text:
# Daylight Calculator

Created by Davide Borchia
Reviewed by Steven Wooding
Last updated: Dec 09, 2022

Use our daylight calculator to find out how many hours of daylight you will have in the summer, particularly if you are tired of those short and dark days in the winter.

• How to calculate daylight hours for any location on Earth (almost!);
• How do the hours of daylight depend on the latitude;
• Some facts on the duration of the day across the Planet;
• and much more!

Jump onboard and discover the science of bright, long days with Omni!

## How do we count daylight? Calculate the length of daylight

Daylight is the portion of the day when the Sun shines on a defined location. Clouds don't count, and even if you were inside, the Sun was above you. According to this definition, daylight is the time that runs between sunrise and sunset. Counting daylight is a pretty straightforward operation: you just need to know the time of the local dawn and twilight and count the hours in between.

Did you know that the effective length of daylight is slightly more than the math suggests to us? The effect of atmospheric refraction, which bends the light rays coming from the Sun in the early and late phases of the day, gives us some minutes of extra Sun every day! We will learn the math behind this phenomenon in the next section.

## How do I calculate the hours of daylight from the latitude?

To calculate the hours of daylight from the latitude, we need to perform a certain number of length calculations to find the time of sunrise and sunset. To do so, we need to know the latitude ($\phi$) and longitude ($\lambda$) of a place on Earth. To calculate the daylight hours for a specific day of the year, we also need to know the day number $n$ ($1$ for the first of January, and so on).

A bit of rather boring math follows: we will write down the steps and comment on them, but they are relatively advanced. You may also find alternative formulas on the Internet: we choose the ones that resulted in the most accurate results, even though they may not be so straightforward in their derivation. Let's start!

🙋 All the following steps use angles in degrees. Don't forget your conversion: our angle converter is your ally!

Since latitude is expressed in degrees, but we instead prefer to measure time, we apply a weird conversion between angle and time:

$\footnotesize\lambda_\mathrm{h} = \frac{\lambda}{15}$

We then find an approximation of the local sunrise and sunset time. Mathematicians call this step "ansatz": an educated guess. In our case, we assume that the sun rises at 6 local time, and sets at 18. Reasonable, right?

$\footnotesize\begin{split} t_\mathrm{rise} &= n+\frac{6-\lambda_\mathrm{h}}{24}\\[1em] t_\mathrm{set} &= n+\frac{18-\lambda_\mathrm{h}}{24} \end{split}$

In the following steps, we will use a generic $t$ to mark the previously found quantities: for a while, it won't make a difference. We'll let you know when to remember about them!

We find the Sun's mean anomaly, a measure of how much the Earth's elliptic orbit overshoots the circular orbit with the same period (we taught you how to calculate this quantity at the orbital period calculator).

$\footnotesize M = (0.9856 \cdot t) - 3.289$

Next, we need to find the coordinates of the Sun in the sky: instead of latitude and longitude, we calculate the declination $\delta$ and right ascension $\mathrm{RA}$.

To find them, we need first to compute the Sun's true latitude:

$\footnotesize \begin{split} L &= M + (1.916 \cdot \sin(M))\\ & + (0.020\cdot \sin(2 \cdot M)) + 282.634 \end{split}$

Calculate $L\mod\ 360\degree$ to ensure that the angle falls in the correct range, $[0\degree,360\degree)$.

For the right ascension:

$\footnotesize \mathrm{RA} = \arctan(0.91764 \cdot \tan(L))$

To make sure that we aligned our coordinates correctly, we check that the true latitude and right ascension fall in the same quadrant of the sky by tuning the latter's value:

$\footnotesize \begin{split} L_\mathrm{quad} &= L\mod 90\degree\\ \mathrm{RA}_\mathrm{quad} &= \mathrm{RA}\mod 90\degree \end{split}$

And use these values to calculate the "updated" right ascension:

$\footnotesize \mathrm{RA} = \frac{RA+L_\mathrm{quad}-\mathrm{RA}_\mathrm{quad}}{15}$

We calculate the Sun's declination, the other coordinate of our Star, according to Earth's reference frame. We directly find the value of its trigonometric functions.

$\footnotesize \begin{split} \sin(\delta) &= 0.39782 \cdot \sin(L)\\[.5em] \cos(\delta) &= \cos(\arcsin(\sin(\delta)) \end{split}$

Then, we find the value of the Sun's local hour angle: this quantity describes the angular position of our star according to the local zenith (the point right above your head):

$\footnotesize\cos(h_\mathrm{a}) = \frac{\cos(z) - \sin(\delta)\cdot\sin(\phi)}{\cos(\delta)\cdot \cos(\phi)}$

Notice that the fraction is not necessarily in the "legal" bounds: we can meet two conditions, with the cosine of $h_\mathrm{a}$ never rising ($\cos(h_\mathrm{a})>1$) or never setting ($\cos(h_\mathrm{a})<-1$). When the value of the cosine is outside of the range, we have full days of sunlight or darkness.

The angle $z$ is the angle of the Sun relative to the zenith. For the "true" sunrise and sunset, it equals $90\degree$. We'll see later the values it can assume.

#### Interlude: how the atmosphere bends light and fights the night

If you want to know the effective length of daylight, you need to consider the refraction of the Sun's rays due to the atmosphere. At sunrise and sunset, the light from the Sun must cross a thicker layer of atmosphere to reach you. The atmosphere has a variable index of refraction, which depends on the density.

Thanks to Snell's law (we explained it thoroughly at the Snell's law calculator), this results in a change in the angle with which the rays cross the atmosphere. At sunrise and sunset, this deflection reaches up to $0.5\degree$.

Considering the Sun's diameter instead of its center, we can reach a correction of $\small 0.833\degree$ to our sunrise/sunset angle, marking effectively the first moment the Sun pokes above the horizon (or when it completely disappears below it). Plug this correction in the formula above by setting:

$z = 90.833\degree$

Now we can convert the hour angle into hours. Remember that we can calculate two values for the hour angle: one using $t_\mathrm{rise}$ for the sunrise, and the other ($t_\mathrm{set}$) for the sunset. Depending on what phenomenon we are considering, we find the hour angle using the inverse cosine in the following equations:

$\footnotesize \begin{split} h_\mathrm{a,\ rise} &= 360\degree - \arccos(\cos(h_\mathrm{a}))\\[.5em] h_\mathrm{a,\ set} &= \arccos(\cos(h_\mathrm{a})) \end{split}$

Convert this angle in hours using the conversion we saw a while ago:

$h_\mathrm{a}= \frac{h_\mathrm{a}}{15}$

We're all set: the UTC time for the phenomenon is, in both cases:

$\footnotesize T = h_\mathrm{a} + \mathrm{RA}- (0.06571 \cdot t) - 6.622,$

where $t$ can be either $t_\mathrm{rise}$ or $t_\mathrm{set}$.

The last passage of our calculations for the hour of sunset and sunlight involves returning the time in UTC to the local time:

$T_\mathrm{local} = T-\lambda_\mathrm{h}+\mathrm{tz}$

If this value is outside the range $(0,24]$, adjust it with the modulo operation. Last (we swear) step: subtract the hours of the sunset and sunrise. If possible, you'll find the length of the day (there may be extreme cases). Here is the final formula, and the only one that you really must know:

$\small t_\mathrm{daylight} = T_\mathrm{local,\ sunset} - T_\mathrm{local,\ sunrise}$

We need to make a couple of remarks:

• The calculated daylight hours depend only on the latitude; they are equal for all longitudes at a fixed latitude value.

• The formula for the daylight time breaks (a bit) for latitudes above the polar circles. In these situations, we need to use the brain to decide if we are witnessing a day of sunlight or a day of darkness.

#### How do the daylight hours change with the latitude?

The animation below shows how the daylight hours vary during the year for all latitudes. The horizontal sections at the extremities correspond to 24 hours of sunlight or darkness. You can see how the curve oscillates around the point $(0\degree,12\ \mathrm{h})$: at the equator, every day has the same length!

Let's fix a point on the horizontal axis. We can see how the daylight hours vary for a specific location: the following image represents the daylight hours across an entire year for the Australian city of Gold Coast.

Since Gold Coast lies in the southern hemisphere, the shortest duration of the day falls in June: in Australia, Christmas doesn't equal snow! Did you notice the colored bands above and below the yellow section representing the daylight hours? Those are the warm hues of sunrise and sunset; believe it or not, we can calculate them!

## More than calculating the daylight hours: twilights and dawns

We can add some complexity by considering the different (and visually distinct) phases of sunrise and sunset: the stages correspond to three angles below the horizon:

• Astronomical dawn/twilight$18\degree$ to $12\degree$ below the horizon;
• Nautical dawn/twilight$12\degree$ to $6\degree$ below the horizon; and
• Civil dawn/twilight$6\degree$ to the horizon.

The duration of the various phases depends on the latitude and the day of the year, but generally, it is longer for higher latitudes. The picture below represents the subdivision of a mid-March day in Gold Coast.

As you can see, the three phases of sunrise and sunset take place in maybe an hour or two. Move to the other side of the globe: we choose the Russian city of Murmansk. There, in mid-March, the daylight hours almost equal the one in Gold Coast, but look at civil, nautical, and astronomical sunrises and sunsets.

They last from the small hours of the night to dawn and from sunset to late in the evening: the night in its darkest form last a mere couple of hour! If you visit a place in the higher latitudes, you may witness almost endless sunrises and sunsets.

## Why does daylight change so much?

Calculating the daylight hours shows how much this quantity varies across latitudes: in this section, we will explain why and give you some examples.

First on the list, we meet a city at the middlemost of the middle latitudes: Rome (but also Chicago, New York, Tbilisi, Sapporo, and Barcelona: only the latitude matters!)

The amount of daylight we calculate depends on the latitude of the location only. This quantity tells us how much the position of the Sun changes in the sky during the year.

For a place close to the equator, the Sun's position oscillates mildly by only $\pm23.44\degree$ from the zenith. The variations in daylight hours are negligible, and you get a climate devoid of seasons dependent on the position of the Earth in its orbit.

On the other extreme, you get locations above the polar circles. At those latitudes, there are days of the year when the Sun never sets and, six months later, never rises. There, you can experience the midnight sun (or a midday night, though fewer people are interested in this one). In the picture below, you can see the amount of daylight calculated for the charming town of Tromsø in Norway.

This image clearly shows that in the summer, you need very dark curtains if you want to sleep there: for a couple of months, the Sun never dips entirely below the horizon. On the other hand... there is never a 24-hour long night: even during the winter solstice, you would still get some light during the four hours long civil sunset.

To see the most authentic example of a long night, we present you with the same diagram for the North Pole (to be honest, a location a degree south of it, at $89\degree$ north).

Now it's clear! If you remember how to read this type of diagram, you will quickly notice that at the poles, the Sun doesn't change height in the sky on the days. The boundary between daylight (yellow) and twilight (orange) happens two times at the equinoxes. On those days, the Sun hovers midway across the horizon, neither rising nor setting. For all the other days, the Sun is either over or below the horizon for 24 hours. Here, for a couple of months a year, you would not even know about the existence of the Sun apart from the magnificent northern (or southern) lights!

The key to these dramatic changes in the duration of the hours of sunlight lies in the angle of the Sun's path in the sky. In the next image, you can see how the Sun moves in the sky during the solstices and the equinoxes for three localities: the mid-latitudes, a location above the polar circle, and on the equator.

You can see so many things here! We will only point out some of them; feel free to explore the rest!

• The Sun is at its highest angle over the horizon on the equator. The opposite is true for polar latitudes.

• However, since the angle of the path of the Sun is more horizontal at the poles, the Sun, during the summer solstice, spends much more time over the horizon.

• Low angle means low energy: this is why there is ice at the poles even though the Sun is out for 24 hours (and also why you can't grow tomatoes in the Nordic countries, even though in the summer the Sun is out for ages!).

• In the polar regions, the Sun needs more time to cross each phase of the sunrise and sunset: this is why dawn and twilight are longer at higher latitudes!

## Calculate the day with Omni!

To calculate the hours of daylight, we had to find the times of sunrise and sunset: you can learn how to do it with Omni's tools:

## FAQ

### How do I calculate the length of the daylight hours?

To calculate the length of the daylight hours, you need to follow some simple steps:

1. Find the time of the local sunrise t(rise).

2. Find the time of the local sunset t(set).

3. Subtract the two values. It may be convenient to convert the time in hh:mm to decimal hours (for example, 18:30 → 18.5). The result is the length of the daylight:
t(daylight) = t(set) - t(rise)

4. Convert back to the desired format if necessary.

### Can a day have 24 hours of sunlight?

Yes. In the regions above the polar circles, there is at least a day when the Sun never sets. This happens starting from the local spring equinox at the pole.

At the poles, the six months between the spring and the autumn equinox never see a sunset, while on the polar circles, only a day during the summer solstices has the famous midnight sun.

The angle of the sunrays remains low, regardless of the time spent by the Sun over the horizon: this contributes to the cold temperatures at high latitudes.

### Why does the sunset is longer at higher latitudes?

The sunset is longer at the higher latitudes since the lower angle of the Sun's path relative to the horizon causes sunrays to cross the regions right below the horizon for a more extended period of time. The thicker atmosphere (due to perspective) is the reason for warmer, reddish-orangish hues. At latitudes high enough, you may witness day-long sunrises and sunsets!

### How many hours of daylight will I receive on an equinox?

During the equinox, you will receive slightly more than 12 hours of sunlight, regardless of your position on the Planet. On those days, the Sun changes hemispheres and everywhere on the Earth; it rises and sets exactly in the east and west and takes 12 hours to cross the sky.

Due to atmospheric refraction, however, you would receive a "late" image of the Sun, even when it is under the horizon. This would give you around 15 minutes of sunlight more for the mid-latitudes!

Davide Borchia
Day
Hemisphere (N/S)
N
Latitude (φ)
°
'
"
Hemisphere (E/W)
W
Longitude (λ)
°
'
"
Time zone
UTC+0 (GMT, WET)
Daylight savings?
No
Uncorrected
Type of dusk
Sunset
☀ Day length
hrs
min
With atmospheric refraction
☀ Day length
hrs
min
People also viewed…

### Black hole collision

The Black Hole Collision Calculator lets you see the effects of a black hole collision, as well as revealing some of the mysteries of black holes, come on in and enjoy!

### Kinematic viscosity of air

Use the kinematic viscosity of air calculator to find out the viscosity of air at given pressure and temperature.

### Secretary problem (Valentine's day)

Use the dating theory calculator to enhance your chances of picking the best lifetime partner.

### Sun angle

Our sun angle calculator helps you estimate the Sun's position in the sky regarding its elevation and azimuth angles.


text:
# Difficult to prove $\lim_ {n \to \infty} x_n - \liminf y_n = \limsup (x_n - y_n)$

I wanted to prove something here about $\liminf$ and $\limsup$, but it is not coming out, I need some help.

Let $\{x_n\}$ and $\{y_n\}$ two sequences of real numbers such that $0\leq y_n\leq x_n$ for all $n$ and $\lim_{n\to\infty}x_n$ exists. I need to prove that

$$\lim_ {n \to \infty} x_n - \liminf y_n = \limsup (x_n - y_n).$$

Ok, I did the following:

The left side can be writen as $$\lim_ {n \to \infty} x_n - \liminf y_n =\lim_ {n \to \infty} x_n - \lim_ {n \to \infty}\inf\{ y_n, y_{n+1},y_{n+2},\ldots\} =$$ $$= \lim_ {n \to \infty}(x_n- \inf\{ y_n, y_{n+1},y_{n+2},\ldots\}) = \lim_ {n \to \infty}\sup \{ x_n-y_n, x_n-y_{n+1},x_n-y_{n+2},\ldots\}$$

while the right side is $$\limsup (x_n - y_n) = \lim_{n\to\infty}\sup\{x_n-y_n, x_{n+1}-y_{n+1}, x_{n+2}-y_{n+2},\ldots\}.$$

I can't see a good way to compare this sets.

Any help is welcome, thanks!

• Let $A_n = \{ x_n - y_k : k \geqslant n\}$ and $B_n = \{ x_k - y_k : k \geqslant n\}$. Let $\varepsilon > 0$. Show that there is an $n(\varepsilon)$ such that $\sup A_n \leqslant \varepsilon + \sup B_n$, and $\sup B_n \leqslant \varepsilon + \sup A_n$ for $n \geqslant n(\varepsilon)$. Jan 18, 2015 at 22:19

Let's define $A_n = \{x_n - y_n, x_n-y_{n+1}, x_n - y_{n+2}, \ldots\}$, $B_n = \{x_n - y_n, x_{n+1} - y_{n+1}, x_{n+2} - y_{n+2}, \ldots\}$. Then, what we wish to prove is that $\lim\sup A_n = \lim\sup B_n$.

Let $z_n$ be an element of $A_n$ with $z_n \ge \sup A_n - \frac{1}{n}$. We'll show that $\lim_{n \to \infty} z_n \le \lim\sup B_n$. Suppose $z_n = x_n - y_m$. Then, there exists a corresponding element $w_n = x_m - y_m$ in $B_n$. We have that $m \ge n$, so for every $\epsilon > 0$, by making $n$ sufficiently large we can guarantee

$$z_n < w_n + \epsilon \le \sup B_n + \epsilon$$

based on the fact that $x_n$ is Cauchy. Since this is true for all $\epsilon > 0$ by making $n$ large enough, it follows that $\lim\limits_{n \to \infty} z_n \le \lim\limits_{n \to \infty} \sup B_n$. But, $\lim_{n\to\infty}z_n = \lim_{n\to\infty}\sup A_n$, so $$\lim_{n \to \infty} \sup A_n \le \lim_{n \to \infty} \sup B_n.$$

Using the same technique, we can similarly show that $\lim\sup B_n \le \lim\sup A_n$. It thus follows that $\lim\sup A_n = \lim\sup B_n$, which is exactly what you wanted to show.

Here's a proof I think is slightly easier. Let's let $x = \lim_{x \to \infty} x_n$. Then, for any $\epsilon > 0$, for $n$ sufficiently large, we have $$x - \epsilon - y_n < x_n - y_n < x + \epsilon - y_n$$ So, taking the $\lim\sup$, $$x - \epsilon + \lim\sup(-y_n) \le \lim\sup (x_n - y_n) \le x + \epsilon + \lim\sup(-y_n)$$

since $\epsilon > 0$ was arbitrary, we have $$x + \lim\sup (-y_n) \le \lim\sup (x_n - y_n) \le x + \lim\sup(-y_n)$$

So, using the well-known identity $\lim\sup (-y_n) = -\lim\inf y_n$ and observing that $x = \lim_{n\to\infty}x_n$, $$\lim_{n\to\infty}x_n - \lim\inf y_n = \lim\sup(x_n-y_n)$$

• Your first prove I'm not sure because you don't know if $z_n$ is in the set, so it is not written as $z_n = x_n - y_m$. The second proof I really appreciate. looks fine and clean. Jan 18, 2015 at 23:33
• @Integral Good point about the first proof. Thankfully, it's not too hard to correct the error you found; I've made the necessary changes so $z_n \in A_n$.
– user88319
Jan 19, 2015 at 0:08

format: plain


text:
# 6 Data documentation

Simply making data available is not sufficient to ensure that it is re-usable (see e.g., Kidwell et al., 2016). Providing documentation (often referred to as ‘metadata’, ‘codebooks’, or ‘data dictionaries’) alongside data files will ensure that other researchers, and future you, can understand what values the data files contain and how the values correspond to findings presented in the research report. This documentation should describe the variables in each data file in both human- and machine-readable formats (e.g., csv, rather than docx or pdf).4 Ideally, codebooks are organized in such a way that each line represents one variable and each information relative to a variable represents a column. Extraneous information, that cannot be read (e.g., colors, formatting), should be be included in the codebook as well. For an example of a codebook based on survey data, see this example by Kai Horstmann (https://osf.io/e4tqy/); for an example based on experimental data see the codebook in our example OSF project (https://osf.io/up4xq/).

Codebooks should include the following information for each variable: the name, description of the variable, units of measurement, coding of values (e.g., “1 = Female”,”2 = Male”), possible options or range in which the data points can fall (e.g., “1 = not at all to 7 = Very much”), value(s) used for missing values, and information on whether and how the variable was derived from other variables in the dataset (e.g., “bmi was derived from body_weight m and body_height l as $$BMI = \frac{m}{l^{2}}$$.”). Other relevant information in a codebook entry can include the source of a measure, instructions for a questionnaire item, information about translation, or scale that an item belongs to.5


text:
# Talk:Discrete Fourier transform (general)

The article "Number-theoretic transform" has been merged into this article: See old talk-page here. Selinger (talk) 01:58, 24 June 2011 (UTC)

• somewhere we need a description of the discrete Fourier transform over any field. This is used, for example, in Reed-Solomon codes.
• the existing main article on the discrete Fourier transform is primarily over the complex numbers (although other fields are briefly mentioned). I think this should stay as it is, because most people who need the DFT only need the complex case, the the added generality of field theory would benefit very few people. Talk:discrete Fourier transform already contains complaints that the content is too abstract.

I also propose that the existing page on the number theoretic transform be merged into this (actually, it should be deleted and redirected here). That page is already in bad shape, and the content is entirely subsumed by this new article.

Selinger (talk) 23:50, 14 March 2008 (UTC)

I have merged in the page on Discrete weighted transform (suggested for two years with no complaints). --Qetuth (talk) 06:47, 21 December 2011 (UTC)

## Requirements for convolution theorem with composite moduli

Regarding this statement:

For the number theoretic transform to work in a useful way when the modulus is composite (for the inverse algorithm, convolution etc. to work), it suffices that the modulus is chosen so that a primitive root of order $n$ exists (where $n$ is the transform length), and such that the multiplicative inverse of $n$ exists. Note that if $\alpha$ is a primitive root of order $n$, then $\alpha$ is automatically invertible with inverse $\alpha^{n-1}$.

I was doing an experiment and when I computed IDFT(DFT([1,0,0,0,0,0,0,0])) in the integers mod 85 using 2 as my primitive 8th root mod 85, the result is [1, 0, 0, 0, 51, 0, 0, 0], not [1,0,0,0,0,0,0,0] as expected (and 8 has an inverse mod 85, 32). Eventually I figured out the problem: 24=16, and 16-1 = 15 is not invertible mod 85, so you can't divide by β-1 in the proof. So a sufficient condition would be: $\alpha^{k}-1$ is invertible for all 0 < k < n (I'm uncertain if this condition is necessary). This obviously applies with prime moduli, but I haven't figured out if it applies with Fermat/Mersenne numbers. Am I correct, and if so, where should I find a reference to back this up? Thanks! Dcoetzee 06:53, 14 December 2011 (UTC)

I found a source and fixed this by generalising the entire article to discuss rings rather than fields. The desired condition to make this work was that $\alpha$ is a principal nth root of unity. Dcoetzee 05:43, 22 December 2011 (UTC)


text:
# The local minima of the function f(x) = x2 – x4 in the range -0.8 ≤ x ≤ 0.8 is located at

This question was previously asked in
GATE PI 2017 Official Paper
View all GATE PI Papers >
1. x = 0
2. x = $$\frac{1}{\sqrt2}$$
3. x = $$-\frac{1}{\sqrt2}$$
4. x = $$1\over 2$$

Option 1 : x = 0

## Detailed Solution

Concept:

Following steps to finding maxima and minima using derivatives.

• Find the derivative of the function.
• Set the derivative equal to 0 and solve. This gives the values of the maximum and minimum points.
• Now we have find second derivative.
1. f(x) is less than 0 then the given function is said to be maxima
2. If f(x) Is greater than 0 then the function is said to be minima

Calculation:

Given

f(x) = x2 – x4 , Range -0.8 ≤ x ≤ 0.8

f'(x) = 2x - 4x3    ----(1)

f''(x) = 2 - 12x2    ----(2)

From equation ( 1 ),

f'(x) = 0

2x - 4x3 = 0

2x[1 - 2x2] = 0

x = 0, $$\frac{1}{\sqrt2}$$$$-\frac{1}{\sqrt2}$$

From equation (2)

f''(0) = 2 (minima)

$$f''\left(\frac{1}{\sqrt2}\right)=f''\left(-\frac{1}{\sqrt2}\right)=-4$$ (maxima)

So, the local minima is at x = 0.


text:
------------------------------------------------------------------------
-- The Agda standard library
--
-- Simple combinators working solely on and with functions
------------------------------------------------------------------------

-- The contents of this file can be accessed from Function.

{-# OPTIONS --without-K --safe #-}

module Function.Base where

open import Level
open import Strict

private
variable
a b c d e : Level
A : Set a
B : Set b
C : Set c
D : Set d
E : Set e

------------------------------------------------------------------------
-- Some simple functions

id : A → A
id x = x

const : A → B → A
const x = λ _ → x

constᵣ : A → B → B
constᵣ _ = id

------------------------------------------------------------------------
-- Operations on dependent functions

-- These are functions whose output has a type that depends on the
-- value of the input to the function.

infixr 9 _∘_ _∘₂_
infixl 8 _ˢ_
infixl 0 _|>_
infix  0 case_return_of_
infixr -1 _$_ _$!_

-- Composition

_∘_ : ∀ {A : Set a} {B : A → Set b} {C : {x : A} → B x → Set c} →
(∀ {x} (y : B x) → C y) → (g : (x : A) → B x) →
((x : A) → C (g x))
f ∘ g = λ x → f (g x)
{-# INLINE _∘_ #-}

_∘₂_ : ∀ {A₁ : Set a} {A₂ : A₁ → Set d}
{B : (x : A₁) → A₂ x → Set b}
{C : {x : A₁} → {y : A₂ x} → B x y → Set c} →
({x : A₁} → {y : A₂ x} → (z : B x y) → C z) →
(g : (x : A₁) → (y : A₂ x) → B x y) →
((x : A₁) → (y : A₂ x) → C (g x y))
f ∘₂ g = λ x y → f (g x y)

-- Flipping order of arguments

flip : ∀ {A : Set a} {B : Set b} {C : A → B → Set c} →
((x : A) (y : B) → C x y) → ((y : B) (x : A) → C x y)
flip f = λ y x → f x y
{-# INLINE flip #-}

-- Application - note that _$_ is right associative, as in Haskell. -- If you want a left associative infix application operator, use -- Category.Functor._<$>_ from Category.Monad.Identity.IdentityMonad.

_$_ : ∀ {A : Set a} {B : A → Set b} → ((x : A) → B x) → ((x : A) → B x) f$ x = f x
{-# INLINE _$_ #-} -- Strict (call-by-value) application _$!_ : ∀ {A : Set a} {B : A → Set b} →
((x : A) → B x) → ((x : A) → B x)
_$!_ = flip force -- Flipped application (aka pipe-forward) _|>_ : ∀ {A : Set a} {B : A → Set b} → (a : A) → (∀ a → B a) → B a _|>_ = flip _$_
{-# INLINE _|>_ #-}

-- The S combinator - written infix as in Conor McBride's paper
-- "Outrageous but Meaningful Coincidences: Dependent type-safe syntax
-- and evaluation".

_ˢ_ : ∀ {A : Set a} {B : A → Set b} {C : (x : A) → B x → Set c} →
((x : A) (y : B x) → C x y) →
(g : (x : A) → B x) →
((x : A) → C x (g x))
f ˢ g = λ x → f x (g x)
{-# INLINE _ˢ_ #-}

-- Converting between implicit and explicit function spaces.

_$- : ∀ {A : Set a} {B : A → Set b} → ((x : A) → B x) → ({x : A} → B x) f$- = f _
{-# INLINE _$- #-} λ- : ∀ {A : Set a} {B : A → Set b} → ({x : A} → B x) → ((x : A) → B x) λ- f = λ x → f {-# INLINE λ- #-} -- Case expressions (to be used with pattern-matching lambdas, see -- README.Case). case_return_of_ : ∀ {A : Set a} (x : A) (B : A → Set b) → ((x : A) → B x) → B x case x return B of f = f x {-# INLINE case_return_of_ #-} ------------------------------------------------------------------------ -- Non-dependent versions of dependent operations -- Any of the above operations for dependent functions will also work -- for non-dependent functions but sometimes Agda has difficulty -- inferring the non-dependency. Primed (′ = \prime) versions of the -- operations are therefore provided below that sometimes have better -- inference properties. infixr 9 _∘′_ _∘₂′_ infixl 0 _|>′_ infix 0 case_of_ infixr -1 _$′_ _$!′_ -- Composition _∘′_ : (B → C) → (A → B) → (A → C) f ∘′ g = _∘_ f g _∘₂′_ : (C → D) → (A → B → C) → (A → B → D) f ∘₂′ g = _∘₂_ f g -- Application _$′_ : (A → B) → (A → B)
_$′_ = _$_

-- Strict (call-by-value) application

_$!′_ : (A → B) → (A → B) _$!′_ = _\$!_

-- Flipped application (aka pipe-forward)

_|>′_ : A → (A → B) → B
_|>′_ = _|>_

-- Case expressions (to be used with pattern-matching lambdas, see

case_of_ : A → (A → B) → B
case x of f = case x return _ of f
{-# INLINE case_of_ #-}

------------------------------------------------------------------------
-- Operations that are only defined for non-dependent functions

infixl 1 _⟨_⟩_
infixl 0 _∋_

-- Binary application

_⟨_⟩_ : A → (A → B → C) → B → C
x ⟨ f ⟩ y = f x y

-- In Agda you cannot annotate every subexpression with a type
-- signature. This function can be used instead.

_∋_ : (A : Set a) → A → A
A ∋ x = x

-- Conversely it is sometimes useful to be able to extract the
-- type of a given expression.

typeOf : {A : Set a} → A → Set a
typeOf {A = A} _ = A

-- Construct an element of the given type by instance search.

it : {A : Set a} → {{A}} → A
it {{x}} = x

------------------------------------------------------------------------
-- Composition of a binary function with other functions

infixr 0 _-⟪_⟫-_ _-⟨_⟫-_
infixl 0 _-⟪_⟩-_
infixr 1 _-⟨_⟩-_ ∣_⟫-_ ∣_⟩-_
infixl 1 _on_ _on₂_ _-⟪_∣ _-⟨_∣

-- Two binary functions

_-⟪_⟫-_ : (A → B → C) → (C → D → E) → (A → B → D) → (A → B → E)
f -⟪ _*_ ⟫- g = λ x y → f x y * g x y

-- A single binary function on the left

_-⟪_∣ : (A → B → C) → (C → B → D) → (A → B → D)
f -⟪ _*_ ∣ = f -⟪ _*_ ⟫- constᵣ

-- A single binary function on the right

∣_⟫-_ : (A → C → D) → (A → B → C) → (A → B → D)
∣ _*_ ⟫- g = const -⟪ _*_ ⟫- g

-- A single unary function on the left

_-⟨_∣ : (A → C) → (C → B → D) → (A → B → D)
f -⟨ _*_ ∣ = f ∘₂ const -⟪ _*_ ∣

-- A single unary function on the right

∣_⟩-_ : (A → C → D) → (B → C) → (A → B → D)
∣ _*_ ⟩- g = ∣ _*_ ⟫- g ∘₂ constᵣ

-- A binary function and a unary function

_-⟪_⟩-_ : (A → B → C) → (C → D → E) → (B → D) → (A → B → E)
f -⟪ _*_ ⟩- g = f -⟪ _*_ ⟫- ∣ constᵣ ⟩- g

-- A unary function and a binary function

_-⟨_⟫-_ : (A → C) → (C → D → E) → (A → B → D) → (A → B → E)
f -⟨ _*_ ⟫- g = f -⟨ const ∣ -⟪ _*_ ⟫- g

-- Two unary functions

_-⟨_⟩-_ : (A → C) → (C → D → E) → (B → D) → (A → B → E)
f -⟨ _*_ ⟩- g = f -⟨ const ∣ -⟪ _*_ ⟫- ∣ constᵣ ⟩- g

-- A single binary function on both sides

_on₂_ : (C → C → D) → (A → B → C) → (A → B → D)
_*_ on₂ f = f -⟪ _*_ ⟫- f

-- A single unary function on both sides

_on_ : (B → B → C) → (A → B) → (A → A → C)
_*_ on f = f -⟨ _*_ ⟩- f

_-[_]-_ = _-⟪_⟫-_
{-# WARNING_ON_USAGE _-[_]-_
"Warning: Function._-[_]-_ was deprecated in v1.4.


text:
# Properties

 Label 3600.2.f.u Level $3600$ Weight $2$ Character orbit 3600.f Analytic conductor $28.746$ Analytic rank $0$ Dimension $2$ CM no Inner twists $2$

# Related objects

## Newspace parameters

 Level: $$N$$ $$=$$ $$3600 = 2^{4} \cdot 3^{2} \cdot 5^{2}$$ Weight: $$k$$ $$=$$ $$2$$ Character orbit: $$[\chi]$$ $$=$$ 3600.f (of order $$2$$, degree $$1$$, not minimal)

## Newform invariants

 Self dual: no Analytic conductor: $$28.7461447277$$ Analytic rank: $$0$$ Dimension: $$2$$ Coefficient field: $$\Q(\sqrt{-1})$$ Defining polynomial: $$x^{2} + 1$$ Coefficient ring: $$\Z[a_1, \ldots, a_{7}]$$ Coefficient ring index: $$2$$ Twist minimal: no (minimal twist has level 90) Sato-Tate group: $\mathrm{SU}(2)[C_{2}]$

## $q$-expansion

Coefficients of the $$q$$-expansion are expressed in terms of $$i = \sqrt{-1}$$. We also show the integral $$q$$-expansion of the trace form.

 $$f(q)$$ $$=$$ $$q + 2 i q^{7} +O(q^{10})$$ $$q + 2 i q^{7} + 6 q^{11} -4 i q^{13} -6 i q^{17} -4 q^{19} -6 q^{29} + 4 q^{31} -8 i q^{37} -8 i q^{43} + 3 q^{49} + 6 i q^{53} -6 q^{59} + 2 q^{61} -4 i q^{67} -12 q^{71} -10 i q^{73} + 12 i q^{77} -4 q^{79} + 12 i q^{83} + 12 q^{89} + 8 q^{91} -2 i q^{97} +O(q^{100})$$ $$\operatorname{Tr}(f)(q)$$ $$=$$ $$2q + O(q^{10})$$ $$2q + 12q^{11} - 8q^{19} - 12q^{29} + 8q^{31} + 6q^{49} - 12q^{59} + 4q^{61} - 24q^{71} - 8q^{79} + 24q^{89} + 16q^{91} + O(q^{100})$$

## Character values

We give the values of $$\chi$$ on generators for $$\left(\mathbb{Z}/3600\mathbb{Z}\right)^\times$$.

 $$n$$ $$577$$ $$901$$ $$2801$$ $$3151$$ $$\chi(n)$$ $$-1$$ $$1$$ $$1$$ $$1$$

## Embeddings

For each embedding $$\iota_m$$ of the coefficient field, the values $$\iota_m(a_n)$$ are shown below.

For more information on an embedded modular form you can click on its label.

Label $$\iota_m(\nu)$$ $$a_{2}$$ $$a_{3}$$ $$a_{4}$$ $$a_{5}$$ $$a_{6}$$ $$a_{7}$$ $$a_{8}$$ $$a_{9}$$ $$a_{10}$$
2449.1
 − 1.00000i 1.00000i
0 0 0 0 0 2.00000i 0 0 0
2449.2 0 0 0 0 0 2.00000i 0 0 0
 $$n$$: e.g. 2-40 or 990-1000 Significant digits: Format: Complex embeddings Normalized embeddings Satake parameters Satake angles

## Inner twists

Char Parity Ord Mult Type
1.a even 1 1 trivial
5.b even 2 1 inner

## Twists

By twisting character orbit
Char Parity Ord Mult Type Twist Min Dim
1.a even 1 1 trivial 3600.2.f.u 2
3.b odd 2 1 3600.2.f.a 2
4.b odd 2 1 450.2.c.a 2
5.b even 2 1 inner 3600.2.f.u 2
5.c odd 4 1 720.2.a.b 1
5.c odd 4 1 3600.2.a.bj 1
12.b even 2 1 450.2.c.d 2
15.d odd 2 1 3600.2.f.a 2
15.e even 4 1 720.2.a.g 1
15.e even 4 1 3600.2.a.ba 1
20.d odd 2 1 450.2.c.a 2
20.e even 4 1 90.2.a.b yes 1
20.e even 4 1 450.2.a.a 1
40.i odd 4 1 2880.2.a.u 1
40.k even 4 1 2880.2.a.bf 1
60.h even 2 1 450.2.c.d 2
60.l odd 4 1 90.2.a.a 1
60.l odd 4 1 450.2.a.e 1
120.q odd 4 1 2880.2.a.k 1
120.w even 4 1 2880.2.a.h 1
140.j odd 4 1 4410.2.a.bf 1
180.v odd 12 2 810.2.e.h 2
180.x even 12 2 810.2.e.e 2
420.w even 4 1 4410.2.a.k 1

By twisted newform orbit
Twist Min Dim Char Parity Ord Mult Type
90.2.a.a 1 60.l odd 4 1
90.2.a.b yes 1 20.e even 4 1
450.2.a.a 1 20.e even 4 1
450.2.a.e 1 60.l odd 4 1
450.2.c.a 2 4.b odd 2 1
450.2.c.a 2 20.d odd 2 1
450.2.c.d 2 12.b even 2 1
450.2.c.d 2 60.h even 2 1
720.2.a.b 1 5.c odd 4 1
720.2.a.g 1 15.e even 4 1
810.2.e.e 2 180.x even 12 2
810.2.e.h 2 180.v odd 12 2
2880.2.a.h 1 120.w even 4 1
2880.2.a.k 1 120.q odd 4 1
2880.2.a.u 1 40.i odd 4 1
2880.2.a.bf 1 40.k even 4 1
3600.2.a.ba 1 15.e even 4 1
3600.2.a.bj 1 5.c odd 4 1
3600.2.f.a 2 3.b odd 2 1
3600.2.f.a 2 15.d odd 2 1
3600.2.f.u 2 1.a even 1 1 trivial
3600.2.f.u 2 5.b even 2 1 inner
4410.2.a.k 1 420.w even 4 1
4410.2.a.bf 1 140.j odd 4 1

## Hecke kernels

This newform subspace can be constructed as the intersection of the kernels of the following linear operators acting on $$S_{2}^{\mathrm{new}}(3600, [\chi])$$:

 $$T_{7}^{2} + 4$$ $$T_{11} - 6$$ $$T_{13}^{2} + 16$$ $$T_{17}^{2} + 36$$

## Hecke characteristic polynomials

$p$ $F_p(T)$
$2$ $$T^{2}$$
$3$ $$T^{2}$$
$5$ $$T^{2}$$
$7$ $$4 + T^{2}$$
$11$ $$( -6 + T )^{2}$$
$13$ $$16 + T^{2}$$
$17$ $$36 + T^{2}$$
$19$ $$( 4 + T )^{2}$$
$23$ $$T^{2}$$
$29$ $$( 6 + T )^{2}$$
$31$ $$( -4 + T )^{2}$$
$37$ $$64 + T^{2}$$
$41$ $$T^{2}$$
$43$ $$64 + T^{2}$$
$47$ $$T^{2}$$
$53$ $$36 + T^{2}$$
$59$ $$( 6 + T )^{2}$$
$61$ $$( -2 + T )^{2}$$
$67$ $$16 + T^{2}$$
$71$ $$( 12 + T )^{2}$$
$73$ $$100 + T^{2}$$
$79$ $$( 4 + T )^{2}$$
$83$ $$144 + T^{2}$$
$89$ $$( -12 + T )^{2}$$
$97$ $$4 + T^{2}$$


text:
# Thread: prove that root 6 is irrational

1. ## prove that root 6 is irrational

plzz prove that root 6 is irrational.

i have to show the whole prove not just say that 6 is not a perfect square so root 6 is irrational. i have to prove it wholly.

2. Suppose $\sqrt{2\cdot3} = n/m$. Then $2\cdot3\cdot m^2 = n^2$. Factor both sides into primes. Then the power of 2 on the RHS is even because it is a square, but the power of 2 on the LHS is odd. This contradicts unique factorization.

3. extremely sryy but i can't understand it properly plzz give me detail.

4. Originally Posted by saha.subham
plzz prove that root 6 is irrational.

i have to show the whole prove not just say that 6 is not a perfect square so root 6 is irrational. i have to prove it wholly.

Or note that $\sqrt{6}$ is a solution to $x^2-6=0$ and the only rational solutions to that are...

P.S. It follows that $\sqrt{2}+\sqrt{3}\notin\mathbb{Q}$

5. srryy but i can;t understand without details

6. But I thought you said you had prove it. What do you understand about proving that roots are not rational?

7. haha! If you square root anything but a perfect square then the answer will be irrational!

PROVEN!!

8. Originally Posted by jgv115
haha! If you square root anything but a perfect square then the answer will be irrational!

PROVEN!!
I have a feeling that if the OP is going to say that then that statement has to be proven in the process.

Assume that $\sqrt{6}$ is rational.

Then $\sqrt{6} = \frac{a}{b}$, where $a, b$ are integers, and $\frac{a}{b}$ is in simplest form (since every rational number can be reduced to simplest form).

Therefore $6 = \frac{a^2}{b^2}$

$6b^2 = a^2$

$2\cdot 3b^2 = a^2$.

Since $a^2$ is even, that means $a$ is even. So write it as $a = 2c$. Then you have

$2\cdot 3b^2 = (2c)^2$

$2\cdot 3b^2 = 4c^2$

$3b^2 = 2c^2$

$b^2 = 2\left(\frac{c^2}{3}\right)$.

Remembering that $b$ is an integer, so is $b^2$. Also, since 2 is a factor, $b^2$ is even, and so $b$ is even. Write it as $b = 2d$.

Remember that $\sqrt{6} = \frac{a}{b}$

but $a = 2c$ and $b = 2d$

So $\sqrt{6} = \frac{2c}{2d}$

$\sqrt{6} = \frac{c}{d}$.

But we said that $\frac{a}{b}$ was already in simplest form.

So we have a contradiction.

This means that $\sqrt{6}$ can not possibly be rational.

Therefore $\sqrt{6}$ is irrational.

,
,

,

,

,

,

,

,

,

,

,

,

# square root of 6 is irrational

Click on a term to search for related topics.


text:
History of the geometric series

I'm interested in understanding the history of the geometric series, especially how it was discovered and whether there exists continued fraction representations for the geometric series, just as there are continued fractions for many elementary functions in mathematics.

I would also appreciate some historical perspective on how geometric series played a role in the development of calculus.

• You can see the following link The Geometric Series Also you can visit mathcs.slu.edu/history-of-math/index.php/9.2_Geometric_series – Sachchidanand Prasad Sep 25 '15 at 9:12
• Apparently, arithmetic/geometric series were already studied in Euclid, though not exactly under this terminology — “continually in proportion”. I don't know (and would like to know) when/where the terminology of arithmetic/geometric series has been invented. – ACL Nov 23 '18 at 9:10

The infinite series had originated in India by the 14th c. An explicit formula for the sum of an infinite (anantya) geometric series is given by the 15th-16th c. Nilkantha in his Aryabhatyabhasya.(Sastri 1970, commentary on Ganita 17, p. 142.)

एवं यासतुल्यच्छेदपरभागपरम्पराया अनन्ताया अपि संयोग

तस्यानंतानांपि कल्प्यामान्स्य योगस्याद्यावयविनः

परंपरांशच्छेदादेकोनच्छेदांशसामयं सर्वत्रापी समानमेव ।

which may be translated:(cf. Sarma 1972, p. 17) The sum of an infinite [anantya] series, whose later terms (after the first) are got by dividing the preceding one by the same divisor everywhere, is equal to the first term multiplied by the common divisor, and divided by one less than the common divisor.

That is a+a + a +···= ad/d-1 .(It is assumed that the divisor d>1, so that the common ratio is less than 1.)

• Well... maybe infinite series do not originated in India, as Archimedes (III century BC) discussed in The Quadrature of the Parabola a geometric series with common ratio 1/4, while Nicole Oresme (1320-1382) discussed in Quaestiones super geometriam Euclidis convergence and divergence conditions for geometric series and obtained the sums of several geometric series (see e.g. here core.ac.uk/download/pdf/144470649.pdf) and also proved the divergence of the harmonic series. – user6530 Jan 5 '20 at 13:33
• To user6530's answer add that the general formula for the sum of a finite geometric series (not just the earlier specific examples from e.g. the Rhind Papyrus) is in Euclid's Elements Prop. IX.35. – David Eppstein Feb 5 '20 at 1:22


text:
Defeated... I give up :S : Share GMAT Experience
Check GMAT Club Decision Tracker for the Latest School Decision Releases https://gmatclub.com/AppTrack

 It is currently 23 Feb 2017, 14:18

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

# Defeated... I give up :S

Author Message
TAGS:

### Hide Tags

Manager
Joined: 04 Oct 2005
Posts: 246
Followers: 1

Kudos [?]: 76 [0], given: 0

Defeated... I give up :S [#permalink]

### Show Tags

15 Nov 2005, 06:52
well, what should I say? topic reveals my feelings at this moment.....
I am though a little suprised, how badly I performed.... gonna make this as short as possible because I kinda feel depressed now.

took my first gmat end of september and score 640.
Study materials prior to first try:
-kaplan
-kaplan800

The kaplan CATS always gave me results between 600 and 700

Then I found this forum and regained hope. Browsing this forum I found out that the OG is essential and lots of other helpful things. I continued to study hard (full-time) to make sure I do a lot better this time. All these 700 posts really motivated me till a few hours ago.
Anways, for my second try I additionally used:
- OG 11th
- 1000 SC (completed 650 questions)
- LSAT CR
- and of course GMAT club!

My performance improvement actually looked quite promising. I found out my weaknesses (RC and CR at first) and worked on them hard. LSAT CR is in my opinion the best training. After you do several hundred questions, you simply know what kinda style the questions are.

One week before my second attempt I took a full length practice test every day (PP and Gmatprep).
In my first try, I got only one mistake in each CR and RC and 8 in SC. I focused in SC in the last days and as it turned out, in my last practise exam I reduced my SC error rate to 3.

Test results in order of occurence:
PP: 740
Gprep: 730
Gprep: 710
PP: 700

Today: 620

This really hits me hard, because I knew I had a realistic chance to cross the 700 barrier. Furthermore, it throws my career planning completely out of line and I need to go plan B (which btw does not really exist yet).
Yet I'd never have pegged that I perform even WORSE than on my first try after studying only kaplan.

I do not have the will to take the GMAT a 3rd time, just cant do it.

oh btw, for AWA I used this pdf file going around, which containts a template for each essay. They are really good and applicable to almost any topic!

Lastly, I would like to thank everyone here in this forum. I really learned a lot here and it was heaps of fun puzzling over some quants or discussing verbal questions. Keep up the good work!
 Kaplan GMAT Prep Discount Codes Economist GMAT Tutor Discount Codes Manhattan GMAT Discount Codes
SVP
Joined: 24 Sep 2005
Posts: 1890
Followers: 19

Kudos [?]: 296 [0], given: 0

### Show Tags

15 Nov 2005, 07:04
Nero44, you have right to believe that you can cross 700 barrier!! I read many of your posts,just wanna say a few words "they're just great!" , especially your maths posts. Maybe yes, as others say GMAT involves some luck . Have you considered to re-take it?! Anyway, don't be so sad ...we here know that you're deserved much more!! Good lucks to you with your endeavor!!
Manager
Joined: 20 Sep 2005
Posts: 55
Followers: 0

Kudos [?]: 3 [0], given: 0

### Show Tags

15 Nov 2005, 09:02
You definitely have potential to score more than 700. Your practice cats consistently show that. I would say spend some time to think about why you couldn't do it on the test day. A lot of things could affect your score - performance anxiety, time of the test, how much sleep you had last day, etc. If I were you I would consider taking it again.
Director
Joined: 21 Aug 2005
Posts: 793
Followers: 2

Kudos [?]: 25 [0], given: 0

### Show Tags

15 Nov 2005, 09:34
Hey nero44! 620 is definitely not you! As your practice scores indicate, you are 700+ material. In which section did you go down this time? If you say Math, then I will be totally surprised. Even 620 and 640, I am surprised. Something should have gone wrong during the test - Definitely not your potential. Were you nervous or tired or did you run out of time by spending too much time in few questions? Retrospect and find out what really caused this. There should be something.
Don't give up yet! Unlike so many other things in life, you get a second (or even a third) chance with GMAT. I understand your frustration. Take a couple of days off, think about and definitely give re-taking the test a thought. You have the goods, so you won't go down everytime. Just think about it.
Good Luck!
VP
Joined: 30 Sep 2004
Posts: 1488
Location: Germany
Followers: 6

Kudos [?]: 332 [0], given: 0

### Show Tags

15 Nov 2005, 09:38
verdammt ! wo mÃ¶chtest du dich bewerben ? was sind deine scores fÃ¼r math und verbal ?
_________________

If your mind can conceive it and your heart can believe it, have faith that you can achieve it.

VP
Joined: 22 Aug 2005
Posts: 1120
Location: CA
Followers: 1

Kudos [?]: 103 [0], given: 0

### Show Tags

15 Nov 2005, 09:53
I will never be contended with your score. Because I know you deserve much better. Gear up my friend and try last time. I am sure you will write another inspirational story for us to follow!!!

_________________

Whether you think you can or think you can't. You're right! - Henry Ford (1863 - 1947)

Senior Manager
Joined: 11 May 2004
Posts: 458
Location: New York
Followers: 1

Kudos [?]: 47 [0], given: 0

### Show Tags

15 Nov 2005, 10:18
Nero,

You def cannot give up. Your scores clearly indicate that you can cross the 700 barrier! As I was reading your posts, I noticed that you too an exam everyday during the week before the test. You probably just got a little burned out. Also, were you taking the practice tests around the same time as your scheduled gmat?

There could be a lot of external factors that affected your score, but reading your posts and contributions clearly shows that you should take it again. Maybe take a few days off, relax, go out, and then come back in a few days full force...

GMAT Club Legend
Joined: 15 Dec 2003
Posts: 4302
Followers: 40

Kudos [?]: 438 [0], given: 0

### Show Tags

15 Nov 2005, 10:26
I know how much effort it takes to undertake the GMAT a third time but I have done it, and so have many of us in this forum. At the end of the day, it depends on how dearly you hold on to your dream of going to your target school. Do not give up. I am a bit of an outlier but I spent about a year, although I actually studied "only" 6 months, in between my second and third try and finally made it. All I can say again is gather yourself up and gear up for another battle. You lost the previous one but you have not lost the war. You will see how great it feels once you will reach the mountain top, I promise.
_________________

Best Regards,

Paul

Manager
Joined: 20 Mar 2005
Posts: 201
Location: Colombia, South America
Followers: 1

Kudos [?]: 16 [0], given: 0

### Show Tags

15 Nov 2005, 11:38
Nero, you are 700+ material, do not even think of giving up, 620 is not good enough for you, everybody here knows you can do it, so give it one more try.

I'm sure GMAT is a bit of luck, and that luck can account for up to 50 points, that is my estimate.

by the way, I never had the chance to see that AWA template you are talking about I wish I would have seen it before I took my test
Senior Manager
Joined: 30 Oct 2004
Posts: 284
Followers: 1

Kudos [?]: 62 [0], given: 0

### Show Tags

15 Nov 2005, 17:27
I understand the frustration... was feeling the same a few weeks back.
Was it test anxiety? What ever the root cause... time to keep it behind and decide what next?

All the best!
_________________

-Vikram

CEO
Joined: 17 Jul 2004
Posts: 3281
Followers: 23

Kudos [?]: 466 [0], given: 0

### Show Tags

15 Nov 2005, 20:53
You should give yourself at least one more try at a higher score- your application score cannot go down from a 640 and based on your recent practice tests it could go significantly higher.

Hjort
SVP
Joined: 05 Apr 2005
Posts: 1731
Followers: 5

Kudos [?]: 74 [0], given: 0

Re: Defeated... I give up :S [#permalink]

### Show Tags

15 Nov 2005, 21:15
nero44 wrote:
I do not have the will to take the GMAT a 3rd time, just cant do it.

nero,

yes, it is understandable that how frustrating it is when one doesnot have the desired score after his/her intense/long preparation.

but this doesnot match to a person like you who is an MBA applicant... you as an mba have to manage/run a company in future and you cannot say leave the battle ground in real world by saying that i tried once/twice/thrice so i donot do it again. this is not a way to manage things in real lief. so take/face it as a challange. you have to do it again and again not only one time but also many times until you get your required/desired score. there are many good people who got their 700 score in their 3rd or 4th attempt.

i always remember this:
giggity wrote:
If you are not going to focus, then go home.
http://www.gmatclub.com/phpbb/viewtopic.php?t=19802

learn from the past.
find your weeakness and heat on it.
note the common mistakes you do.

sucess is there...
good luck...

note: btw, what is your score breakdown?
Current Student
Joined: 29 Jan 2005
Posts: 5238
Followers: 26

Kudos [?]: 381 [0], given: 0

### Show Tags

15 Nov 2005, 22:56
Nero,

Forget about ETS for a moment. There could have been a myriad of outside factors that affected your performance. Were you extremely stressed out? Did you get a full nights sleep the night before the exam? Did you pace yourself correctly?? Were you a bit hyper-critical? Also what was your Q/V breakdown?

Remember that a third of the test is experimental, so perhaps you may have spent a disproportional amount of time on those questions only to cut yourself short on time for the ones that really mattered. I completely understand your frustration. I too took the same Kaplan course, used the same Kaplan Practice tests, studied the same white Kaplan course book, only to score a 570!!

At all costs, do not give up- at least not now. You have replied to so many of my "difficult posts" in the math forum, the same posts that other members missed and still managed to score above a 700. We all know you have the potential to shatter the 700 ceiling, and you will, but just not this time.

Shake off your disappointment and reengage after a couple days of introspection. Deconstruct the test if you have to (thats my plan B), but do not let go of your dreams. Keep striking while the iron is still hot.

See you in the forum buddy.
Intern
Joined: 27 Mar 2005
Posts: 5
Followers: 0

Kudos [?]: 0 [0], given: 0

### Show Tags

15 Nov 2005, 22:58
take it from a 4 time test taker, you should take it again and hit your target score....
Senior Manager
Joined: 04 May 2005
Posts: 282
Location: CA, USA
Followers: 1

Kudos [?]: 52 [0], given: 0

### Show Tags

15 Nov 2005, 23:58
I am guessing your issue is not preparation or ability; it could be your testing skill or the ability to overcome anxiety. If that is the case, then you might want to focus on that - instead of the math or verbal knowledge preparation.

I do believe PP is a good gauge of your score. Your score is quite good by that standard.

take care and cheer up !
SVP
Joined: 03 Jan 2005
Posts: 2243
Followers: 16

Kudos [?]: 329 [0], given: 0

### Show Tags

18 Nov 2005, 06:23
Nero, I can feel how disappointed you are now. Don't be too hard on yourself. Give yourself a break, say wth and relax. Then when you can gether the strength again, I'd join the others and say you can give it another try! If you believe you definitely have the potential of a higher score, don't take what it gave you as an answer. You know just by statistics if you got a score that is lower than your true potential, the next time it's more likely you'll have a good score that reflect your potential, and hey maybe even higher. So yeah, go ahead and take a rest, and then come back sometime, and see what you feel. We are all behind you.
_________________

Keep on asking, and it will be given you;
keep on seeking, and you will find;
keep on knocking, and it will be opened to you.

Manager
Joined: 04 Oct 2005
Posts: 246
Followers: 1

Kudos [?]: 76 [0], given: 0

### Show Tags

18 Nov 2005, 09:25
Hey all,
@ everyone: thank you so much for your words. Really cheered me up big time.

I relaxed pretty well the last days to re-think everything.

I still do not plan to retake it a third time at this point. I'll give it some days maybe some weeks to see what happens. I had an interview last week for an intern position just to keep me busy til I have new plans.

I still do not know what was happening at test day. Some of you asked for my score breakup:
1st: 640 Q44 V34
2nd 620 Q44 V32

It seems as if I had not studied at all between my first and my second attempt... I usually am very solid with testing experience.. never had any trouble with nervousness during my undergrad (graduated this august, so it's all quite recent). That's what basically puzzles me, because I'm the kinda secure guy. I usually always know what I can expect and what not, EXCEPT for this time...

@ Christoph: I was planning to apply for an MSc finance study in the UK. LSE was upon my top-shots, but my GMAT score put me out of race there.

@ Hjort: you got a real good point there.. 640 down to 620 looks really bad and does not reflect my character....

Thing is, the application round began in mid October and goes till summer. The longer you wait, the lower your chaces for admission. Sending your application after new years, is already considered as late. The rest of your profile must be extremely strong in order to have a reasonable chance. Intake rate is about 3-5% ....

My initial plan was to get everything ready by november, but apparently this plan is worth Â§\$%& now. I just don't know if it's worth to invest so much time to boost gmat. The delay in time may weaken my chances for my other picks even more. Additionally I may start the intern from january on, so there hardly isn't even time to do so....

maybe anyone here has some ideas bout this... d'be glad to hear your opinion. gonna start on my PS now and have a cold

18 Nov 2005, 09:25
Similar topics Replies Last post
Similar
Topics:
MY 7TH GMAT: I GIVE UP :( 6 29 Aug 2014, 07:26
1 Failure in first attempt. Devastated but i won't give up 9 23 Apr 2014, 06:50
GMAT won but I am not ready to give up 5 14 Aug 2010, 01:03
2 Disappointing score- but i wnt give up 10 28 Sep 2008, 17:26
I'm not going to get what I want OR how to give up 8 25 Aug 2007, 21:20
Display posts from previous: Sort by


text:
# Tag Info

5

It seems me that the answers of mathe and Yves Klett do not meet expectations of the author. The latter is as much as I have got it, to have a short analytical expression for the solution. Probably the author has an intention to use the result further in some analytical calculations, or to do something comparable. Am I right? If yes, one should first of ...

5

Solve[L == (3 W)/2 + 3/2 Sqrt[4 A^2 Pi^2 + W^2] - Sqrt[ 6 A^2 Pi^2 + 3 W^2 + 5 W Sqrt[4 A^2 Pi^2 + W^2]]/Sqrt[2], W, Quartics -> False] or Solve[L == (3 W)/2 + 3/2 Sqrt[4 A^2 Pi^2 + W^2] - Sqrt[ 6 A^2 Pi^2 + 3 W^2 + 5 W Sqrt[4 A^2 Pi^2 + W^2]]/Sqrt[2], W, Reals]

1

One shotgun approach is to sic Simplify or FullSimplify onto your solution: sol1 = Solve[ L == (3 W)/2 + (3 Sqrt[4 A^2 Pi^2 + W^2])/2 - Sqrt[6 A^2 Pi^2 + 3 W^2 + 5 W Sqrt[4 A^2 Pi^2 + W^2]]/Sqrt[2], W]; sol2 = Simplify[sol1]; LeafCount /@ {sol1, sol2} ByteCount /@ {sol1, sol2} {3849, 3077} {111720, 92840} (Note: FullSimplify is still ...

3

I wanted to let everyone know what happened to me in case it prevents the stress I'm feeling right now. I was worried about the changes in the way dates are handled as I have many lines of code with date features as I am backtesting investments. I decided to see what happens and if I could fix any issues with great inputs above. I made sure not to ...

5

Description ? at the beginning of a line is the short from of Information. Given a pattern that matches multiple Symbols Information returns a list of them in alphabetical (canonical) order, in columns top to bottom and left to right. You are therefore asking how you can modify the behavior of this System function. You would need some way of storing the ...

2

First, I should say that I could find no examples of using RegionPlot3D with regions in the documentation. It works on some regions, not on others, and in this case runs longer than one wants to wait. It runs nonstop because Reduce[Exists[{u, v}, x - Cos[u] == 0 && y - Cos[v] - Sin[u] == 0 && z - Sin[v] == 0 && 0 <= u ...

0

I think you may find value in this rule: sequentialQ[x_List] := x == Range[x[[1]], x[[-1]]] rule = HoldPattern[Plus[s : F_[_, j___] ..]] /; sequentialQ @ {s}[[All, 1]] :> (HoldForm[Sum[F[i, j], {i, ##}]] & @@ {Min@#, Max@#} &[{s}[[All, 1]]]); Test: F[1.0, q, r] + F[1.5, q, r] + F[2, q, r] + F[2.7, q, r] + F[3.0, q, r] /. rule

2

Here is a different sort of answer, but very V10-style. The only logical expression however is Element, so I'm afraid this will fall short. Clear[regFn, regFnmesh]; regFnmesh[polyh_] := regFnmesh[polyh] = ConvexHullMesh@PolyhedronData[polyh, "VertexCoordinates"] regFn[polyh_] := With[{region = regFnmesh[polyh]}, {##} ∈ region &] This ...

2

I suppose they are nice....They check that a point is inside each facet plane. # -> PolyhedronData[#, "RegionFunction"] & /@ {"Octahedron", "Dodecahedron", "Icosahedron"} (* {"Octahedron" -> (2 (#1 + #3) <= Sqrt[2] + 2 #2 && 2 (#1 + #2 + #3) <= Sqrt[2] && 2 (#2 + #3) <= Sqrt[2] + 2 #1 && 2 #3 <= ...

0

You can use Mod with a non-integer second argument: f[a_] := Ceiling[Mod[a, 2 \[Pi]]/\[Pi]] To generalize you can make a Periodic operator as follows: Periodic[T_, offset_: 0][f_] := f@*((Mod[#, T] - offset) &) f2pi = Periodic[2 \[Pi]][Ceiling[#/\[Pi]] &] f3pi = Periodic[3 \[Pi], 0.1][Ceiling[#/\[Pi]] &] quadratic[x_] := (x/\[Pi])^2; parabola ...

2

You can use a combination of Floor, Floor[x, a] gives the greatest multiple of a less than or equal to x and Mod, Mod[m, n] gives the remainder on division of m by n to get f[x_] := 1 + Floor[Mod[x, 2 π]/π]

3

You can also use the built-in function SquareWave: ClearAll[f]; f[x_] := SquareWave[{2, 1}, x/(2 Pi)]; Plot[f[x], {x, 0, 10 Pi}, Ticks -> {Pi Range[10], Automatic}, ExclusionsStyle -> Automatic, PlotRange -> {0, 3}, ImageSize -> 400] Update: another alternative: Use ListInterpolation with options InterpolationOrder->0 and ...

1

Here is a method that uses UnitStep and Sin to generate the square wave. Sin produces the periodicity and UnitStep maps the sinusoid into a square-wave. This method will be faster than Which. f[x_] := 1 + UnitStep[Sin[x + Pi]] Plot[f[x], {x, -10, 10}, Exclusions -> None, PlotStyle -> Thick]

1

Here's one way to make your function periodic: f[t_] := Which[0 <= t < Pi, 1, Pi <= t < 2 Pi, 2, t < 0, f[t + 2 Pi], t > 2 Pi, f[t - 2 Pi]] For example: Plot[f[t], {t, -10, 10}] This method works well for any function you care to use (not just square waves). For instance, f[t] can be linear in one half and quadratic in the second ...

2

In mathematics, a set can be write in two(or more) form: one is {3x | x in [0,1]}, the other {x | P(x)}, where P is called a propositional function or predicate. Since Mathematica have a great power of dealing with quantifiers, once you can write a set in the second form above, you can do plenty of things and tricks that you would have thought to have ...

6

The answer is 0, if the question is what is the value of Mean[fit["FitResiduals"]] and fit is a linear, least-squares fitted model. data = WeatherData["London", "Temperature", {{2004, 1, 1}, {2013, 12, 31}, "Day"}]; normaldata = Partition[Reverse[data["Values"][[All, 1]]], 2, 1]; fit = LinearModelFit[SetPrecision[normaldata, Infinity], x, x, ...

1

Something like this: m4 = Quiet[Table[a[[i, j]], {i, 4}, {j, 4}]]; soln = With[{det = Simplify[Det[m4]]}, Compile[{{a, _Real, 2}}, det]]; soln[RandomReal[1., {4, 4}]] -0.036117495644621564 But Det is quite efficient as it is, I would think.

11

foo = With[{f = #0}, (# /. {p___, a, b, c, q___} :> Join[{p, "abc"}, f @ {q}])] & lst = {1, 2, a, b, c, 3, {4, a, b, c}, 5}; foo@lst (* {1, 2, "abc", 3, {4, "abc"}, 5} *) This works because With automatically renames the patterns used in RuleDelayed since both are scoping constructs. Other constructs can be used as well such as RuleDelayed itself: ...

3

You can "inject" that variable by grabbing it with With. It then gets put into the body of the With verbatim, hence it gets inside the Function in the desired form. f2[expr_] := With[{var = First@Variables[expr]}, expr /. var -> # &] f2[x^2] (* Out[79]= x^2 /. x -> #1 & *) This can also be done in the way you had tried, with a bit of work ...

1

f[a_, b_, c_] = a + b + c; list = {2, 3}; f[a, Sequence @@ list] (* 5 + a *)

2

I really don't know what all your complicated code is supposed to do and I'm too tired to find out. However from your problem statement this should help: f = Sin[#] Sin[#2] &; a = Array[f, {6, 6}, 0]; min = Min[a] Position[a, min] Sin[2] Sin[5] {{3, 6}, {6, 3}} For values rather than coordinates consider SortBy or for somewhat better ...

11

Let's start by taking a look at the compiled form of one of our queries: DatasetCompileQuery[Query @ First @ spans] (* DatasetWithOverrides@*Checked[Slice[205 ;; 313], Identity] *) We can see that the operation is not implemented directly in terms of part. Indeed, there are three components: DatasetWithOverrides, GeneralUtilitiesChecked and ...

1

As defined, eislminx2 returns a list rather than a value. I modified its definition to use First (only?) value. I rationalized your equations so that they do not limit the precision of the calculations. In addition to specifying a value for epsilon, h needs a value. Do you know any constraints on values of n1 or n2 (region of interest)? eislexpl[epsilon_, ...

1

Updating mutually dependent functions at each iteration It seems like one important aspect which leads to the circular dependency problem here (pointed out earlier) is that you want to update your function definitions for tmed, tmedpred, ttf, med, and tpred at each iteration. Your code uses global rules to create and define your functions (e.g. f[x_] := ...

3

t4[x_,t_]:=Evaluate[t1[x,t]+t2[x,t]]

3

The following set of expressions, which is taken from the relevant part of your code and simplified for discussion purposes, causes infinite recursion. tpred[x_, t_] := .6 tmedpred[x, t] + .4 med[x, t]; tmed[x_?NumericQ, t_?NumericQ] := tpred[x, t]; tmedpred = tmed tpred is defined in terms of tmedpred. tmed is defined in terms of tpred. Then tmedpred ...

5

It's fun to use Associations as a circular linked list, which automatically handles the cyclic nature of these ranges: Clear@monthRange monthRange[start_, end_] := Module[{monthsLL, head}, monthsLL = Fold[ <|#2 -> #|> &, Reverse[DataPacletsCalendarDataDump`MonthList["Gregorian"] ~Join~ {monthsLL}] ]; head = ...

12

The separation-of-variables solution you quoted has two indices appearing in it: n and j (the subscripts of the coefficient $A_{nj}$). Here, n is azimuthal mode order, i.e. it counts the number of nodes along the direction in which the polar-angle $\theta$ varies (divided by 2). The index j is needed because the wave is supposed to satisfy the boundary ...

5

1) Get the list of days and months. Note that days are represented in MMA as symbols (and not strings as months are), hence the use of ToString to make them in a consistent type with the list of months (credit to @Mr.Wizard for this tip). monthList=DateValue[{2014,#,1},"MonthName"]&/@Range[12] (* ...

10

For full ranges There is the function DayRange that can be used for this purpose, but not in the same simple way like CharacterRange. For the days: DayName /@ DayRange[Today, Today ~DatePlus~ {{1, "Week"}, {-1, "Day"}}] {Wednesday, Thursday, Friday, Saturday, Sunday, Monday, Tuesday} For the months: DateValue[#, "MonthName"] & /@ DayRange[Today, ...

3

If you use ToString, make sure you specify InputForm (compare ToString[1/x] v.s. ToString[1/x, InputForm]). Why not something like this? pureify[f_, x_] := Function @@ {f /. x -> #} Table[InverseFunction[pureify[foo, x]][x], {foo, {Sin[x], Log[x], Sqrt[x]}}] (* {ArcSin[x], E^x, x^2} *)

2

Here's my approach to determine the number of slots in the pure function without counting those in any sub-functions. That means MapIndexed[#1^#2 &, #1] & (per @TeakeNutma's comment in @JohnMcGee's answer) will only return one slot. I do this by deleting all inner Functions before counting the slots: Clear[slotCount] ...

1

Thanks to @hieron suggestion to start with the "pure" heads I found this solution: GetHeads[fun_] := ToExpression@First@StringSplit[ToString[fun], "["] f = GetHeads /@ {Sin[x], Log[x], Sqrt[x]} {Sin, Log, Sqrt} Table[InverseFunction[foo[#] &][x], {foo, f}] {ArcSin[x], E^x, x^2}

1

Without "Head-Substitution", you may achieve it: inv = InverseFunction /@ {Sin, Log, Sqrt} x // inv // Through out: {ArcSin[x], E^x, x^2} Edit1: Yes, to get rid of the arguments I just wanted to suggest you inv = InverseFunction /@ ToExpression /@ (StringTake[#, {1, -4}] & /@ToString /@ {Sin[x], Log[x], Sqrt[x]}) x // inv // Through But I ...

3

It's difficult to answer this question definitively, without more detail on the type of function that you're after. As you refer to the PiecewiseExpand function, I'm rather guessing that you're looking for a single, piecewise linear function that passes through the points. If so, perhaps this works: affineFormula[{{x1_, y1_}, {x2_, y2_}}, x_] := y1 + ...

6

The docs specify that the domain should (usually) be Reals or Integers. These are keywords. You probably want the "domain" to be specified as a constraint. Maximize[{ Abs[f[{p1, p2, p3, p4, p5}, {0.5, l2}]], {p1 + p2 + p3 + p4 + p5 == 1 && p1 >= 0 && p2 >= 0 && p3 >= 0 && p4 >= 0 && p5 >= 0 ...

5

is it possible to specify which range that slot takes from? The short answer is "no" (not counting solutions with named arguments since they aren't Slots) but if you really, really want to do it you can write it like I've done below, then #1 refers to the inner Map and #2 refers to the outer Map: Map[ Map[ {Sin[#], Cos[#2]} & @@ # &, ...

6

You can nest Functions to accomplish what I believe you want to: f = i \[Function] {Sin[i], Cos[#]} & /@ list[[i ;;]]; Array[f, 3] // Column {{Sin[1],Cos[a]},{Sin[1],Cos[b]},{Sin[1],Cos[c]}} {{Sin[2],Cos[b]},{Sin[2],Cos[c]}} {{Sin[3],Cos[c]}} This looks nicer in the Notebook: Other methods include: # /. i_ :> ({Sin[i], Cos[#]} & /@ ...

7

My first suggestion would be to localize the outer Slot, like this: With[{indx = #}, {Sin[indx], Cos[#]} & /@ list[[indx ;;]]] & /@ Range[3] (* {{{Sin[1], Cos[a]}, {Sin[1], Cos[b]}, {Sin[1], Cos[c]}}, {{Sin[2], Cos[b]}, {Sin[2], Cos[c]}}, {{Sin[3], Cos[c]}}} *) You could also rework the process to get the full set of tuples of ...

6

As @Mr.Wizard has pointed out, your question isn't well specified. However, a pure function always needs a minimum number of arguments, otherwise an error message is thrown: #1 + #2 &[a] Function::slotn: Slot number 2 in #1+#2& cannot be filled from (#1+#2&)[a] >> a + #2 So finding the minimum number of required arguments of a pure ...

13

Your question is not well specified for several reasons: Pure functions accept a flexible number of arguments #1 + #2 &[a, b, c, d] a + b It is common for some arguments to not be used: #1 + #3 & @@ {a, b, c, d} a + c SlotSequence includes all arguments after the given position: +##3 & @@ {a, b, c, d} c + d Without ...

1

Try Length@Union@Cases[f, _Slot, Infinity] 2 The Union eliminates duplicates

1

Because the latter is equivalent to {1,2,3,4,5}[[2]] = 0 which is invalid. The former stays as listCopy[[2]] = 0 You can't change (i.e. assign to) parts of a literal like this. You can change (i.e. assign to) a variable. Generally, f[x_] := x[[2]] will substitute the value of x directly. So will With[{x=...}, x[[2]] ]. In contrast, ...

8

This is not an answer. It is just a very long comment. Both a simple manually operated drill press and a computer-controlled five-axis omni-mill can drill a hole through a piece of bar stock. And both will do the actual drilling in about the same amount of time. If one hole in one bar is all you want, then you will accomplish the job much faster with the ...

4

Setting b = 1 from the start, your equation can be rewritten as (3 k)/(3 + a k^2) == Tan[k] This has multiple solutions for k for any given value of a. Here's e.g. the plot of the LHS and the RHS for a = 1; the intersections are the solutions: Plot[ { (3 k)/(3 + k^2), Tan[k] }, {k, -10, 10}, Exclusions -> Tan[k] == 0, PlotPoints -> 500 ] ...

4

I stand by @MrWizard's answer. The natural way is with named functions. If it is evaluated, an alternative is anything that prevents Slot[1] from appearing literally. Personally I like Identity[Slot][1] more than Slot@@{1}. If you insist on "protecting" a slot, the only way I know is with another function, so this would work. {#, First[# &]} & /@ ...

5

You can also use Block for this: Block[{slot = #}, {#, slot} & /@ {1, 2}] {{1, #1}, {2, #1}} This works because Block implements dynamic scoping, so the symbol slot is not replaced by #1 until slot is evaluated (i.e. after the Function has been created and mapped over the list). As pointed out by Mr. Wizard, this means that if slot is never ...

11

Using Function with a named parameter, as halmir showed, is the standard way to do this, however anything that prevents a literal Slot[1] from appearing in the body of the Function will work. Inactive as chuy showed is one possibility, but I find this cleaner: {#, Slot @@ {1}} & /@ {1, 2} {{1, #1}, {2, #1}} If the body will not be evaluated you ...

10

Maybe something like (using Inactive): Activate[{#, Inactive[Slot][1]}] & /@ {1, 2} (* {{1, #1}, {2, #1}} *)

12

It depends on what you want to do with result, but you could try like this: Function[{x}, {x, #}] /@ {1, 2} {{1, #1}, {2, #1}}

Top 50 recent answers are included


text:
# Math Help - graphing rational function

1. ## graphing rational function

just wondering if anyone knows how to determine the inflection points,max/min and concavity of the rational function:

f(x)=x2+4/x2-9

just finished finding the vertical and horizontal asymptotes, symmetry, x and y intercepts, and domain, but can't recall the meathods for determining the above. I used to know how to do this like the back of my hand!

2. Originally Posted by dcfi6052
just wondering if anyone knows how to determine the inflection points,max/min and concavity of the rational function:

f(x)=x2+4/x2-9

just finished finding the vertical and horizontal asymptotes, symmetry, x and y intercepts, and domain, but can't recall the meathods for determining the above. I used to know how to do this like the back of my hand!
You do it just like you do it for any other function.

Relative max/min:
Set f'(x) = 0. The values of x where this happens are your candidates.

$f'(x) = \frac{(2x)(x^2-9) - (x^2+4)(2x)}{(x^2-9)^2}$

$f'(x) = -\frac{26x}{(x^2-9)^2}$

So f'(x) = 0 only when x = 0.

Now is this a relative max or a relative min? You can find out by graphing, but we need the second derivative to get the other stuff anyway, so let's do the second derivative test on this.

$f''(x) = \frac{(-26)(x^2-9)^2 - (-26x)(2)(x^2-9)(2x)}{(x^2-9)^4}$

$f''(x) = \frac{78(x^2+3)}{(x^2-9)^3}$ (after a bit of factoring)

So what is f''(x) for x = 0? $f''(0) = -\frac{26}{81}$. Since f''(0) is negative, the point (0, -4/9) is a relative maximum.

As it happens, the second derivative IS the concavity, so you have already found this.

An inflection point is where f''(x) = 0. So this happens when $x^2 + 3 = 0$, which has no real solutions for x. Thus there are no inflection points.

-Dan


text:
anonymous one year ago help me please

1. anonymous

@JoannaBlackwelder @mathmate @phi

2. anonymous

@sammixboo @ganeshie8

3. anonymous

@isaac4321 @Nnesha

4. anonymous

@ckallerid

5. anonymous

@rishavraj

6. anonymous

@arindameducationusc

7. mathmate

@Tazmaniadevil Do you know the size of the product matrix, the result of a $$A_{3\times3}B_{3\times3}=C_{?\times?}$$

8. anonymous

what what is that

9. mathmate

Have you worked with matrices before?

10. anonymous

i have not

11. mathmate

Here are some videos that you can go through before you work on the problem, or else you would not be learning anything from the exercise: https://www.khanacademy.org/math/precalculus/precalc-matrices https://en.wikipedia.org/wiki/Matrix_(mathematics) The first one is especially good, but you need to be patient and go through all the videos.

12. anonymous

13. anonymous

can you still answer this for me imm in a quiz and i have like 10 mins left

14. anonymous

@mathmate

15. anonymous

@LynFran

16. anonymous

@imqwerty

17. LynFran
18. anonymous

thnx buddy


text:
Document Type : Original Manuscript

Authors

1 Department of Mathematics, University of Semnan, P.O. Box 35195-363, Semnan, Iran.

2 Faculty of Engineering- East Guilan, University of Guilan, P.O. Box 44891-63157, Rudsar, Iran.

Abstract

In this paper we define $\varphi$-Connes module amenability of
a dual Banach algebra $\mathcal{A}$ where $\varphi$ is a bounded $w_{k^*}$-module
homomorphism from $\mathcal{A}$ to $\mathcal{A}$. We are mainly
concerned with the study of $\varphi$-module normal
virtual diagonals. We show that if $S$ is a weakly cancellative
inverse semigroup with subsemigroup $E$ of idempotents, $\chi$
is a bounded $w_{k^*}$-module homomorphism from $l^1(S)$ to $l^1(S)$ and $l^1(S)$
as a Banach module over $l^1(E)$ is $\chi$-Connes module amenable, then it has a $\chi$-module normal virtual
diagonal. In the case $\chi=id$, the converse holds

Keywords

###### ##### References
1. M. Amini, Module amenability for semigroup algebras, Semigroup Forum, 69 (2004), 243–254.
2. M. Amini, A. Bodaghi, D. Bagha and D. Ebrahimi, Module amenability of the second dual and module topological center of semigroup algebras, Semigroup
Forum, 80 (2010), 302–312.
3. M. Amini and R. Rezavand, Module operator amenability of the Fourier algebra of an inverse semigroup, Semigroup Forum, 92 (2014), 45–70.
4. M. Amini and R. Rezavand, Module operator virtual diagonals on the Fourier algebra of an inverse semigroup, Semigroup Forum, 97 (2018), 562–570.
5. M. Lashkarizadeh Bami, M. Valaei and M. Amini, Super module amenability of inverse semigroup algebras, Semigroup Forum, 86 (2013), 279–283.
6. H. G. Dales, A. T. Lau and D. Strauss, Banach algebras on semigroups and their compactifications, Mem. Amer. Math. Soc., 205 (2010), 1–165.
7. H. G. Dales, Banach algebras and automatic continuity, London Math. Soc. Monogr. ser., Clarendon Press (2000)
8. M. Daws, Connes amenability of bidual and weighted semigroup algebras, Math. Scand., 99 (2006), 217–246.
9. M. M. Day, Amenable Semigroups, Illinois J. Math., 1 (1957), 509–544.
10. E. G. Effros , Amenability and virtual diagonals for Von Neumann algebras, J. Funct. Anal., 78 (1988), 137–156.
11. A. Ghaffari and S. Javadi, φ-Connes amenability of dual Banach algebras, Bull. Iranian Math. Soc., 43 (2017), 25–39.
12. B. E. Johnson, Cohomology in Banach algebras, Mem. Amer. Math. Soc., 127
(1972).
13. W. D. Munn, A class of irreducible matrix representations of an arbitrary inverse semigroup, Proc. Glasgow Math. Assoc., 5 (1961), 41–48.
14. T. W. Palmer, Arens multiplication and a characterization of W-algebras, Proc. Amer. Math. Soc. 44 (1974), 81–87.
15. M. A. Rieffel, Induced Banach representations of Banach algebras and locally compact groups, J. Func. Analysis, 1 (1967), 443–491.
16. R. Rezavand, M. Amini, M. H. Sattari and D. Ebrahimi Bagha, Module Arens regularity for semigroup algebras, Semigroup Forum, 77 (2008), 300–305.
17. W. Rudin, Functional Analysis, McGraw Hill, New York, 1991.
18. V. Runde, Lectures on Amenability, Lecture Notes in Mathematics 1774, Springer Verlag, 2002.
19. V. Runde, Amenability for dual Banach algebras, Studia Math., 148 (2001), 47–66.
20. V. Runde, Connes amenability and normal virtual diagonals for measure algebras I, J. London Math. Soc., 67 (2003), 643–656.
21. V. Runde, Connes amenability and normal virtual diagonals for measure algebras II, Bull. Austral. Math. Soc., 68 (2003), 325–328.


text:
Hardcover | $22.95 Trade | £15.95 | ISBN: 9780262072892 | 136 pp. | 5.375 x 8 in | 62 b&w illus.| August 2007 Ebook |$15.95 Trade | ISBN: 9780262251976 | 136 pp. | 5.375 x 8 in | 62 b&w illus.| August 2007

# The Plenitude

Creativity, Innovation, and Making Stuff

## Overview

We live with a lot of stuff. The average kitchen, for example, is home to stuff galore, and every appliance, every utensil, every thing, is compound—composed of tens, hundreds, even thousands of other things. Although each piece of stuff satisfies some desire, it also creates the need for even more stuff: cereal demands a spoon; a television demands a remote. Rich Gold calls this dense, knotted ecology of human-made stuff the "Plenitude." And in this book—at once cartoon treatise, autobiographical reflection, and practical essay in moral philosophy—he tells us how to understand and live with it.

Gold writes about the Plenitude from the seemingly contradictory (but in his view, complementary) perspectives of artist, scientist, designer, and engineer—all professions pursued by him, sometimes simultaneously, in the course of his career. "I have spent my life making more stuff for the Plenitude," he writes, acknowledging that the Plenitude grows not only because it creates a desire for more of itself but also because it is extraordinary and pleasurable to create.

Gold illustrates these creative expressions with witty cartoons. He describes "seven patterns of innovation"—including "The Big Kahuna," "Colonization" (which is illustrated by a drawing of "The real history of baseball," beginning with "Play for free in the backyard" and ending with "Pay to play interactive baseball at home"), and "Stuff Desires to Be Better Stuff" (and its corollary, "Technology Desires to Be Product"). Finally, he meditates on the Plenitude itself and its moral contradictions. How can we in good conscience accept the pleasures of creating stuff that only creates the need for more stuff? He quotes a friend: "We should be careful to make the world we actually want to live in."

Rich Gold (1950-2003) was an artist, composer, designer, inventor, lecturer, and writer. Equally at home in the worlds of avant-garde art, academia, and business, he worked at various times for Sega, Mattel, and Xerox PARC.

## Reviews

"... a wonderful read if you have any interest at all in art or science or design or engineering or creativity or innovation or the morality of our material culture or managing any of those people or processes... Some books are so good that you know you'll read them again, and you know this by the time you finish. With this book, I knew by page 25." Computerworld"—

"Gold displays casual insightssuch as illustrating the sheer abundance of the plenitude by pointing out the variety of shirts in an audience and the work that went into eachand pads this very skinny book with his own goofy cartoons. The result is a fun splash in some of the important ideas behind modern consumption." Publisher's Weekly"—

"Gold's nimble mind unpacks the contradictions and consequences of our stuff-clotted world, quoting a friend's warning: "We should eb careful to make the world we actually want to live in.' This is not the traditional anti-materialist rantand that helps make it a valuable rumination on a prevailing 21st-century condition. The Morning News"—

"This little book, with its simple logic and language and unforgettable, whimsical drawings, will change the way its readers look at the world around them." Susan Salter Reynolds LA Times"—

## Endorsements

"Rich Gold was one of the most creative and unusual minds of our era. His unique vision lives on in this book."Alan Kay"—

"This is a gem that will shape your way of seeing and thinking about the world forever. Rich was one of the true visionaries of Xerox PARC and this unique book, in both its form and content, provides a window into a brilliant and incredibility imaginative mind at work."John Seely Brown , Former Chief Scientist, Xerox Corp and former director of its Palo Alto Research Center (PARC), and co-author of The Social Life of Information"—


text:
# BinaryFormatter serialization methods are obsolete and prohibited in ASP.NET apps

Serialize and Deserialize methods on BinaryFormatter, Formatter, and IFormatter are now obsolete as warning. Additionally, BinaryFormatter serialization is prohibited by default for ASP.NET apps.

## Change description

Due to security vulnerabilities in BinaryFormatter, the following methods are now obsolete and produce a compile-time warning with ID SYSLIB0011. Additionally, in ASP.NET Core 5.0 and later apps, they will throw a NotSupportedException, unless the web app has re-enabled BinaryFormatter functionality.

The following serialization methods are also obsolete and produce warning SYSLIB0011, but have no behavioral changes:

5.0

## Reason for change

These methods are marked obsolete as part of an effort to wind down usage of BinaryFormatter within the .NET ecosystem.

• Stop using BinaryFormatter in your code. Instead, consider using JsonSerializer or XmlSerializer. For more information, see BinaryFormatter security guide.

• You can temporarily suppress the BinaryFormatter compile-time warning, which is SYSLIB0011. We recommend that you thoroughly assess your code for risks before choosing this option. The easiest way to suppress the warnings is to surround the individual call site with #pragma directives.

// Now read the purchase order back from disk
using (var readStream = new FileStream("myfile.bin", FileMode.Open))
{
var formatter = new BinaryFormatter();
#pragma warning disable SYSLIB0011
return (PurchaseOrder)formatter.Deserialize(readStream);
#pragma warning restore SYSLIB0011
}

You can also suppress the warning in the project file.

<PropertyGroup>
<OutputType>Exe</OutputType>
<TargetFramework>net5.0</TargetFramework>
<!-- Disable "BinaryFormatter is obsolete" warnings for entire project -->
<NoWarn>\$(NoWarn);SYSLIB0011</NoWarn>
</PropertyGroup>

If you suppress the warning in the project file, the warning is suppressed for all code files in the project. Suppressing SYSLIB0011 does not suppress warnings caused by using other obsolete APIs.

• To continue using BinaryFormatter in ASP.NET apps, you can re-enable it in the project file. However, it's strongly recommended not to do this. For more information, see BinaryFormatter security guide.

<PropertyGroup>
<TargetFramework>net5.0</TargetFramework>
<!-- Warning: Setting the following switch is *NOT* recommended in web apps. -->
<EnableUnsafeBinaryFormatterSerialization>true</EnableUnsafeBinaryFormatterSerialization>
</PropertyGroup>

For more information about recommended actions, see Resolving BinaryFormatter obsoletion and disablement errors.


text:
#jsDisabledContent { display:none; } My Account |  Register |  Help

# Isotonic regression

Article Id: WHEBN0002836143
Reproduction Date:

 Title: Isotonic regression Author: World Heritage Encyclopedia Language: English Subject: Collection: Publisher: World Heritage Encyclopedia Publication Date:

### Isotonic regression

An example of isotonic regression

In numerical analysis, isotonic regression (IR) involves finding a weighted least-squares fit x\in \Bbb{R}^n to a vector a\in \Bbb{R}^n with weights vector w\in \Bbb{R}^n subject to a set of non-contradictory constraints of the kind x_i \ge x_j.

Such constraints define partial order or total order and can be represented as a directed graph G=(N,E), where N is the set of variables involved, and E is the set of pairs (i, j) for each constraint x_i \ge x_j. Thus, the IR problem corresponds to the following quadratic program (QP):

\min \sum_{i=1}^n w_i (x_i - a_i)^2 \text{subject to }x_i\ge x_j~ \text{ for all } (i,j)\in E.

In the case when G=(N,E) is a total order, a simple iterative algorithm for solving this QP is called the pool adjacent violators algorithm (PAVA). Best and Chakravarti (1990) have studied the problem as an active set identification problem, and have proposed a primal algorithm in O(n), the same complexity as the PAVA, which can be seen as a dual algorithm.[1]

IR has applications in statistical inference, for example, to fit of an isotonic curve to mean experimental results when an order is expected. A benefit of isotonic regression is that it does not assume any form for the target function, such as linearity assumed by linear regression.

Another application is nonmetric multidimensional scaling,[2] where a low-dimensional embedding for data points is sought such that order of distances between points in the embedding matches order of dissimilarity between points. Isotonic regression is used iteratively to fit ideal distances to preserve relative dissimilarity order.

Isotonic regression is also sometimes referred to as monotonic regression. Correctly speaking, isotonic is used when the direction of the trend is strictly increasing, while monotonic could imply a trend that is either strictly increasing or strictly decreasing.

Isotonic regression under the L_p for p>0 is defined as follows:

\min \sum_{i=1}^n w_i |x_i - a_i|^p \mathrm{subject~to~}x_i\ge x_j~ \text{ for all } (i,j)\in E.

## Simply ordered case

To illustrate the above, let x_1 \leq x_2 \leq \ldots \leq x_n, and f(x_1) \leq f(x_2) \leq \ldots \leq f(x_n), and w_i \geq 0 .

The isotonic estimator, g^*, minimizes the weighted least squares-like condition:

\min_g \sum_{i=1}^n w_i (g(x_i) - f(x_i))^2

Where g is the unknown function we are estimating, and f is a known function.

Software has been developed in the R statistical package for computing isotone (monotonic) regression.[3]

## References

1. ^ Best, M.J.; & Chakravarti N. (1990). "Active set algorithms for isotonic regression; a unifying framework". Mathematical Programming 47: 425–439.
2. ^
3. ^ De Leeuw, Jan; K. Hornik; P. Mair (2009). "Isotone Optimization in R: Pool-Adjacent-Violators Algorithm (PAVA) and Active Set Methods". Journal of statistical software 32 (5): 1.

• Robertson, T.; Wright, F. T.; Dykstra, R. L. (1988). Order restricted statistical inference. New York: Wiley.
• Barlow, R. E.; Bartholomew, D. J.; Bremner, J. M.; Brunk, H. D. (1972). Statistical inference under order restrictions; the theory and application of isotonic regression. New York: Wiley.
• Shively, T.S., Sager, T.W., Walker, S.G. (2009). "A Bayesian approach to non-parametric monotone function estimation".
• Wu, W. B.; Woodroofe, M.; & Mentz, G. (2001). "Isotonic regression: Another look at the changepoint problem". Biometrika 88 (3): 793–804.
This article was sourced from Creative Commons Attribution-ShareAlike License; additional terms may apply. World Heritage Encyclopedia content is assembled from numerous content providers, Open Access Publishing, and in compliance with The Fair Access to Science and Technology Research Act (FASTR), Wikimedia Foundation, Inc., Public Library of Science, The Encyclopedia of Life, Open Book Publishers (OBP), PubMed, U.S. National Library of Medicine, National Center for Biotechnology Information, U.S. National Library of Medicine, National Institutes of Health (NIH), U.S. Department of Health & Human Services, and USA.gov, which sources content from all federal, state, local, tribal, and territorial government publication portals (.gov, .mil, .edu). Funding for USA.gov and content contributors is made possible from the U.S. Congress, E-Government Act of 2002.

Crowd sourced content that is contributed to World Heritage Encyclopedia is peer reviewed and edited by our editorial staff to ensure quality scholarly research articles.

By using this site, you agree to the Terms of Use and Privacy Policy. World Heritage Encyclopedia™ is a registered trademark of the World Public Library Association, a non-profit organization.


text:
#### Vol. 15, No. 5, 2022

 Recent Issues
 The Journal About the Journal Editorial Board Editors’ Interests Subscriptions Submission Guidelines Submission Form Policies for Authors Ethics Statement ISSN: 1948-206X (e-only) ISSN: 2157-5045 (print) Author Index To Appear Other MSP Journals
A homogenized limit for the 2-dimensional Euler equations in a perforated domain

### Matthieu Hillairet, Christophe Lacave and Di Wu

Vol. 15 (2022), No. 5, 1131–1167
##### Abstract

We study the motion of an ideal incompressible fluid in a perforated domain. The porous medium is composed of inclusions of size $a$ separated by distances $\stackrel{~}{d}$, and the fluid fills the exterior. We analyze the asymptotic behavior of the fluid when $\left(a,\stackrel{~}{d}\right)\to \left(0,0\right)$.

If the inclusions are distributed on the unit square, this issue was studied recently when $\stackrel{~}{d}∕a$ tends to zero or infinity, leaving aside the critical case where the volume fraction of the porous medium is below its possible maximal value but nonzero. We provide the first result in this regime. In contrast with former results, we obtain an Euler-type equation where a homogenized term appears in the elliptic problem relating the velocity and the vorticity.

Our analysis is based on the so-called method of reflections whose convergence provides novel estimates on the solutions to the div-curl problem which is involved in the 2-dimensional Euler equations.

##### Keywords
ideal incompressible fluid, porous medium, homogenization, method of reflections
##### Mathematical Subject Classification 2010
Primary: 35B27, 35J25, 76B03
Secondary: 35B25, 35C20, 35Q31
##### Milestones
Revised: 10 September 2020
Accepted: 26 November 2020
Published: 29 September 2022
##### Authors
 Matthieu Hillairet Institut Montpelliérain Alexander Grothendieck CNRS, Université de Montpellier Montpellier France Christophe Lacave Université Grenoble Alpes, CNRS Institut Fourier Grenoble France Di Wu School of Mathematics South China University of Technology Guangzhou China


text:
### ACEL, ${g}_{x}$, ${g}_{y}$, ${g}_{z}$

Defines the Earth’s gravitational acceleration. If no parameters are entered (i.e., ACEL), then the following setting is used: ${g}_{x}={g}_{y}=0$ , ${g}_{z}=-9.81m{s}^{-2}$.


text:
# How is a signal multiplied with the window functions?

How the input signal get multiplied with the window functions like Hamming, Hanning, Blackman, etc?

I have tried to multiply the input signal of $10\textrm{ MHz}$ sampled at $100\textrm{ MHz}$ using the Blackmann window function. The output graph is

• How is the multiplication perfomed ?

The input signal is $1\times 64$ row matrix, the Blackman window generates its samples in $64\times 1$ column matrix. The multiplication generates a $64\times 64$ matrix i.e one sample of input signal is multiplied to 64 samples of the window function. For a particular point sample there are 64 sample value. I intend to calculate 128-point FFT with padding zeros.

• So for a particular sample point which value should be considered to calculate FFT?

Here is my code :

N = 64; % signal length (power of 2)
T = 10*(10^-9) ; % sampling period (and rate) is set to 1
A = 1; % sinusoid amplitude
phi = 0; % phase of zero
f = 10*(10^6); % frequency (under Nyquist limit)
nT = [0:N-1]*T; % discrete time axis
w = blackman(N,'periodic');
x1 = A*cos(2*pi*f*nT);
x = w*x1  ; % windowed signal
subplot (311);
stem (x1);
wp = [w;zeros((zpf-1)*N,1)];
subplot (313);
stem (x);
subplot (312);
stem (w);
disp (x1);
disp (w);
disp (x);

• Welcome to SE.DSP! You appear to have done the right thing. What do you think is wrong? Please edit your question with the extra information. – Peter K. Jul 25 '17 at 12:48
• I would say his problem is that he wanted to obtain his original signal weighted by the window, but instead he is obtaining a set of signals corresponding to his window function weighted by each sample of his signal. In my opinion it's just a problem of Matlab code, he did a matrix product instead of a dotted product. – Florent Jul 25 '17 at 23:46
• Husain Asif : you should post your code, I'm pretty sure it's just a simple syntax issue – Florent Jul 25 '17 at 23:46
• @HusainAsif with a dot... A .* B. As said earlier, post your code if you want more than guesses. source – Florent Jul 26 '17 at 4:42
• not here, in your question, with proper formatting please. – Florent Jul 26 '17 at 4:51

Matlab window functions such as hamming(N) returns the $N$ point Hamming window sequence $w[n]$ in a $N \times 1$ matrix (a column vector).

Therefore if your sequence $x[n]$ is represented in a row vector of size $1 \times N$ then for obtaining the sample-by-sample product $x[n]w[n]$, you should either transpose $w[n]$ or the other $x[n]$, so that their dimensions permit the correct sample-by-sample multiplication in between and produce the output sequence $v[n]=w[n]x[n]$ represented in a matrix of the size $N \times 1$ or $1 \times N$ depending on which one you have transposed.

v0 = x.*w      % x: Nx1, w: Nx1, v0: Nx1
v1 = x .* w'   % x: 1xN, w: Nx1, transpose w => v1: 1xN
v3 = x' .* w   % x: 1xN, w: Nx1, transpose x => v3: Nx1

Note that if the transposed signal is complex valued, then you should use the conjugate transpose instead :

v1 = conj(x').* w  % x[n] was complex valued...

Note also that the sample by sample product is perfomed by the $.*$ operation under Matlab;

  v = x .* w    % is a sample by sample product implementing v[n] = x[n]w[n]

If you omit the period and use the $*$ alone, you get matrix product or the dot product in case of vectors involved.

For this case when you have two vectors, $x$ of size $N \times 1$ and $w$ of size $1 \times N$, then their matrix product would produce a matrix of size $N \times N$

v = x * w    % x: N x 1, w: 1 x N => v: N x N