2L^2}} \\ \end{align}

We’ve got it to a point where it is as about as simple as you can write it! Next up, let’s substitute $$\left| \frac{V_{SCOPE}}{V_{SIG}} \right|$$ with $$\frac{1}{2}$$), as this was the condition we were tweaking the frequency for (peak-to-peak voltage measured across inductor by the oscilloscope was half the peak-to-peak voltage generated by the signal generator).

\begin{align} \frac{1}{2} &= \frac{\omega L}{\sqrt{R^2 + \omega^2L^2}} \nonumber \\ \frac{1}{4} &= \frac{\omega^2 L^2}{R^2 + \omega^2L^2} \nonumber \\ R^2 + \omega^2 L^2 &= 4\omega^2 L^2 \nonumber \\ R^2 &= 3 \omega^2 L^2 \nonumber \\ L^2 &= \frac{1}{3} \frac{R^2}{\omega^2} \nonumber \\ L &= \sqrt{\frac{1}{3}} \frac{R}{\omega} \end{align}

Now we just sub in $$2\pi f = \omega$$ to get the final equation:

\begin{align} L &= \sqrt{\frac{1}{3}} \frac{R}{2\pi f} \end{align}

## Manufacturer Part Numbers

• DR1050: Series of inductors by Eaton (previously Cooper Bussmann).
• NRx: Popular series of inductors from Taiyo Yuden. x denotes the coating resin, options are NR, NRH, NRS and NRV. The series contains a number of different SMD package sizes and inductances.

## Inductor Packages

Unfortunately for the PCB designer, almost all SMD inductor packages are non-standard and unique to the manufacturer and series.

## References

1. Wiktionary. henry. Retrieved 2022-07-10, from https://en.wiktionary.org/wiki/henry↩

2. Vishay Dale. _IHLP-1212BZ-11: IHLP Commercial Inductors, Low DCR Series (datasheet). Retrieved 2022-07-11, from https://www.vishay.com/docs/34289/lp12bz11.pdf↩ ↩ ↩

3. everythingRF. What is Self Resonant Frequency (SRF) in Inductors?. Retrieved 2022-07-11, from https://www.everythingrf.com/community/what-is-self-resonant-frequency↩

4. Coilcraft (2003, Sep 16). Document 363-1: Measuring Self Resonant Frequency. Retrieved 2022-07-11, from https://www.coilcraft.com/getmedia/8ef1bd18-d092-40e8-a3c8-929bec6adfc9/doc363_measuringsrf.pdf↩

5. Ronald Dekker. A Simple Method to Measure Unknown Inductors. DOS4Ever. Retrieved 2022-07-06, from https://www.dos4ever.com/inductor/inductor.html↩

## Authors

### Geoffrey Hunter

Dude making stuff.


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# Math Help - Definition of Sin or Cos

1. ## Definition of Sin or Cos

Where do sin and cos come from.

I use them on my caluclator but I never really understand where they are getting the number from. My teacher keeps mentioning that when he was a student a hundred years ago they had to look up all of the values in a book.

so where do the values come from. could I derive them myself?

2. See MathWorld - Sine and MathWorld - Cosine.

One can express Sine and Cosine in natural exponents or infinite series.

3. Originally Posted by Math Dude
Where do sin and cos come from.

I use them on my caluclator but I never really understand where they are getting the number from. My teacher keeps mentioning that when he was a student a hundred years ago they had to look up all of the values in a book.

so where do the values come from. could I derive them myself?
For sure!
There are formulas that do the work very well, but are quite difficult to explain expecially if you have not studied calculus.
The name is Taylor's formula
http://www.math.jhu.edu/courses/107/...es/node34.html

But here is an example
you get something like
0.09983341664

Now try with this:
x - x^3/6
if you put 0.1 = 1/10 in place of x you get
1/10 -1/6000 ~ 0.09983333333
how many decimal digit are correct? Six!

Better results (and longer calculation) are obtained if you use
x^5/120 - x^3/6 + x
0.1^5/120 - 0.1^3/6 + 0.1 ~ 0.09983341666
Ten decimal digits are correct!
They are long calculation to be made by hand, aren't they?

Once matematicians made all by hand. Look at you calculator with more respect from now on

bye

format: plain


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### C-16, r. 10 - Regulation respecting the standards of equivalence for the issuance of a permit of the Ordre des chiropraticiens du Québec

4.02. In order to determine whether a candidate shows that he has the level of knowledge required under section 4.01, the board of directors shall take the following factors into account:
(a)  the total number of academic years;
(b)  the courses taken;
(c)  the nature and duration of experience;
(d)  the training periods served;
(e)  the fact that the candidate holds one or several diplomas obtained in Québec or elsewhere.
Where the appreciation made under the first paragraph does not allow a decision to be made, the board of directors may require an examination or training period to complete the said appreciation.
Decision 81-09-22, s. 4.02.


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TL;DR

Where you are patient enough to continue on the Bayes rabbit hole.

In last post Some Bayes helps we took a look at how we can use a campaign of measurements to get a table that helps us understanding how to infer information from more future measurements.

# (re-)Starting point

We restart from the estimated conditional probabilities table:

$X_1$ $X_2$ $X_M$
C $P_{X_1|C} = \frac{\Gamma_{X_1} }{\sum_X \Gamma_X}$ $P_{X_2|C} = \frac{\Gamma_{X_2} }{\sum_X \Gamma_X}$ $P_{X_M|C} = \frac{\Gamma_{X_M} }{\sum_X \Gamma_X}$
D $P_{X_1|D} = \frac{\Delta_{X_1} }{\sum_X \Delta_X}$ $P_{X_2|D} = \frac{\Delta_{X_2} }{\sum_X \Delta_X}$ $P_{X_M|D} = \frac{\Delta_{X_M} }{\sum_X \Delta_X}$

Remember that we also have some a-priori estimation of absolute probability of a contact $P_C$, and that the maximum likelihood principle led us to establish the following inference algorithm:

• if $P_C \cdot P_{X|C} \geq (1 - P_C) \cdot P_{X|D}$ we infer $C$ from $X$, otherwise
• if $P_C \cdot P_{X|C} < (1 - P_C) \cdot P_{X|D}$ we infer $D$ from $X$.

This inference algorithm partitions the set of all measured classes $X_1 .. X_M$ into two sets:

$\mathscr{C} = \{X : X \Rightarrow C\} \\ \mathscr{D} = \{X : X \Rightarrow D\}$

# Can we be wrong?

You bet we can. The fact itself that a column in our table can have two non-zero values means that we will be wrong in some cases - the whole point of the maximum likelihood approach is to try and choose the option that makes us fail less often.

There are two kind of errors that we can make:

• saying that a contact happened, when it didn’t - this is called a false positive;
• saying that a contact did not happen, when it did - this is called a false negative.

Each of the possible classes from our measurement will contribute to at most one of these errors (it might contribute to neither if the column has one $1$ and one $0$, but this is the real world).

In particular:

• elements in $\mathscr{C}$ all lead to inferring a contact, and can thus contribute to false positives only;
• elements in $\mathscr{D}$ can only contribute to false negatives.

## False positives

Let’s start with false positives. As we saw, we chose to infer $C$ from event $X$ because this applies:

$P_C \cdot P_{X|C} \geq (1 - P_C) \cdot P_{X|D}$

which is equal to say:

$P_{C|X} = \frac{P_C \cdot P_{X|C}}{P_X} \geq \frac{(1 - P_C) \cdot P_{X|D}}{P_X} = P_{D|X}$

This basically means that, when $X$ happens, we will be right in inferring $C$ with probability $P_{C|X}$, and fail otherwise.

For this reason, the term on the right is exactly our false positive probability, conditioned to event $X$:

$P_{fp|X} = P_{D|X}$

It’s easy at this point to calculate the overall probability of a false positive, adding up all terms that can lead to this kind of error weighting them with the probability of each event $X$:

$P_{fp} = \sum_{X \in \mathscr{C}}(P_X \cdot P_{fp|X}) = \sum_{X \in \mathscr{C}}(P_X \cdot P_{D|X}) = \sum_{X \in \mathscr{C}}P_{DX} = \sum_{X \in \mathscr{C}}((1 - P_C) \cdot P_{X|D}) \\ P_{fp} = (1 - P_C) \sum_{X \in \mathscr{C}}P_{X|D}$

The summatory takes all items from the table corresponding to the row for $D$ and to those columns that provide an inference to $C$; this sum is weighted with the absolute probability of distancing, which makes sense because a false positive means being distanced (which happens with probability $(1 - P_C)$) but having ended up with classes that we consider for a contact.

## False Negatives

Calculating the probability of false negatives is the dual of what we considered in the previous section, so we end up with:

$P_{fn} = P_C \sum_{X \in \mathscr{D}}P_{X|C}$

# Not all wrongs are created equal

As we saw, for each measured event $X$ with non-zero probability of being associated to $C$ and $D$ (i.e. $X$’s column in the table has both entries greater than zero) we end up with some probability of doing errors (respectively false positives and false negatives).

Do we think they are the same? Maybe not.

Just as you would probably pick your umbrella if the wheather forecasts tell you there’s a 40% probability of rain, at the risk of carrying it unnecessarily most of the times, we might want to be conservative and prefer false positives over false negatives.

For this reason, we might set a higher limit to the probability of false negatives, and move all classes that would exceed this limit from $\mathscr{D}$ to $\mathscr{C}$. This would be against the maximum likelihood principle and would mean being more wrong in average, but at least it would be our favourite flavor of being wrong.

Note that it’s still meaningful to allow for some false negatives, though: in most cases, bringing this number down to zero would mean that every Bluetooth exchange is interpreted as a contact, which might not be economically feasible (e.g. for lack of sufficient testing capabilities).

# Fancier uses?

The mathematical model we ended up with can also be used for a hybrid solution. Many things in life are not black and white, so why should this be different?

At the basic level, we can consider that measuring event $X$ gives us two probabilities, i.e. one for inferring a contact ($C$) and one for inferring correct distancing ($D$). We already saw these two probabilities, expressed in terms of quantities we have or have assumed:

$P_{C|X} = \frac{P_C \cdot P_{X|C}}{P_X} = \frac{P_C \cdot P_{X|C}}{P_C \cdot P_{X|C} + (1 - P_C) \cdot P_{X|D}} \\ P_{D|X} = \frac{(1 - P_C) \cdot P_{X|D}}{P_X} = \frac{(1 - P_C) \cdot P_{X|D}}{P_C \cdot P_{X|C} + (1 - P_C) \cdot P_{X|D}}$

So, if our health testing capabilities are of $K$ tests per day, we might do like this:

• record all occurrences of every event $X$, without removing any;
• sort them by inverse $P_{C|X}$, i.e. from greater probability of a contact to lower ones;
• each day, remove the top $K$ and apply the health test to them.

This would allow prioritizing the most probable contacts, remain within the limits of the SSN checking capabilities, and decide where to draw the line of excessive testing when there are more available data.


text:
# Matching Supply with Demand

## Quiz 1 :The Process View of the Organization

Question Type
(Dell) What percentage of cost of a Dell computer reflects inventory costs Assume Dell's yearly inventory cost is 40 percent to account for the cost of capital for financing the inventory, the warehouse space, and the cost of obsolescence. In other words, Dell incurs a cost of $40 for a$100 component that is in the company's inventory for one entire year. In 2001, Dell's 10-k reports showed that the company had $400 million in inventory and COGS of$26,442 million.
Free
Essay

Inventory Cost is the Cost associated with Ordering, Holding, Obsolescence Pilferage cost of the Inventory which is expressed as a percentage of the Inventory Valuation
Consider the following data given in the question:
Inventory in D's 10-k = $400 Cost of goods sold in D's 10-k =$26,442
Calculate the Inventory Turns for company D for the year 2001 by using the formula as follows:
Calculate the cost of inventory per unit from the formula as follows:
Therefore, the percentage of cost a D computer reflects inventory costs is

Tags
(Airline) Consider the baggage check-in of a small airline. Check-in data indicate that from 9 a.m. to 10 a.m., 255 passengers checked in. Moreover, based on counting the number of passengers waiting in line, airport management found that the average number of passengers waiting for check-in was 35. How long did the average passenger have to wait in line
Free
Essay

Little's Law gives us a relationship between Average Inventory, Flow rate and the Flow Time. Arrival rate is the flow rate of the passengers. The average time the passenger spends in the system is the Flow Time.
Consider the following data given in the question:
Arrival rate or Flow rate = 255 Passenger per hour
Average Inventory of passengers = 35 passengers
Calculate the flow time, using Little's law, as follows:
Therefore, the average passenger has to wait in line is

Tags
(Inventory Cost) A manufacturing company producing medical devices reported $60,000,000 in sales over the last year. At the end of the same year, the company had$20,000,000 worth of inventory of ready-to-ship devices. a. Assuming that units in inventory are valued (based on COGS) at $1,000 per unit and are sold for$2,000 per unit, how fast does the company turn its inventory The company uses a 25 percent per year cost of inventory. That is, for the hypothetical case that one unit of $1,000 would sit exactly one year in inventory, the company charges its operations division a$250 inventory cost. b. What-in absolute terms-is the per unit inventory cost for a product that costs $1,000 Free Essay Answer: Answer: Inventory Turns is the inverse of the Flow time. Cost of the Goods Sold is the Cost of the production that company incurs in making the goods when it is sold. Inventory turns is the Number of times the inventory is sold or used in a year or a period of time. Consider the following data given in the question: Cost of Goods Sold as$1000 per unit for a Sales Cost of $2000 per unit. Note: Since$60,000,000 sales were realized in the last year, it can be concluded that the COGS for this value will be half of the sales.
Calculate the Cost of Goods Sold from sales of $60,000,000 as$30,000,000 as follows:
Therefore, the company turns its inventory in
b)
Calculate the cost of inventory per unit from the formula as follows:
Therefore, for a product costing $1000 the Inventory Cost would be 16.66% of$1000 is

Tags
(Apparel Retailing) A large catalog retailer of fashion apparel reported $100,000,000 in revenues over the last year. On average, over the same year, the company had$5,000,000 worth of inventory in their warehouses. Assume that units in inventory are valued based on cost of goods sold (COGS) and that the retailer has a 100 percent markup on all products. a. How many times each year does the retailer turn its inventory b. The company uses a 40 percent per year cost of inventory. That is, for the hypothetical case that one item of $100 COGS would sit exactly one year in inventory, the company charges itself a$40 inventory cost. What is the inventory cost for a $30 (COGS) item You may assume that inventory turns are independent of the price. Essay Answer: Tags (LaVilla) LaVilla is a village in the Italian Alps. Given its enormous popularity among Swiss, German, Austrian, and Italian skiers, all of its beds are always booked in the winter season and there are, on average, 1,200 skiers in the village. On average, skiers stay in LaVilla for 10 days. a. How many new skiers are arriving-on average-in LaVilla every day b. A study done by the largest hotel in the village has shown that skiers spend on average$50 per person on the first day and $30 per person on each additional day in local restaurants. The study also forecasts that-due to increased hotel prices-the average length of stay for the 2003/2004 season will be reduced to five days. What will be the percentage change in revenues of local restaurants compared to last year (when skiers still stayed for 10 days) Assume that hotels continue to be fully booked! Essay Answer: Tags (Highway) While driving home for the holidays, you can't seem to get Little's Law out of your mind. You note that your average speed of travel is about 60 miles per hour. Moreover, the traffic report from the WXPN traffic chopper states that there is an average of 24 cars going in your direction on a one-quarter mile part of the highway. What is the flow rate of the highway (going in your direction) in cars per hour Essay Answer: Tags (Industrial Baking Process) Strohrmann, a large-scale bakery in Pennsylvania, is laying out a new production process for their packaged bread, which they sell to several grocery chains. It takes 12 minutes to bake the bread. How large an oven is required so that the company is able to produce 4,000 units of bread per hour (measured in the number of units that can be baked simultaneously) Essay Answer: Tags (Mt. Kinley Consulting) Mt. Kinley is a strategy consulting firm that divides its consultants into three classes: associates, managers, and partners. The firm has been stable in size for the last 20 years, ignoring growth opportunities in the 90s, but also not suffering from a need to downsize in the recession at the beginning of the 21st century. Specifically, there have been-and are expected to be-200 associates, 60 managers, and 20 partners. The work environment at Mt. Kinley is rather competitive. After four years of working as an associate, a consultant goes "either up or out"; that is, becomes a manager or is dismissed from the company. Similarly, after six years, a manager either becomes a partner or is dismissed. The company recruits MBAs as associate consultants; no hires are made at the manager or partner level. A partner stays with the company for another 10 years (a total of 20 years with the company). a. How many new MBA graduates does Mt. Kinley have to hire every year b. What are the odds that a new hire at Mt. Kinley will become partner (as opposed to being dismissed after 4 years or 10 years) Essay Answer: Tags (Major U.S. Retailers) The following table shows financial data (year 2004) for Costco Wholesale and Walmart, two major U.S. retailers. Source: Compustat, WRDS. Assume that both companies have an average annual holding cost rate of 30 percent (i.e., it costs both retailers$3 to hold an item that they procured for $10 for one entire year). a. How many days, on average, does a product stay in Costco's inventory before it is sold Assume that stores are operated 365 days a year. b. How much lower is, on average, the inventory cost for Costco compared to Walmart of a household cleaner valued at$5 COGS Assume that the unit cost of the household cleaner is the same for both companies and that the price and the inventory turns of an item are independent.
Essay
(McDonald's) The following figures are taken from the 2003 financial statements of McDonald's and Wendy's. 1 Figures are in million dollars. a. In 2003, what were McDonald's inventory turns What were Wendy's inventory turns b. Suppose it costs both McDonald's and Wendy's $3 (COGS) per their value meal offerings, each sold at the same price of$4. Assume that the cost of inventory for both companies is 30 percent a year. Approximately how much does McDonald's save in inventory cost per value meal compared to that of Wendy's You may assume the inventory turns are independent of the price.


text:
# How do you divide (1.1x10^22)/(6.022x10^23)?

Mar 29, 2015

You can split the fraction as it follows:
$\setminus \frac{1.1 \cdot {10}^{22}}{6.022 \cdot {10}^{23}} = \setminus \frac{1.1}{6.022} \setminus \setminus \frac{{10}^{22}}{{10}^{23}}$

and deal with the two parts separately.

$\setminus \frac{1.1}{6.022}$ is a common numeric division, and the result is $0.1827$

As for $\setminus \frac{{10}^{22}}{{10}^{23}}$, you can see it in two ways: either you expand the powers, having a fraction of the form
$\setminus \frac{10 \cdot 10 \cdot \ldots \cdot 10}{10 \cdot 10 \cdot 10 \cdot \ldots \cdot 10}$, with 22 "tens" at the numerator and 23 at the denominator. So, you can simplify all the "tens", except one at the denominator, obtaining ${10}^{22} / {10}^{23} = \frac{1}{10}$.

In a more formal fashion, you have that ${10}^{22} / {10}^{23} = {10}^{- 1}$, which is again $\frac{1}{10}$.

So, the final answer is $\setminus \frac{1.1 \cdot {10}^{22}}{6.022 \cdot {10}^{23}} = 0.1827 \cdot {10}^{- 1}$.


text:
Mathematics

# Write the quadratic equation whose roots are 2 and -6, and whose leading coefficient is 1 .(Use the letter  to represent the variable.)  =0

$(x-2)(x+6)=\\ x^2+6x-2x-12=\\ x^2+4x-12$
$roots\ are\ 2\ and\ -6\ and\ a=1\ then:\\\\1(x-2)(x+6)=0\\\\x\cdot x+x\cdot6-2\cdot x-2\cdot6=0\\\\x^2+6x-2x-12=0\\\\x^2+4x-12=0$


text:
# Minimum problem...with distances.

• January 4th 2009, 05:31 PM
Jukixel
Minimum problem...with distances.
This one requires some drawing...I think. A diagram maybe? ^^"

Town A is 11 miles from a straight river and town B is 6 miles from the same river. The distance from town A to town B is 13 miles. A pumping station is to be built along the river to supply water to both towns. Where should the pumping station be built so that the sum of the distances from the pumping station to the town is a minimum?

Thanks for the help. =)
• January 4th 2009, 06:14 PM
Soroban
Hello, Jukixel!

Quote:

Town A is 11 miles from a straight river and town B is 6 miles from the same river.
The distance from town A to town B is 13 miles.
A pumping station is to be built along the river to supply water to both towns.
Where should the pumping station be built so that the sum of the distances
from the pumping station to the towns is a minimum?

There are two stages to this problem . . .

First, find the "horizontal" distance between the two towns.

In the diagram below, right triangle $ABE$ has side 5 and hypotenuse 13.
. . Hence: . $AB \,=\,CD \,=\,12$
Code:

    A * - - - - - - - - - - - - - * B       |                          :       |        *                : 5       |            13    *      :   11 |                          * E       |                          |       |                          |       |                          | 6       |                          |     - * - - - - - - - - - - - - - * -       C                          D

Then the diagram looks like this . . .
Code:

    A *       | *       |  *       |    *   11 |      *                  * E       |        *              * |       |          *          *  |       |            *      *    | 6       |              *  *      |       * - - - - - - - - *- - - - *       C        x        P  12-x  D

The pumping station is at $P.$

Let $CP \,=\,x \quad\Rightarrow\quad PD \,=\,12-x$

In right triangle $ACP\!:\;\;AP \,=\,\sqrt{x^2+11^2}$

In right triangle $EDP\!:\;\;PE \,=\,\sqrt{(12-x)^2+6^2}$

The total distance from $P$ to the towns is:

. . $D \:=\:\sqrt{x^2+121} + \sqrt{x^2 - 24x + 180}$

And that is function you must minimize . . .

• January 4th 2009, 06:15 PM
mr fantastic
Quote:

Originally Posted by Jukixel
This one requires some drawing...I think. A diagram maybe? ^^" Mr F says: You think??

Town A is 11 miles from a straight river and town B is 6 miles from the same river. The distance from town A to town B is 13 miles. A pumping station is to be built along the river to supply water to both towns. Where should the pumping station be built so that the sum of the distances from the pumping station to the town is a minimum?

Thanks for the help. =)

A diagram is essential. Draw one.

Now find the horizontal distance between each town. From you diagram you should see that the line segment joining A and B is the hypotenuse of a right-triangle. Use Pythagoras' theorem to get that the horizontal distance between A and B is 12 miles (alternatively, you are familiar with your Pythagorean triples and will have no trouble recognising the 5-12-13 right-triangle).

Let the pumping station P be x miles along the river from the perpendicular line joing A to the river.

Then $d\overline{AP}^2 = 11^2 + x^2$ and $d\overline{BP}^2 = 6^2 + (12 - x)^2$.

Therefore $L = d\overline{AP} + d\overline{BP} = \, ....$

Edit: Is this really a pre-calculus question, that is, you're expected to use technology to find the value of x that minimises the distance ....?

Otherwise it requires calculus .....
• January 5th 2009, 11:18 AM
HallsofIvy
Quote:

Originally Posted by Jukixel
This one requires some drawing...I think. A diagram maybe? ^^"

Town A is 11 miles from a straight river and town B is 6 miles from the same river. The distance from town A to town B is 13 miles. A pumping station is to be built along the river to supply water to both towns. Where should the pumping station be built so that the sum of the distances from the pumping station to the town is a minimum?

Thanks for the help. =)

I see that both Soroban and Mr. Fantastic are assuming the two towns are on the same side of the river. With that, it follows that distance "along the river" from A to B (from the point where the perpendicular from A meets the river to the point where the perpendicular is 12 miles.

But I would do this in a completely different way (especially if you are cannot use "Calculus"): using a "reflection". That is, imagine that B' is 6 miles from the river but on the opposite side of the river from B. Obviously (I hope it is obvious!) the shortest distance from A to B' is a straight line. We can determine where that line crosses the river by "similar triangles"- the triangle formed by A, foot of perpendicular from A to the river, point where the straight line from A to B' crosses the river is similar to the corresponding B' triangle. That follows from the "vertical angles" theorem- where lines cross (the line from A to B' and the river) angles across from one another are congruent.

If we let the distance from where the perpendicular from A hits the river to the point where the line AB' crosses the river be "x" then the distance from that point to where the perpendicular from B hits the river is 12- x. So we have $\frac{11}{x}= \frac{6}{12-x}$ so 11(12- x)= 132- 11x= 6x or 5x= 132 and x= 132/5= 26.4 miles.

Now you can show that this point is also correct for the shortest total distance between A and B. The two triangles, with B' and B are congruent so the distance from that crossing point on the river to B' is the same as from that point to B.
• January 5th 2009, 12:39 PM
mr fantastic
Quote:

Originally Posted by HallsofIvy
I see that both Soroban and Mr. Fantastic are assuming the two towns are on the same side of the river. With that, it follows that distance "along the river" from A to B (from the point where the perpendicular from A meets the river to the point where the perpendicular is 12 miles.

But I would do this in a completely different way (especially if you are cannot use "Calculus"): using a "reflection". That is, imagine that B' is 6 miles from the river but on the opposite side of the river from B. Obviously (I hope it is obvious!) the shortest distance from A to B' is a straight line. We can determine where that line crosses the river by "similar triangles"- the triangle formed by A, foot of perpendicular from A to the river, point where the straight line from A to B' crosses the river is similar to the corresponding B' triangle. That follows from the "vertical angles" theorem- where lines cross (the line from A to B' and the river) angles across from one another are congruent.

If we let the distance from where the perpendicular from A hits the river to the point where the line AB' crosses the river be "x" then the distance from that point to where the perpendicular from B hits the river is 12- x. So we have $\frac{11}{x}= \frac{6}{12-x}$ so 11(12- x)= 132- 11x= 6x or 5x= 132 and x= 132/5= 26.4 miles.

Now you can show that this point is also correct for the shortest total distance between A and B. The two triangles, with B' and B are congruent so the distance from that crossing point on the river to B' is the same as from that point to B.

The old unconscious assumption trick. Second time this week I've fallen for it.


text:
## arXiv preprint: The Instability of a Quantum Superposition of Time Dilations

Title: The Instability of a Quantum Superposition of Time Dilations

Abstract: Using the relativistic concept of time dilation we show that a superposition of gravitational potentials can lead to nonunitary time evolution. For sufficiently weak gravitational potentials one can still define, for all intents and purposes, a global coordinate system. A probe particle in a superposition of weak gravitational fields will, however, experience dephasing due to the different time dilations. The corresponding instability timescale is accessible to experiments, and can be used as a degree of macroscopicity. Finally, we suggest an experiment with smoothly tunable amplification in a microwave interferometer that allows a quantitative study of the quantum to classical boundary.

## arXiv preprint: How bilayer excitons can greatly enhance thermoelectric efficiency

Title: How bilayer excitons can greatly enhance thermoelectric efficiency

Abstract: Currently, one of the major nanotechnological challenges is to design thermoelectric devices that have a high figure of merit. To that end, we propose to use bilayer excitons. Bilayer exciton systems are shown to have an improved thermopower and an enhanced electric counterflow and thermal conductivity, with respect to regular semiconductor-based thermoelectrics. Here we present a roadmap towards experimental realization of a bilayer exciton thermocouple. A bilayer exciton heterostructures of p- and n-doped Bi2Te3 can have a figure of merit zT∼60. Another material suggestion is to make a bilayer out of electron-doped SrTiO3 and hole-doped Ca3Co4O9.

## arXiv preprint: Exciton condensation in strongly correlated electron bilayers

Title: Exciton condensation in strongly correlated electron bilayers

Abstract:We studied the possibility of exciton condensation in Mott insulating bilayers. In these strongly correlated systems an exciton is the bound state of a double occupied and empty site. In the strong coupling limit the exciton acts as a hard-core boson. Its physics are captured by the exciton t-J model, containing an effective XXZ model describing the exciton dynamics only. Using numerical simulations and analytical mean field theory we constructed the ground state phase diagram. Three homogeneous phases can be distinguished: the antiferromagnet, the exciton checkerboard crystal and the exciton superfluid. For most model parameters, however, we predict macroscopic phase separation between these phases. The exciton superfluid exists only for large exciton hopping energy. Additionally we studied the collective modes and susceptibilities of the three phases. In the superfluid phase we find the striking feature that the bandwidth of the spin-triplet excitations, potentially detectable by resonant inelastic x-ray scattering (RIXS), is proportional to the superfluid density. The superfluid phase mode is visible in the charge susceptibility, measurable by RIXS or electron energy loss spectroscopy (EELS).

## arXiv preprint: Determinant quantum Monte Carlo study of exciton condensation in the bilayer Hubbard model

Title: Determinant quantum Monte Carlo study of exciton condensation in the bilayer Hubbard model

Abstract: We studied the possibility of exciton condensation in a strongly correlated bilayer extended Hubbard model using Determinant Quantum Monte Carlo. To model both the onsite repulsion U and the interlayer interaction V we introduced a novel update scheme extending the standard Sherman-Morrison update. We observe that the sign problem increases dramatically with the inclusion of the interlayer interaction V, which prohibits at this stage a unequivocal conclusion regarding the presence of exciton condensation. However, enhancement of the interlayer tunneling results suggest that the strongest exciton condensation tendency lies around 10-20 % p/n-doping. Magnetic properties and conductivity turn out to be relatively independent of the interlayer interaction.

## arXiv preprint: Influence of long-range interactions on charge ordering phenomena on a square lattice

Title: Influence of long-range interactions on charge ordering phenomena on a square lattice

Abstract: Usually complex charge ordering phenomena arise due to competing interactions. We have studied how such ordered patterns emerge from the frustration of a long-ranged interaction on a lattice. Using the lattice gas model on a square lattice with fixed particle density, we have identified several interesting phases; such as a generalization of Wigner crystals at low particle densities and stripe phases at densities in between rho = 1/3 and rho = 1/2. These stripes act as domain walls in the checkerboard phase present at half-filling. The phases are characterised at zero temperatures using numerical simulations, and mean field theory is used to construct a finite temperature phase diagram.

## arXiv preprint: Enhancement of spin propagation due to interlayer exciton condensation

Title: Enhancement of spin propagation due to interlayer exciton condensation

Abstract: We show that an interlayer exciton condensate doped into a strongly correlated Mott insulator exhibits a remarkable enhancement of the bandwidth of the magnetic excitations (triplons). This triplon is visible in the dynamical magnetic susceptibility and can be measured using resonant inelastic x-ray scattering. The bandwidth of the triplon scales with the exciton superfluid density, but only in the limit of strong correlations. As such the triplon bandwidth acts as a probe of exciton-spin interactions in the condensate.

## arXiv preprint: Dynamics of a single exciton in strongly correlated bilayers

Title: Dynamics of a single exciton in strongly correlated bilayers

Abstract: We formulated an effective theory for a single interlayer exciton in a bilayer quantum antiferromagnet, in the limit that the holon and doublon are strongly bound onto one interlayer rung by the Coulomb force. Upon using a rung linear spin wave approximation of the bilayer Heisenberg model, we calculated the spectral function of the exciton for a wide range of the interlayer Heisenberg coupling \alpha=J_{\perp}/Jz. In the disordered phase at large \alpha, a coherent quasiparticle peak appears representing free motion of the exciton in a spin singlet background. In the N\'{e}el phase, which applies to more realistic model parameters, a ladder spectrum arises due to Ising confinement of the exciton. The exciton spectrum is visible in measurements of the dielectric function, such as c-axis optical conductivity measurements.

## arXiv preprint: The dynamical frustration of interlayer excitons delocalizing in bilayer quantum antiferromagnets

Last week I put my work on exciton dynamics in strongly correlated bilayers onto the arXiv!

Title: The dynamical frustration of interlayer excitons delocalizing in bilayer quantum antiferromagnets

Abstract: Using the self-consistent Born approximation we study the delocalization of interlayer excitons in the bilayer Heisenberg quantum antiferromagnet (see figure above). Under realistic conditions we find that the coupling between the exciton motion and the spin system is strongly enhanced as compared to the case of a single carrier, to a degree that it mimics the confinement physics of carriers in Ising spin systems. We predict that the ‘ladder spectrum’ associated with this confinement physics should be visible in the c-axis exciton spectra of insulating bilayer cuprates such as YBa2Cu3O6. Our discovery indicates that finite density systems of such excitons should show very rich physical behavior.

## arXiv preprint: Prediction of the quantization of magnetic flux in double layer exciton superfluids

I just posted a preprint paper on the arXiv, with the following abstract.

Currently a way is lacking to detect unambiguously the possible phase coherence of an exciton condensate in an electron-hole double layer. Here we show that despite the fact that excitons are charge-neutral, the double layer exciton superfluid exhibits a diamagnetic response. In devices with specific circular geometry the magnetic flux threading between the layers must be quantized in units of $latex \frac{h}{e} \chi_m$ where $latex \chi_m$ is the diamagnetic susceptibility of the device. We discuss possible experimental realizations of the predicted unconventional flux quantization.

See the whole article at arXiv:1009.1793.


text:
Hecke operators acting as correspondences?

This question is inspired by Relation between Hecke Operator and Hecke Algebra

I remember having heard of yet another way of looking at Hecke operators acting on the spaces of modular forms for classical congruence subgroups of $SL_2(\mathbf{Z})$, and I would like to ask if anyone knows a reference for that. Here are some details and a related question.

There is a universal elliptic curve $U$ over $\mathbf{H}$, the upper half-plane. It is an analytic manifold obtained by taking the quotient of $\mathbf{C}\times\mathbf{H}$ by the action of $\mathbf{Z}^2$ given by $(n,m)\cdot (z,\tau)=(z+n+m\tau,\tau)$ where $n,m\in\mathbf{Z},z\in \mathbf{C}$ and $\tau\in\mathbf{H}$.

A torsion free finite index subgroup $\Gamma$ of $SL_2(\mathbf{Z})$ (congruence or not) acts on $U$. Here is the first question: for which $\Gamma$ is the quotient $U/\Gamma$ algebraic?

In any case $U/\Gamma$ is algebraic if $\Gamma=\Gamma(N)=\ker (SL_2(\mathbf{Z})\to SL_2(\mathbf{Z}/N))$ and $N\geq 3$. Denote the corresponding quotient $U/\Gamma(N)$ by $U(N)$. This is the universal elliptic curve with a level $N$ structure. (The notation $U(N)$ may be non-standard, in which case please let me know.)

There is a natural map $U(N)\to Y(N)=\mathbf{H}/\Gamma(N)$. Now take the $n$-th fibered cartesian power $U^n(N)$ of $U(N)$ over $Y(N)$ and let $p:U^n(N)\to Y(N)$ be the projection.

All derived direct images of the constant sheaf under $p$ decompose as direct sums of the Hodge local systems, which correspond to the symmetric powers of the standard representation of $\Gamma(N)$. Using the Eichler-Shimura isomorphism one can construct classes in $H^1(Y(N),V_k)$ from modular forms of weight $k+2$ for $\Gamma(N)$ where $V_k$ is the local system on $Y(N)$ that comes the $k$-th symmetric power of the standard representation. (In fact these classes can be interpreted in terms of the Hodge theory, see Eichler-Shimura isomorphism and mixed Hodge theory) So modular forms give elements in the $E_2$ sheet of the Leray spectral sequence for $p$.

I would like to ask: can one interpret the Hecke operators acting on modular forms as correspondences acting on $U^n(N)$ over $Y(N)$ (and hence, on the Leray spectral sequence for $p$)?

-

From these two maps, one gets two different families of elliptic curves over Y(N,p). Explicitly, the pullback of Un(N) along q1 is isomorphic to the n:th fibered power of the universal elliptic curve over Y(N,p); let us denote it Un(N,p). Let Q(N,p) be the quotient of U(N,p) by the cyclic subgroup C, and take its n:th fibered power as well. Then similarly Qn(N,p) is the pullback of Un(N) along q2. Finally, the quotient map gives us $\phi : U^n(N,p) \to Q^n(N,p)$.
But this data gives us the right correspondence on Un(N) over Y(N), namely, one takes the composite $q_{1\ast} \phi^\ast q_2^\ast$ (where I use q1 and q2 also for the induced maps on fibered powers of universal curves).


text:
# Histogram fill with logarithmic scale

I face a "filling" problem when I run one of the pgfplots example but with y-log scale. Here is the MWE

\documentclass{standalone}
\usepackage{pgfplots}

\begin{document}
\begin{tikzpicture}
\begin{semilogyaxis}[ymin=0,ymax=1,enlargelimits=false]
[const plot,fill=blue,draw=black]
coordinates
{(0,0.1)    (0.1,0.15)  (0.2,0.5)   (0.3,0.62)
(0.4,0.56) (0.5,0.58)  (0.6,0.65)  (0.7,0.6)
(0.8,0.58) (0.9,0.55)  (1,0.52)}
\closedcycle;
\end{semilogyaxis}
\end{tikzpicture}
\end{document}

Here is the output result

I also try to change the minimal value of y since logarithm of 0.0 is undefined but it does not help.

-
Logarithmic bar plots are a Bad Idea: The zero point on a logarithmic plot is arbitrary, which defeats the whole idea of representing quantities using lengths. There's a nice discussion of this on bzintelguru.com/blog/…. A simple line plot is often more appropriate. –  Jake Nov 7 '12 at 17:55
@Jake I saw your answer in another topic. I get the point raised in the link but it really depends on what you want to see and to emphasize. In high energy particle physics, it is quite usual to show both the linear and logarithmic scale especially for low probability event. –  Xavier Garrido Nov 7 '12 at 23:37
Ah, that's interesting! Would you happen to have a reference to a paper at hand where a logarithmic bar chart is used, by any chance? –  Jake Nov 8 '12 at 7:36
arxiv.org/abs/1104.3716 but maybe this is what you call a simple line plot –  Xavier Garrido Nov 8 '12 at 8:46

You can set the origin for logarithmic plots to always be at the bottom edge of the plot, by using log origin y=infty. This will affect the path drawn by \closedcycle, and define where plot styles like ybar and ycomb start. By default, the origin is set to log origin y=0.

\documentclass{standalone}
\usepackage{pgfplots}

\begin{document}
\begin{tikzpicture}
\begin{semilogyaxis}[
ymin=0.05,
ymax=1,
log origin=infty,
enlargelimits=false
]


text:
There have been several different articles that advocate for the use of denotational semantics, where we give a function from the syntax to another language that we already understand. In Coq, the target language is often Gallina. However, because all Gallina programs terminate, we can’t give a direct interpretation of non-terminating programs. An alternative is to give a denotation of your source language through a co-inductive definition of traces. The problem with this is that co-recursion is notoriously non-compositional in Coq. The paco work provides a compositional proof rule for co-inductive types, but the is no similar definition for co-inductive definitions. In this post I’m going to explain how to build a general recursion combinator in Coq that is completely composable- a surprisingly difficult problem that I pondered for a while in grad school.

The implementation of these ideas can be found in the coq-interaction-trees library. EDIT: Since the original posting, I have started collaborating on these ideas in the InteractionTrees repository repository.

## The Co-inductive Definition

One way to represent non-terminating computations is as a co-inductive data type. Adam Chlipala addresses this in his book, but the solution there isn’t satisfying because it can’t be written in a general way. Adam gives the following example (notations removed), trying to define the Fibonacci function co-inductively1.

CoInductive thunk (A : Type) : Type :=
| Answer : A -> thunk A
| Think : thunk A -> thunk A.

CoFixpoint fib (n : nat) : thunk nat :=
match n with
| _ =>
TBind (fib (pred n)) (fun n1 =>
TBind (fib (pred (pred n))) (fun n2 =>
end.

Gallina rejects this definition by saying that the recursive call to fib is not “guarded”. Intuitively, this means that Gallina can not guarantee that there is a finite amount of computation between two adjacent constructors. In the definition of fib above, it turns out that there is, but the syntactic check that Coq does is too conservative to prove this.

## Enter Extensible Effects

To get around the guardedness check we are going to generalize the thunk type above into a free monad. Here’s the new definition.

Section Eff.
Variable eff : Type -> Type.

CoInductive Eff (T : Type) : Type :=
| Pure : T -> Eff T
| Delay : Eff T -> Eff T
| Interact : forall {U}, eff U -> (U -> Eff T) -> Eff T.
End Eff.

Here, the entire Eff definition is parameterized over an algebra of effects (eff. The first two constructors are exactly the same as Answer and Think, and the Interact constructor allows us to appeal to eff.

On top of this definition, it isn’t too difficult to define monadic bind by co-induction.

Section bind.
Context {A B} (k : A -> Eff B).
CoFixpoint bind' (c : Eff a) : Eff B :=
match c with
| Pure v => k v
| Delay d => Delay (bind' d)
| Interact e k' => Interact e (fun x => bind' (k' x))
end.
End bind.

Definition bind {A B} c k := @bind' A B k c.

It is important, but subtle, that we lift the continuation to not be a parameter to the co-fixpoint due to Coq’s rule for reducing co-fixpoints2. In essence, without this lifting, Coq’s guardedness check (like its well-founded check) can not infer that the k argument is constant across co-recursive calls (similarly for the well-founded check on fixpoints).

### The Problem with mfix

Using this definition (even the previous definition of thunk) it is trivial to implement “diverging” computations, i.e. ones that never produce a Pure (or Answer). So, one might assume that we could implement a fix-point combinator that wraps up this recursion appropriately. In Gallina, this function would have the following type3.

CoFixpoint mfix_attempt (f : (a -> Eff eff b) -> a -> Eff eff b)
: a -> Eff eff b.

Naively, one would expect this function to have a very simple implementation:

CoFixpoint mfix_attempt (f : (a -> Eff eff b) -> a -> Eff eff b)
: a -> Eff eff b :=
f (mfix f).

However, Coq rightfully rejects this definition because it isn’t guaranteed to be productive. To give a simple example for the problem, suppose Coq accepted this definition and we called mfix with the identity function.

mfix_attempt id = id (mfix_attempt id) = id (id (mfix_attempt id)) = ...

As Coq tries to reduce this term it never finds a constructor so, by our intuition above, the definition is not guarded. One might (accurately) say that we shouldn’t call mfix with id, but instead with Delay. Doing this would cause us to have an infinite tower of Delays, which is exactly what we want. However, it isn’t sufficient to show that the function is guarded for some calls, it has to be guarded for any f.

One might think that we can address this problem by inserting the Delay into the definition itself.

CoFixpoint mfix_attempt2 (f : (a -> Eff eff b) -> a -> Eff eff b)
: a -> Eff eff b :=
Delay (f (mfix f)).

Now, if we instantiate f with id we get the nice (guarded) tower of Delays that we expect. However, Coq still rightfully rejects this definition because not all uses of mfix are guarded. Here’s a simple definition that breaks guardedness.

Definition a_little_faster (rec : a -> Eff eff b) (x : a) : Eff eff b :=
match rec x with
| Pure y => Pure y
| Interact e k => Interact e k
| Delay y => y
end.

Just to demonstrate the problem, let’s look at the unrolling.

  mfix_attempt2 a_little_faster x
= Delay (a_little_faster (mfix_attempt2 a_little_faster) x)
= Delay (match mfix_attempt2 a_little_faster x with
| ...
| Delay y => y
end)
= Delay (match Delay (a_little_faster (mfix_attempt2 a_little_faster) x) with
| ...
| Delay y => y
end)
= Delay (a_little_faster (mfix_attempt2 a_little_faster) x) (* cycle *)

We do get the head Delay like we expect, but after that a_little_faster takes a function that appears guarded and makes it unguarded by dropping the head constructor that is actually doing the guarding.

### An Attempt with Parametricity

The problem with a_little_faster is that it inspects the rec function rather than simply calling it. The above definition of mfix would be great if we could require that f uses the recursive call only by calling it and not pattern matching on the result. If Coq had parametricity built in then we might be able to achieve this by generalizing over the use of Eff eff in the body of the fixpoint. Something like:

CoFixpoint mfix_param
(f : forall eff' (inj : forall a, eff a -> Eff eff' a),
(a -> Eff eff' b) -> (a -> Eff eff' b))
: a -> Eff eff b :=
fun x => @f eff (fun _ e => Interact e ret) (mfix f).

Here the f becomes a little bit more complicated because we parameterize over m but we also want a way to do all the effects that we used to be able to do, i.e. those allowed by Eff eff, which is what the inj argument does. While we can actually implement this function, the naive definition above does not work. Coq again rejects it due to guardedness, though one could argue that, assuming parametricity, this definition is guarded, it would be difficult to capture the property algorithmically. For example, if f also takes a function forall a, m a -> Eff eff a, then parametricity no longer guarantees the guardedness condition.

## Enforcing Parametricity with Effects

Since Coq doesn’t have parametricity built in, we need to enforce it ourselves. Essentially, we need to build a sub-language of Gallina that represents functions that use their argument in a limited way (essentially it only binds their result). This may seem like a tall order, but in actuality it isn’t as difficult as one might think; essentially the entire problem lies in the use of the “middle” arrow in the type of f, i.e.

CoFixpoint mfix_attempt
(f : (a -> Eff eff b) -> (* <-- this arrow! *)
a -> Eff eff b)
: a -> Eff eff b.

This arrow says that we can use all of Gallina to implement the body of the recursive function. If we want to restrict the things that the body can do, we simply need to replace this arrow with a more restricted representation of functions. For our purposes, one easy thing to do is to represent the idea of making a recursive call as an effect. Here, we could write the following4:

Inductive fixpoint a b : Type -> Type :=
| call (_ : a) : fixpoint a b b.

CoFixpoint mfix
(f : a -> Eff (bothE eff (fixpoint a b)) b)
: a -> Eff eff b.

In this phrasing of the “function,” we offer access to the argument (which is the recursive call) only through the effect layer of Eff (namely the fixpoint/call effect). Thus, if the f function can not inspect the body of the recursive call, it only sees the constructor.

The implementation of mfix is surprisingly simple. Essentially, all we need to do is to interpret the added fixpoint effect using the usual interpretation mechanism (which I am inling here).

Section mfix.
Variable f : forall x : a, Eff (bothE eff fixpoint) (b x).

CoFixpoint finterp {T : Type}
(c : Eff (bothE eff fixpoint) T)
: Eff eff T :=
match c with
| Pure x => Pure x
| Interact (Left e) k =>
Interact e (fun x => finterp (k x))
| Interact (Right e) k =>
match e in fixpoint X return (X -> _) -> _ with
| call x => fun k =>
(* interpret the call by running f on x *)
Delay (finterp (bind (f x) k))
end k
| Delay x => Delay (finterp x)
end.

Definition mfix (x : a) : Eff eff b :=
finterp (f x).
End mfix.

Note that we are interpreting f x by replacing all occurrences of Right (call x) with finterp (f x). In the code, since Interact carries its continuation (k which also needs to be interpreted), this is written finterp (bind (f x) k). The dependent pattern match on e is simply necessary to justify the fact that the type of k accepts the value returned by the recursive call.

We can actually use this implementation to implement mfix_param.

CoFixpoint mfix_param
(f : forall eff' (inj : forall a, eff a -> Eff eff' a),
(a -> Eff eff' b) -> (a -> Eff eff' b))
: a -> Eff eff b :=
fun x =>
mfix (f (bothE (Eff eff) (fixpoint a b))
(fun _ e => bind (Left e) ret)
(fun x => Interact (Right (Call x))).

Interaction trees provide the underlying semantics for the new version of Vellvm, and seem to be a topic of interest in DeepSpec, especially the work being done at UPenn.

Alternative styles of defining semantics can be used to get around this problem. Classic small-step operational semantics are the swiss-army-knife that solve just about every semantic problem though purely syntactic methods. Obviously there are huge benefits to this approach in terms of generality, but re-usability seems to be much harder since each semantics is specialized to its own syntax. Arthur Charguéraud’s Pretty-Big-Step Semantics provides a way to encode the semantics of non-terminating programs using a relational representation.

The standard trick that has been proposed for solving this problem functionally is to use fuel. However, fuel still only allows us to talk about terminating programs- we need an alternative definition to describe the semantics of non-terminating programs. Similarly, in the presence of input, fuel breaks down. Consider, for example, the program that reads an input integer and loops that many times before returning. There is no finite amount of fuel that you know will be sufficient to run the program to completion before you read the input. Thus, to use fuel you need to give a “fuel until next event” as a function from the variable context which requires a resumption monad which seems more complex, though I haven’t looked deeply into it.

## Conclusions

I’ve never seen this encoding of fixpoints through effects before, though I’d be interested if it exists somewhere. Any pointers in the comments would be greatly appreciated.

Moving forward, I’m interested to understand if I can define a generic logic, such as the step-indexed work of the Iris project on top of this semantic definition. If so, it might provide a common ground for leveraging Iris (and the sophisticated techniques that it has built) to reason about new programming languages.

#### Footnotes

1. Note that the Fibonacci function should be defined inductively, and Coq will gladly accept the basic definition.

2. This gist gives a concrete example where the version using a Section (i.e. a regular lambda abstraction) succeeds and the version that simply makes them an argument to the cofix fails.

3. Unlike Haskell’s mfix from MonadFix, classical recursion combinators expose the functional form. This mimics the fact that Haskell values actually include non-termination, as opposed to the Haskell function space (which is total).

4. It is easy to generalize this to dependent functions, where the type of b is a -> Type


text:
# 4.4 Poisson distribution -- rrc math 1020

 Page 1 / 18

There are two main characteristics of a Poisson experiment.

1. The Poisson probability distribution gives the probability of a number of events occurring in a fixed interval of time or space if these events happen with a known average rate and independently of the time since the last event. For example, a book editor might be interested in the number of words spelled incorrectly in a particular book. It might be that, on the average, there are five words spelled incorrectly in 100 pages. The interval is the 100 pages.
2. The Poisson distribution may be used to approximate the binomial if the probability of success is "small" (such as 0.01) and the number of trials is "large" (such as 1,000). You will verify the relationship in the homework exercises. n is the number of trials, and p is the probability of a "success."

The random variable X = the number of occurrences in the interval of interest.

The average number of loaves of bread put on a shelf in a bakery in a half-hour period is 12. Of interest is the number of loaves of bread put on the shelf in five minutes. The time interval of interest is five minutes. What is the probability that the number of loaves, selected randomly, put on the shelf in five minutes is three?

Let X = the number of loaves of bread put on the shelf in five minutes. If the average number of loaves put on the shelf in 30 minutes (half-hour) is 12, then the average number of loaves put on the shelf in five minutes is $\left(\frac{5}{30}\right)$ (12) = 2 loaves of bread.

The probability question asks you to find P ( x = 3).

## Try it

The average number of fish caught in an hour is eight. Of interest is the number of fish caught in 15 minutes. The time interval of interest is 15 minutes. What is the average number of fish caught in 15 minutes?

$\left(\frac{15}{60}\right)$ (8) = 2 fish

A bank expects to receive six bad checks per day, on average. What is the probability of the bank getting fewer than five bad checks on any given day? Of interest is the number of checks the bank receives in one day, so the time interval of interest is one day. Let X = the number of bad checks the bank receives in one day. If the bank expects to receive six bad checks per day then the average is six checks per day. Write a mathematical statement for the probability question.

P ( x <5)

## Try it

An electronics store expects to have ten returns per day on average. The manager wants to know the probability of the store getting fewer than eight returns on any given day. State the probability question mathematically.

P ( x <8)

You notice that a news reporter says "uh," on average, two times per broadcast. What is the probability that the news reporter says "uh" more than two times per broadcast.

This is a Poisson problem because you are interested in knowing the number of times the news reporter says "uh" during a broadcast.

a. What is the interval of interest?

b. What is the average number of times the news reporter says "uh" during one broadcast?

b. 2

c. Let X = ____________. What values does X take on?

c. Let X = the number of times the news reporter says "uh" during one broadcast.
x = 0, 1, 2, 3, ...

d. The probability question is P (______).

d. P ( x >2)

what is the stm
is there industrial application of fullrenes. What is the method to prepare fullrene on large scale.?
Rafiq
industrial application...? mmm I think on the medical side as drug carrier, but you should go deeper on your research, I may be wrong
Damian
How we are making nano material?
what is a peer
What is meant by 'nano scale'?
What is STMs full form?
LITNING
scanning tunneling microscope
Sahil
how nano science is used for hydrophobicity
Santosh
Do u think that Graphene and Fullrene fiber can be used to make Air Plane body structure the lightest and strongest. Rafiq
Rafiq
what is differents between GO and RGO?
Mahi
what is simplest way to understand the applications of nano robots used to detect the cancer affected cell of human body.? How this robot is carried to required site of body cell.? what will be the carrier material and how can be detected that correct delivery of drug is done Rafiq
Rafiq
what is Nano technology ?
write examples of Nano molecule?
Bob
The nanotechnology is as new science, to scale nanometric
brayan
nanotechnology is the study, desing, synthesis, manipulation and application of materials and functional systems through control of matter at nanoscale
Damian
Is there any normative that regulates the use of silver nanoparticles?
what king of growth are you checking .?
Renato
What fields keep nano created devices from performing or assimulating ? Magnetic fields ? Are do they assimilate ?
why we need to study biomolecules, molecular biology in nanotechnology?
?
Kyle
yes I'm doing my masters in nanotechnology, we are being studying all these domains as well..
why?
what school?
Kyle
biomolecules are e building blocks of every organics and inorganic materials.
Joe
anyone know any internet site where one can find nanotechnology papers?
research.net
kanaga
sciencedirect big data base
Ernesto
Introduction about quantum dots in nanotechnology
what does nano mean?
nano basically means 10^(-9). nanometer is a unit to measure length.
Bharti
do you think it's worthwhile in the long term to study the effects and possibilities of nanotechnology on viral treatment?
absolutely yes
Daniel
how to know photocatalytic properties of tio2 nanoparticles...what to do now
it is a goid question and i want to know the answer as well
Maciej
Abigail
for teaching engĺish at school how nano technology help us
Anassong
How can I make nanorobot?
Lily
Do somebody tell me a best nano engineering book for beginners?
there is no specific books for beginners but there is book called principle of nanotechnology
NANO
how can I make nanorobot?
Lily
what is fullerene does it is used to make bukky balls
are you nano engineer ?
s.
fullerene is a bucky ball aka Carbon 60 molecule. It was name by the architect Fuller. He design the geodesic dome. it resembles a soccer ball.
Tarell
what is the actual application of fullerenes nowadays?
Damian
That is a great question Damian. best way to answer that question is to Google it. there are hundreds of applications for buck minister fullerenes, from medical to aerospace. you can also find plenty of research papers that will give you great detail on the potential applications of fullerenes.
Tarell
how did you get the value of 2000N.What calculations are needed to arrive at it
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text:
# 198. House Robber

## The Problem

Link to original problem on Leetcode.

You are a professional robber planning to rob houses along a street. Each house has a certain amount of money stashed, the only constraint stopping you from robbing each of them is that adjacent houses have security systems connected and it will automatically contact the police if two adjacent houses were broken into on the same night.

Given an integer array nums representing the amount of money of each house, return the maximum amount of money you can rob tonight without alerting the police.

Examples

Example 1:

Input: nums = [1,2,3,1]
Output: 4
Explanation: Rob house 1 (money = 1) and then rob house 3 (money = 3).
Total amount you can rob = 1 + 3 = 4.

Example 2:

Input: nums = [2,7,9,3,1]
Output: 12
Explanation: Rob house 1 (money = 2), rob house 3 (money = 9) and rob house 5 (money = 1).
Total amount you can rob = 2 + 9 + 1 = 12.

Constraints
• 1 <= nums.length <= 100
• 0 <= nums[i] <= 400

## My Solution

A leetcode problem with a story! This is quite rare, and a real treat!1 This is a dynamic programming problem, and as such we can solve it either recursively or iteratively.

### Recursive Solution

To solve this problem recursively, we should think about what sub-problem we're trying to solve. Whenever I'm at a house at index, I have to decide: am I better off robbing this house, or waiting until I get to the next house? If I rob this house, I get the value of its possessions nums[index], plus the value of all the houses I robbed before if the last house I robbed was the house from two before it at nums[index - 2]. If I skip this house, I get the value of all the houses I robbed before this house ending at the house immediately before it at nums[index - 1]. If index starts at the last value of nums, then we can write a recursive function to figure out the sum of the best sequence of houses by working back from index. I can work backward until I'm out of houses to compare—that is to say, the index is less than 0. That's our base case.

// This function is modified to accept a second argument// from how it's scaffolded in leetcode: index. We set a// default index value of nums.length - 1 to signify we// are starting at the last value of nums.function rob(nums: number[], index = nums.length - 1): number {	if (index < 0) {		// Indices less than zero are the base case; we are		// out of houses to rob! Non-existant houses have		// a value of zero.		return 0;	} else {		// If we're not yet out of houses to rob, we return the		// maximum of this house plus the house two before, or		// the house one before. We recursively call the function		// to get the sum of other house's values.		return Math.max(			rob(nums, index - 2) + nums[index],			rob(nums, index - 1)		);	}};

This function is correct, but it's not performant. Because we call the function again twice for each comparison, our time complexity is $O(2^{n})$! If you try running it in leetcode, it will time out. We need to improve it with memoization.

### Recursive Solution + Memoization

To improve on the previous solution, we add memoization to remember values we've previously computed. This dramatically speeds up our solution and gives us a time complexity of $O(n)$. Here, we pass around a Map with the index as key and the result of our comparisons as the value.

// We've once again modified the function. Now it also takes// a JavaScript Map to use for our memoization, which we// initialize as a new empty Map.function rob(nums: number[], index = nums.length - 1, memo = new Map<number, number>()): number {	if (index < 0) {		return 0;	} else if (memo.has(index)) {		// Instead of always computing non-base cases, we'll first		// check to see if they exist in our memo Map and return		// that if we find it.		return memo.get(index);	} else {		const result = Math.max(			rob(nums, index - 2, memo) + nums[index],			rob(nums, index - 1, memo)		);		// Instead of returning the result of comparisons right		// away, we first add it to the memo Map so we can find		// it later if we need it.		memo.set(index, result);		return result;	}};

### Iterative Solution

Our recursive solution was top-down. That is to say, we started at the farthest house and worked backward. For our iterative solution, we'll instead go bottom-up. We'll calculate answers to sub-problems near the beginning in order to answer more sub-problems as we go along. The same sub-problem logic still holds: for each house we're at, we want to know if we're better off taking the sum of this house and all the houses previously robbed as of two house before, or just the sum of all house robbed previously as of one house before.

We could use an array to remember the optimal possible result for any given index in nums, but we don't have to. All we really need are two variables to remember the sums for one house back and two houses back. With each iteration across nums, we can just update those two variables. So not only do we get time complexity of $O(n)$, but we also get space complexity of $O(1)$!.

function rob(nums: number[]): number {	if (nums.length === 0) return 0;	let oneHouseBack = 0, twoHousesBack = 0, temp = oneHouseBack;	for (let thisHouse of nums) {		// We're about to reassign the value of oneHouseBack,		// so we want to store the current value somewhere.		// We put it in this temp variable, because we're going		// to update twoHousesBack to be the current value of		// oneHouseBack, and oneHouseBack to the max of our		// comparison.		temp = oneHouseBack;		oneHouseBack = Math.max(twoHousesBack + thisHouse, oneHouseBack)		twoHousesBack = temp;	}	return oneHouseBack};

1. If you like tough data structures and algorithms problems with a story, check out Advent of Code for more and better puzzles. ↩


text: Image text transcribed for accessibility: A CMOS amplifier is shown in Figure Ql. 1 where all (W/L) = 20 except (W/L) 4 = 10 Use mu C = 90 mu ?/V2, mu rho C = 30 mu ?/V2, V = - V = 0.5V. lambda = lambda rho = 0 and lambda = lambda rho = 0.04 V - 1. Gate of MS is biased with a DC voltage of 2 .0V Figure Q l 1 Determine the drain currents in devices Ml - M5 and the DC voltage at node A. Determine Gm. Rout and hence, the small-signal, low-frequency gain of the amplifier (vout/vin). Determine the output voltage swing for the amplifier.


text:
Mathematics
OpenStudy (anonymous):

integral question?!?!?hellpp

OpenStudy (anonymous):

if it is given that $\int\limits_{2}^{4} x ^{3}dx=60 and \int\limits_{2}^{4}xdx=6 and \int\limits_{2}^{4}dx=2$

OpenStudy (anonymous):

if u had 2 find....a.$\int\limits_{2}^{2}x ^{3}dx? and \int\limits_{2}^{4}15dx? and \int\limits_{2}^{4}(x ^{4}+4)?$

OpenStudy (anonymous):

the second one is 15*2=30

OpenStudy (anonymous):

I'm not sure I understand the question.

OpenStudy (anonymous):

i think u got the 2nd one right...its jus the top n bottom parts tht change tht confuse me..

OpenStudy (anonymous):

basically..use only the given info,,n sum rules to find it

OpenStudy (anonymous):

the third one you have to separate the integrals into x^4 and 4.... then you do 60*6 + 2*4 = 368

OpenStudy (anonymous):

how do u put numbers on brackets after u integrate?

OpenStudy (anonymous):

oh! first one equals 0 because there's a rule that says if the upper and lower bounds are the negative opposites of each other, the integral equals 0.

OpenStudy (anonymous):

ohhok..i think u got all of them..thank u soo much!!

OpenStudy (anonymous):

:)

OpenStudy (anonymous):

first one is a zero, next one is 30

OpenStudy (anonymous):

last one: 1032/5

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text:
# Expectation Value for system of identical particles

1. May 11, 2012

### kenphysics

Dear All:

I have a quite mysterious and cumbersome question concerning with the expectation values for a system of identical particles. For example, suppose I have a system of N identical bosons given by the wavefunction ψ(x1,x2,...xN), which is of course symmetrized. My concern is:

1. What is <x1> really means? I know mathematically it is given by:

∫dx1dx2...dxN x1 ψ*(x1,x2,...xN)ψ(x1,x2,...xN)

but I think this is unphysical? We cannot measure this value, right? since physical observable for identical particles should be symmetric, like x1+x2+x3, right? But can I say <x1>=<x2>=<x3> = 1/3<x1+x2+x3> due to symmetric argument?

2. Suppose I really want to compute the above quantity, how to write it in second quantized form, that is using field operators? Is it given by:

<x1> = ∫ x1 <ψdagger(x1)ψ(x1)> dx1 ? where ψ is now a field annihilation operator that annihilate a particle at the point x1. If yes, can someone give me some ideas on how to show it starting from the wavefunction description?

3. Continue the same idea, then what is <(x2-x1)^2> means? Also, how to write it in second quantized form?

Thank you for all

2. May 11, 2012

### Jano L.

Hello kenphysics,

welcome to PF.

Why do you think so? If we used this wave function to describe the ensemble of helium atoms, the expression you wrote for <x1> would most probably serve to calculate the expected average of the position of any helium atom. Due to symmetric property of the wave function, we would have <x1> = <x2> = ... and so on.

3. May 12, 2012

### kenphysics

OK! In fact, what trouble me a lot is that for second quantization, we always consider operators like:

$\sum_{i=1}^N f(x_i)$ (Sinlge-body operator)
$\sum_{i<j}^N V(x_i,x_j)$ (Two-body operator)

But we seldom consider operators, liked x1, x2 or (x2-x1)^2, (x3-x2)^2. So what I actually want to know is can I still write down <x1> and <(x2-x1)^2> in second quantized form?


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# How do you find the product (x^2+5x-1)(5x^2-6x+1)?

Feb 12, 2017

To multiply these two terms you multiply each individual term in the left parenthesis by each individual term in the right parenthesis.

$\left(\textcolor{red}{{x}^{2}} + \textcolor{red}{5 x} - \textcolor{red}{1}\right) \left(\textcolor{b l u e}{5 {x}^{2}} - \textcolor{b l u e}{6 x} + \textcolor{b l u e}{1}\right)$ becomes:

$\left(\textcolor{red}{{x}^{2}} \times \textcolor{b l u e}{5 {x}^{2}}\right) - \left(\textcolor{red}{{x}^{2}} \times \textcolor{b l u e}{6 x}\right) + \left(\textcolor{red}{{x}^{2}} \times \textcolor{b l u e}{1}\right) + \left(\textcolor{red}{5 x} \times \textcolor{b l u e}{5 {x}^{2}}\right) - \left(\textcolor{red}{5 x} \times \textcolor{b l u e}{6 x}\right) + \left(\textcolor{red}{5 x} \times \textcolor{b l u e}{1}\right) - \left(\textcolor{red}{1} \times \textcolor{b l u e}{5 {x}^{2}}\right) + \left(\textcolor{red}{1} \times \textcolor{b l u e}{6 x}\right) - \left(\textcolor{red}{1} \times \textcolor{b l u e}{1}\right)$

$5 {x}^{4} - 6 {x}^{3} + {x}^{2} + 25 {x}^{3} - 30 {x}^{2} + 5 x - 5 {x}^{2} + 6 x - 1$

We can now group and combine like terms:

$5 {x}^{4} - 6 {x}^{3} + 25 {x}^{3} + {x}^{2} - 30 {x}^{2} - 5 {x}^{2} + 5 x + 6 x - 1$

$5 {x}^{4} + \left(- 6 + 25\right) {x}^{3} + \left(1 - 30 - 5\right) {x}^{2} + \left(5 + 6\right) x - 1$

$5 {x}^{4} + 19 {x}^{3} - 34 {x}^{2} + 11 x - 1$