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## Tuesday, January 27, 2009

### Quick text compression

Recently I faced a problem related to short text compression. The idea was to reduce the space needed to save SMS messages. Since each SMS contains up to 160 characters the classic LZW or Huffman methods won't work out-of-the-box, mostly because of the size of the resulting dictionary. It is even worse if we consider that most messages are less than 80 characters long.

Finally I decided to use a fixed dictionary and a sort of hybrid Huffman code compression (which isn't Huffman at all, but it retains some similarity -somehow). Then I started looking for letter, digraph and trigraph probability in words and texts. There are many resources on the net like this and this one where symbol probability is listed for different languages, including Spanish which was the one I used.

The compression algorithm tries to use the minimum number of bits to encode frequent letters/digraphs/trigraphs. It is not optimal since the dictionary is fixed but it does a good job, reducing message size to 60-70% on most cases. If the right text is picked then it can be reduced to 30% but that is cheating, of course! The space character is one of the most used ones along with the 'e' letter (Spanish at least). Trigraphs and digraphs also play an important role in compression.

Lower/uppercase letters is another issue, but since SMS messages are mostly written in lowercase or uppercase that is not a problem. A trick is to invert the whole text to see which text gets the best compression ratio. This is quite fast since the algorithm is simple and the strings are short. Another trick would be to provide more than one dictionary (maybe three or four) and see which one does better with the desired message. The resulting space overhead is about two or three bits which should be acceptable for long messages.

Another possibility is to compress many messages together with Huffman or any other compression method. The drawback is that the message won't be a unit itself and then message management becomes messy.

## Friday, January 16, 2009

### Dirty little bugs - a pin/macro approach

Program testing can be used to show the presence of bugs, but never to show their absence!

Edsger W. Dijkstra

/-

Measuring and observing usually (if not always) change what one is measuring. I've never seen Schrödinger's cat, at least not when leading with embedded systems nor in my garden, but the former implication causes many problems when trying to debug time-critical functions/operations in embedded software and hardware.

Usually a printf-like UART output works but that would slow down or even hang code which can't wait for the UART to stream its data out. Even though with a substantial buffer it will come a time where it will get stuck.

There is a nice, well known techinque, that uses output pins to reflect some state, such as which function or interrupt handler the microcontroller is doing at the time. Since setting the output value requires only a few instructions, this approach suits better the time-critical scenarios. This technique is helpful in many ways: one may measure from interrupt latency and cpu usage to interrupt or thread deadlock.

That's how the following code came up to mind, here is a simple header file which is based on the macros I posted some time ago here.

#ifndef _dbgpin_h_#define _dbgpin_h_#include "portutil.h"/* Enables or disables debugging */#if    DBGPIN_ENABLED 1/* Debug pins */#define DBGPIN_01    0,2#define DBGPIN_02    0,3//---- etc/* Initialization */#define DBGPIN_INIT( pin )    do { FPIN_AS_OUTPUT(pin); FPIN_CLR(pin);  } while(0);/*** Beware of returns, exclude them!*/#if    DBGPIN_ENABLED#define DBGPIN_BLOCK( pin, block ) \           FPIN_SET_( pin ); \           do { \               block \           } while(0); \           FPIN_CLR_( pin );#else#define    DBGPIN_BLOCK( pin, block ) block#endif#endif /* _dbgpin_h_ */

Here is an example:

DBGPIN_BLOCK( DBGPIN_01,a = b;callSomething(););

There is a small drawback: the do-while structure won't let any variable declared inside 'block' to be visible outside. That can be solved by removing the do-while, if you need it that way just go ahead and delete those two lines.

Also notice that a return in 'block' won't allow the macro to clear the corresponding pin, so applying the macro to a function with many return points is messy. The alternative is to change the original function's name and call it through a sort of wrapper which does it inside a DBGPIN_BLOCK macro. It's not pure beauty but it works nice, and once you've done the debugging it can be disabled by changing DBGPIN_ENABLED to 0.

## Sunday, January 4, 2009

### Qt-Embedded and busy-waiting

Qt-embedded is astonishing. I'm currently developing a product which uses embedded linux, color 800x480 LCD and touchscreen. Qt-embedded does really nice, faster than gtk on DirectFB and much better-looking.

However, qt won't show any notification to the user when it is busy, be it refreshing an window, loading a frame, etc. This behaviour is somewhat expected, but on a product I'm developing the user needs to be notified (somehow) that the application is doing alright but is busy, so the [in]famous sand clock jumps in.

Since it's a touchscreen device the mouse cursor was disabled when compiling Qt, and in the end it was better since it looks like only B/W cursors are supported on embedded linux (or at least in this platform?).

I thought of a trick by using QTimer, and later I found out some posts when googling. Here is what I finally did:

#ifndef _busy_notify_h_#define _busy_notify_h_#include <QTimer>#include <QApplication>/*** Class to show a bitmap inside a widget when* the event loop is busy.*/class BusyNotify : QObject{  Q_OBJECTpublic:  // Call to initialise an instance  static void init();  BusyNotify();  /**  * Show the busy cursor,  * it will disappear once the event loop is  * done and the timer times out */  void showBusyCursor();private:  bool waiting;  int screenWidth, screenHeight; //cachedpublic slots:  void TimerTimeout();};// This should be the one usedextern BusyNotify *BusyNotifier;#endif

#include "BusyNotify.h"#include <QPixmap>#include <QPainter>#include <QDesktopWidget>#include <QLabel>BusyNotify *BusyNotifier;static QPixmap *waitPixmap;static QTimer *timer;static QWidget *widget;void BusyNotify::init(){  BusyNotifier  = new BusyNotify();  //load sand clock image  waitPixmap  = new QPixmap( "icons/png/sand_clock.png" );  timer = new QTimer();  timer->setInterval( 100 ); //this could be less..  timer->setSingleShot(false);  Qt::WindowFlags flags = Qt::Dialog;  //flags |= Qt::FramelessWindowHint;  flags |= Qt::CustomizeWindowHint | Qt::WindowTitleHint;  flags |= Qt::WindowStaysOnTopHint;  widget = new QWidget( NULL, flags );  widget->setWindowTitle("Wait...");  QLabel *lbl = new QLabel( widget );  lbl->setText("");  lbl->setPixmap( *waitPixmap );  lbl->setAlignment( Qt::AlignCenter );  widget->resize( waitPixmap->width() + 2, waitPixmap->height() + 2 );  widget->move( screenWidth/2 - widget->width()/2, screenHeight/2 - widget->height()/2 );  connect( timer, SIGNAL(timeout()), BusyNotifier, SLOT(TimerTimeout()) );}BusyNotify::BusyNotify() : QObject(){  waiting = false;  screenWidth = QApplication::desktop()->screenGeometry(0).width();  screenHeight = QApplication::desktop()->screenGeometry(0).height();}void BusyNotify::showBusyCursor(){  if ( !waiting )  {      widget->setVisible(true);      // this will force the widget to show up      // as fast as possible      QApplication::processEvents();      waiting = true;      timer->start();  }}void BusyNotify::TimerTimeout(){  widget->setVisible(false);  timer->stop();  waiting = false;}

After defining the source and header files above one just needs to call BusyNotify::init() once, at program startup, and then BusyNotifier->showBusyCursor() any time the sand clock is needed. 'showBusyCursor()' comes from an old version based on the mouse busy cursor.

How it works

This class works fine when processing or loading (widgets, etc.) is done in the main loop which is also Qt's event processor. When calling showBusyCursor() Qt will be forced to show the sand clock. Then processing can be done and the QTimer will timeout whenever Qt is able to process events once again (since the 100mseg delay is small enough), so the sand clock will remain on the screen as long as needed. Of course, if processing is done in another thread this won't work, but that's not what it is intended for.


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2019
Том 71
№ 4

# Zabutnaya V. I.

Articles: 1
Article (Russian)

### On the best polynomial approximation in the space L2 and widths of some classes of functions

Ukr. Mat. Zh. - 2012. - 64, № 8. - pp. 1025-1032

We consider the problem of the best polynomial approximation of $2\pi$-periodic functions in the space $L_2$ in the case where the error of approximation $E_{n-1}(f)$ is estimated in terms of the $k$th-order modulus of continuity $\Omega_k(f)$ in which the Steklov operator $S_h f$ is used instead of the operator of translation $T_h f (x) = f(x + h)$. For the classes of functions defined using the indicated smoothness characteristic, we determine the exact values of different $n$-widths.

format: plain


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# zbMATH — the first resource for mathematics

On modification of the $$q$$-$$L$$-series and its applications. (English) Zbl 1025.11030
In an earlier paper [Proc. Japan Acad., Ser. A 75, 23–25 (1999; Zbl 0931.11031)] the author introduced a version of the $$q$$-Riemann $$\zeta$$-function. Now he extends his results introducing and studying the $$q$$-Hurwitz $$\zeta$$-function and $$q$$-$$L$$-series (for another approach see J. Satoh [J. Number Theory 31, 346–362 (1989; Zbl 0675.12010)]). For $$q\to 1$$, some relations for the ordinary Dirichlet $$L$$-series are obtained. In particular, a new proof is given for Katsurada’s formula for the values of the Dirichlet $$L$$-series at positive integers [M. Katsurada, Acta Arith. 90, 79–89 (1999; Zbl 0933.11042)].

##### MSC:
 11M41 Other Dirichlet series and zeta functions 11B68 Bernoulli and Euler numbers and polynomials 05A30 $$q$$-calculus and related topics
##### Citations:
Zbl 0931.11031; Zbl 0675.12010; Zbl 0933.11042
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Browse Questions

# A plant which is cultivated by tissue culture technique is

$\begin{array}{1 1}(A)\;\text{Citrus}\\(B)\;\text{Apple}\\(C)\;\text{Pear}\\(D)\;\text{Guava}\end{array}$

Can you answer this question?

A plant which is cultivated by tissue culture technique is citrus
Hence (A) is the correct answer.
answered Jul 15, 2014


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Cash-to-cash cycle

The cash-to-cash cycle represents the amount of time, in days, between the firm paying cash for input materials (e.g., raw materials for a manufacturer; finished resale product for a retailer) and receiving cash from customers. As seen in the formula, the C2C cycle is comprised of three elements: cash used in inventory and in accounts receivable and cash that is free due to nonpayment of accounts payable. Specifically:

$\mbox{C2C Cycle = days funds are used for inventory + days funds are used for receivables + days funds are available due to unpaid payables}\,\!$

This is opertaionalized from an accounting perspective as:

$\mbox{C2C Cycle}=\frac{\mbox{Average inventory level}}{\mbox{CoGs/Days in period}}+\frac{\mbox{Average receivabes}}{\mbox{Sales/Days in period}}+\frac{\mbox{Average payables}}{\mbox{CoGs/Days in period}}$

The C2C cycle value is an indicator of leanness when reduced inventory results from more efficient and effective processes and flows. It is also an indicator of leanness if days sales outstanding are reduced from more efficient processes. For example, a rewrite of ordering systems may reduce billing errors, which in turn reduces disputes and shortens the order-to-cash cycle. Everyday low pricing (EDLP) may also reduce errors and billing disputes. Arbitrarily reducing the credit cycle to customers or extending the payables cycle will reduce the C2C cycle. However, these efforts are not a result of leanness and thus are not viewed as sustainable nor as reflecting improvements in supply chain productivity. The formula for estimating the C2C cycle is provided below.

Example

Suppose that data in the Table was generated from the records of an actual business. The data provides the required input for evaluating the lengh of the cash-to-cash cycle using the above C2C formula. The cycle values for inventory, accounts payable, and accounts receivable are shown separately. The C2C cycle increased from 20,35 to 22,30 during the period. Managers, however, are cautioned that more periods are required to detect trends and that relative performance against industry norms is the key.

$\mbox{January C2C Cycle}=\frac{\mbox{(480+320)/2}}{\mbox{750/31}}+\frac{\mbox{(420+610)/2}}{\mbox{1120/31}}+\frac{\mbox{(-305-200)/2}}{\mbox{750/31}}$

• January = [ (480+320)/2 ]/(750/31) + [ (420+610)/2 ]/( 1120/31 ) + [ (-305-200)/2 ]/(750/31) = 16.53+14.25-10.44 = 20.35
• February =[(320+270)/2]/(705/28) + [(610+580)/2]/(950/28) + [(-200-150)/2]/(705/28) = 11.72+17.54-6.95 = 22.30

$\mbox{January C2C Cycle}=\frac{\mbox{(480+320)/2}}{\mbox{750/31}}+\frac{\mbox{(420+610)/2}}{\mbox{1120/31}}+\frac{\mbox{(-305-200)/2}}{\mbox{750/31}}$

Day / Period 1.1 31.1 28.2
Days in period 31 28
Profit / loss statement Revenue (for the month) 1120 950
Cost of goods sold (for the month) 750 705
Gross margin (for the month) 370 245
Balance sheet items Accounts receivable 420 610 580
Accounts payable -305 -200 -150
Inventory: Raw, in-process, finished goods 480 320 270
Due to inventory = 16.53 11.72
Accounts receivable = 14.25 17.54
Accounts payable = -10.44 -6.95
C2C Total = 20.35 22.30


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# Find the Root of the problem

Calculus Level 3

Consider the polynomial $$f(x) = x^3 + x +1$$ find the real value of $$x$$ such that $$f(x) = 0$$. Answer correct to two decimal places

×

Problem Loading...

Note Loading...

Set Loading...


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## Triangle

Subject: Compulsory Maths

#### Overview

Triangles are three-sided closed figures which have three straight sides joined at three vertices and have three angles enclosed within the figure at the vertices. There are several types of triangles based on the lengths of its sides and the angles they contain. What many do not know is that a triangle is a three-sided polygon.

#### Basic concepts of geometry

The basic concepts of geometry which have been dealt with previous classes and necessary to study in this class are listed below. A definite image which has no length, breadth and thickness is called a point. It is denoted by symbol dot (.) and capital letters A, B, C etc.

Straight line: The shortest locus which passes through two fixed points is called the straight line. It can be extended on both sides.

Line segment: A certain portion of a straight line is called the line segment. It has a fixed measurement.

Curved line: The line joining two fixed points without fixed direction is called the curved line.

Parallel lines: Any two or more than two straight lines which intersect each other after extending or the length of the perpendicular distance between them is always equal are parallel lines.

Angle: When any two straight lines or line segments meet at a point, they form a corner which is called an angle.

### Relation between the pair of angles

A pair of angles having same vertex and a common side is called adjacent angles. If the exterior sides of both angles lie on a straight line then their sum is two right angles.

2 . Vertically opposite angles
When two straight line segments intersect at a point then a pair of angles formed in opposite each other is called vertically opposite angles.

#### Classification of triangles on the basis of sides

1. Equilateral triangles

A triangle whose all sides are equal is known as an equilateral triangle. Each angle of an equilateral triangle are 60o. So, an equilateral triangle is also called as equiangular i.e all there internal angles are also congruent to each other and are each 60o. They are regular polygons.

2. Isosceles triangle

A triangle whose (at least) two sides are equal is called an isosceles triangle. Base angles of an isoceles triangle are equal. An isoceles triangle, therefore, has both two equal sides and two equal angles. In $\triangle$ABC, ∠B = ∠C ∴ $\triangle$ABC is an isosceles triangle.

3. Scalene triangle

A triangle whose all three sides are unequal is called a scalene triangle. In $\triangle$XYZ, none of the sides are equal. ∴ $\triangle$XYZ is a scalene triangle.

#### Classification of triangles on the basis of angles

1. Acute-angled triangle: A triangle whose all angles are acute angles (less than 90o) is called an acute-angled triangle. In the given figure, all angles are less than 90o so it is an acute-angled triangle.

2. Obtuse-angled triangle: A triangle whose one angle is obtuse (greater than 90o) is called an obtuse-angled triangle. In the given figure, ∠Y is 120(greater than 90o) ∴ $\triangle$XYZ is an obtuse-angled triangle.

3. Right-angled triangle: A triangle whose one angle is right angle (90o) is called right-angled triangle. In $\triangle$ABC, ∠B = 90o, so $\triangle$ABC, is a right-angled triangle.

Properties of triangles

• The sum of interior angles of a triangle is two right angles or 180.
• The exterior angle formed by producing a side of a triangle is equal to the sum of the two non-adjacent interior angles.
• The sum of any two sides of the greater angle of a triangle is longer than the opposite side of the smaller angle.
• The base angles of an isosceles triangle are equal
• All angles of an equilateral triangle are equal.

Triangles are three-sided closed figures which have three straight sides joined at three vertices and have three angles enclosed within the figure at the vertices. There are several types of triangles based on the lengths of its sides and the angles they contain. What many do not know is that a triangle is a three-sided polygon.

#### Axioms and Postulates

Axoims

An axiom is a self-evident truth which is well-established, that accepted without controversy or question. Some of the axioms are presented in the following table with their conditions:

 Axoims Conditons Equality of addition When the equal quantities are added to both sides of equal quantities, the sum is also equal. e.g if a=b then a+c=b+c Equality of subtraction When the equal quantites are subtracted from both sides of equal quantities, the difference is also equal. e.g. if a=b then a-c=b-c Equality of multiplication When the equal quantites are multiplied by the same quantity, the product is also equal. e.g. if a=b then a*c=b*c Equality of division When the equal quantites are divided by the same quantity, the quotient is also equal. e.g. if a=b then a/c = b/c Equality axiom When two separate quantites are equal to a quantity the they is also equal to each other, If a=c and b=c then a=c Whole part of axiom A whole quantity is always equal to the sum of its all parts and the whole quantity is always greater than each of its parts. e.g. AD=AB+BC+CD and AD>AB or AD>BC or AD>CD Substitution axiom A quantity can be replaced by another equal quantity. This doesn't alter the final result. If, a=b then ax+c can be expressed as bx+c.

Postulates

A statement which is taken to be true without proof is called a postulate. Postulates are the basic structure from which lemmas and theorems are derived. Some of the postulates that we need in our geometry are listed below.

1. There is only one straight line that you can draw between any two points.

2. An infinite number of straight lines can be drawn through a point.

3. A line contains exactly at least two points.

4. Through any three noncollinear points, there exist exactly one plane.

5. Only one line can be the bisector of a given angle.

6. A straight line can be produced on either side infinitely.

7. Through one point, there is only line parallel to the first line.

8. The length of perpendicular means the distance between a point and a line.

#### Properties of angles when two parallel lines are intersected by a transversal line:

Let's draw at least two parallel lines AB and XY. The transversal line CD intersects AB at E and XY at F. We can establish the following relations among the angles formed:

a) Alternate angles are equal: If two parallel lines are intersected by a transversal line, then the alternate angles so formed are equal. ∴ ∠AEF = ∠EFY and ∠BEF = ∠EFX

b) Corresponding angles are equal: If two parallel lines are intersected by a transversal line, then the corresponding angles so formed are equal.

∴ ∠AED = ∠EFX, ∠AEF = ∠XFC

∠DEB = ∠EFY, ∠BEF = ∠YFC

c) The sum of co-interior angles is 180o: If two parallel lines are intersected by a transversal line, then the sum of the co-interior angles so formed is 180o.

∴ ∠AEF + ∠EFX = 180o

∠BEF + ∠EFY = 180o

d) The sum of co-exterior angles is 180o: If two parallel lines are intersected by a transversl line, then the sum of the co-exterior angles so formed is 180o.

∴ ∠AED + ∠XEC = 180and ∠DEB + ∠YEC = 180o

#### Theorem 1 : The sum of the angles of any triangle is equal to two right angles.

Experimental Verification:

Step1: Let's draw three triangles of different orientations sizes. And name them as $\triangle$ABC each. (Draw the triangles in such a way that you can measure each angle by using a protractor.)

Step 2: Measure each angle of these three triangles with the help of protractor and fill up the following table.

 Fig. ∠A ∠B ∠C Results (i) ∠A +∠B +∠C = (ii) ∠A +∠B +∠C = (iii) ∠A +∠B +∠C =

Conclusion :

Theoretical proof:

Given: ∠ABC, ∠BCA and ∠BAC are three angles of triangle ABC.

To proof: ∠ABC + ∠BCA + ∠BAC = 180o

Construction: Let's draw a straight line parallel to BC through A.

Proof:

 S.N. Statements Reasons 1 ∠XAB = ∠ABC XY $\parallel$ BC and being alternate angles. 2 ∠YAC = ∠ACB XY $\parallel$ BC and being alternate angles. 3 ∠XAB + ∠BAC + ∠YAC = ∠XAY According to whole-part axiom. 4 ∠ABC + ∠BAC + ∠ACB = ∠XAY From statement (1), (2) and (3). 5 ∠XAY = 1800 A straight angle constitutes 180o (two right angles) 6 ∴ ∠ABC + ∠BAC + ∠ACB = 180o or 2 right angles. Substituting the value of ∠XAY into a statement (4).

Proved

#### Theorem 2: The exterior angle formed by extending a side of a triangle is equal to the sum of other two non-adjacent angles.

Experimental verification:

Step 1: Let's draw three triangles of different size in different orientations. Produce side BC to D in each figure. In each figure, ∠ACD is an exterior angle and ∠A and ∠B are two non-adjacent angles inside the triangle.

Step 2: Measure the size of exterior angle and other two non-adjacent angles in each figure and complete the table.

 Fig. Exterior angle ∠ACD ∠A ∠B Results (i) (ii) (iii)

Conclusion:

Theoretical proof:

Given: ABC is a triangle whose side BC is extended upto D such that ∠ACD ia an external angle.

To prove: ∠ACD = ∠ABC + ∠BAC

Proof:

 S.N. Statement Reasons 1 ∠ACD + ∠BCA = 180o The sum of adjacent angles in a straight line is 180o. 2 ∠ABC + ∠BCA + ∠BAC = 180o The sum of angles in a triangle is 180o. 3 ∠ACD + ∠BCA = ∠ABC + ∠BCA + ∠BAC From statements (1) and (2). 4 ∠ACD = ∠ABC + ∠BAC Cancelling ∠BCA on both sides of the statement (3).

Proved
Alternative method:

Given: ABC is a triangle whose side BC is extended up to D such that ∠ACD in an external angle.

To prove: ∠ACD = ∠ABC + ∠BAC

Construction: Let us draw CE parallel to BA.

Proof:

 S.N. Statements Results 1 ∠ACD = ∠ACE + ∠DCE Whole part axiom. 2 ∠DCE = ∠ABC CE $\parallel$ BA being corresponding angles. 3 ∠ACE = ∠BAC CE $\parallel$ BA being alternate angles. 4 ∠ACD = ∠BAC + ∠ABC From statements (1), (2) and (3).

Proved

#### Verification of properties of Isosceles triangle

Isosceles triangle is that triangle whose two sides are equal. Triangle is symmetric being two sides equal hence it possesses different properties. For example, the base angles of an isosceles triangle are equal. This kind of properties is proved as theoretical proof here which duly needs the conditions of congruency of triangles. Before this, we discuss the different conditions of congruency of triangles in a brief.

### Congruency of triangles

There are 3 sides and 3 angles in a triangle. Two triangles are congruent when 3 parts (out of 6 parts) of one triangle are equal to 3 corresponding parts of another triangle

as per the following conditions. We take these conditions as axioms.

#### i) S.S.S. axiom:

The two triangles are said to be congruent when three sides of a triangle are respectively equal to three corresponding sides of another triangle under S.S.S. axiom.

In $\triangle$ABC and $\triangle$MNO ,

a) AB = MN (S)

b) BC = NO (S)

c) AC = MO (S)

∴ $\triangle$ABC ≅ $\triangle$MNO [S.S.S. axiom]

Corresponding parts of congruent triangles are also equal. i.e. ∠A =∠M, ∠B = ∠N and ∠C =∠O.

#### ii) S.A.S. axiom:

The two triangles are said to be congruent when two sides and an angle made by them of a triangle are respectively equal to the corresponding sides and an angle of another triangle, under S.A.S. axiom.

In $\triangle$ABC and $\triangle$PQR,

a) AB = PQ (S)

b) ∠B = ∠Q (A)

c) BC = QR (S)

∴ $\triangle$ABC = $\triangle$PQR [S.A.S. axiom]

Now, ∠C = ∠R and ∠A = ∠P [corresponding angles of congruent triangles]

AC = PR [Corresponding sides of congruent triangles]

#### iii) A.S.A. axiom:

The two of are said to be congruent when two angles and their adjacent side of one triangle are respectively equal to the corresponding angles and side of another triangle, under A.S.A. axiom.

In $\triangle$ABC and $\triangle$DEF,

a) ∠B = ∠E (A)

b) BC = EF (S)

c) ∠C = ∠F (A)

∴ $\triangle$ABC ≅ $\triangle$DEF [A.S.A. axiom]

Now, AC = DF and AB = DE

[Correspondong sides of congruent triangles]

∠A =∠D [Corresponding angles of congruent triangles]

#### iv) R.H.S. axiom:

The two right-angled triangles are said to be congruent when the hypotenuse and one of the remaining sides of both triangles are respectively equal, under R.H.S. axiom.

In right angled $\triangle$ABC and $\triangle$MNO,

a)∠B = ∠N (R)

b) AC = MO (H)

c) BC = NO (S)

∴ $\triangle$ABC ≅ $\triangle$MNO [ R.H.S. axiom]

Now, ∠C = ∠O and ∠A = ∠M [Corresponding angles of congruent triangles]

AB = MN [Corresponding sides of congruent triangles]

#### v) S.A.A. axiom:

The two triangles are said to be congruent when two angles and a side of one triangle are respectively equal to the corresponding angles and side of another triangle, under S.A.A. axiom. This axiom can be verified by using A.S.A. axiom.

Here,

a) ∠A = ∠D [Given]

b) ∠B = ∠E [Given]

c) ∠A + ∠B + ∠C = ∠D + ∠E + ∠F [Sum of the angles of any triangle is 180o]

d) ∠C = ∠F [From (a), (b) and (c)]

Now, in $\triangle$ABC and $\triangle$DEF,

e) ∠B = ∠F (A) [Given]

f) BC = EF (S) [Given]

g) ∠C = ∠F (A) [From (d)]

h) $\triangle$ABC ≅ $\triangle$DEF [A.S.A. axiom]

Now, AB = DE and AC = DF [Corresponding sides of congruent triangles]

### Isosceles triangle

#### Theorem 3:

Experimental verification:

Step 1: Draw three isosceles triangle ABC with different shapes and sizes in different orientation where AB = AC.

Step 2: Measure the angles opposite to the equal sides of each triangle and tabulate them in the table.figure

Conclusion:

Theoretical proof:

Given: $\triangle$ABC is an isosceles triangle where AB = AC.

To prove: ∠ABC = ∠ACB

Construction: Let us draw AD⊥BC, from the vertex A.

Proof:

 S.N. Statement Reasons 1. In $\triangle$ABD and $\triangle$ACD i) ∠ADB = ∠ADC (R) Both angles are right angle. ii) AB = AC (H) Given iii) AD = AD (S) Common side 2. ∴ $\triangle$ABD ≅ $\triangle$ACD By RHS axiom. 3. ∠ABD = ∠ACD i.e. ∠ABC = ∠ACB Corresponding angles of congruent triangles are equal.

Proved

#### Theorem 4: Conversion of theorem 3

Experimental verification:

Step 1: Draw three line segments BC of different lengths in different positions.

Step2: Equal sizes of angles at both points B and C on each line segment is drawn. Mark the points as A where the arms of these angles meet. Now three $\triangle$ABC are formed.

Step 3: Measure the length of each side opposite to the equal angles in each triangle (i.e. AB and AC) and tabulate them in the table.

figure

Conclusion:

Theoretical proof:

Given: In $\triangle$ABC, base angles are equal i.e. ∠B =∠C.

To prove: AB = AC, $\triangle$ABC is an isosceles triangle.

Construction: From the vertex A, let us draw AD⊥BC.

Proof:

 S.N. Statement Reasons 1. In $\triangle$ABD and $\triangle$ACD i) AD = AD (S) Common side ii) ∠ADB = ∠ADC (A) By construction, AD⊥BC, both angles are equal. iii) ∠ABD = ∠ACD (A) Given 2. ∴ $\triangle$ABD ≅ $\triangle$ACD By S.A.A. axiom. 3. AB = AC i.e. $\triangle$ABC is an isosceles triangle. Corresponding sides of the congruent triangles are equal.

Proved

#### Theorem 5:

Experimental verification:

Step 1: Draw three isosceles triangle ABC with different positions and sizes such that AB = AC in each triangle.

Step 2: Draw the bisector of the vertex angle ∠A in each triangle. The bisector meets BC at D.

Step 3: Measure the lengths of BD and DC and the angles ADB and ADC, then tabulate them in the table.

Conclusion:

Theoretical proof:

Given: $\triangle$ABC is an isosceles triangle where AB = AC. AD is the bisector of∠BAC.

To prove: AD⊥BC and BD = DC.

Proof:

 S.N. Statement Reasons 1. In $\triangle$ADB and $\triangle$ACD i) AB = AC (S) Given ii) ∠BAD = ∠CAD (A) Given iii) AD = AD (S) Common side 2. ∴ $\triangle$ABD ≅ $\triangle$ACD S.A.S. axiom 3. ∠ADB =∠ADC Corresponding angles of congruent triangles are equal. 4. AD⊥BC Equal adjacent angles in a linear pair mean the line is perpendicular. 5. BD = DC Corresponding sides of congruent triangles are equal. 6. AD is the perpendicular bisector of BC. From statement 4 and 5.

Proved

#### Theorem 6: Converse of theorem 5

Experimental verification:

Step 1: Draw three isosceles triangles ABC with different positions and size in a different position such that AB = AC in each triangle.

Step 2: Mark the mid-point of BC at D. Join A and D in each triangle.

Conclusion:

Theoretical proof:

Given: $\triangle$ABC is an isosceles triangle where AB = AC.

AD joins the vertex A and mid-point D of the base BC.

 S.N. Statements Reasons 1. In $\triangle$ADB and $\triangle$ACD i) AB = AC (S) Given ii) AD = AD (S) Common side iii) BD = DC (S) Given 2. ∴ $\triangle$ABD ≅ $\triangle$ACD By S.S.S. axiom 3. ∠ADB = ∠ADC Corresponding angles of congruent triangles are equal. 4. AD⊥ BC Each angle of adjacent angles in a linear pair are 900 both. 5. ∠BAD = ∠CAD Corresponding angles of congruent triangles are equal.

Proved

#### Experiment no.1: The sum of two sides of a triangle is greater than the third side.

Experimental verification:

Step 1: Three triangles ABC with different positions and sizes in different orientation are drawn.

Step 2: Measure all three sides of each triangle and fill up the table.

 Figure AB BC CA AB + BC BC + CA AB + CA Result i) ii) iii)

Conclusion:The sum of two sides of a triangle is greater than the third side.

#### Experiment no.2: In any triangle, the angle opposite to the longer side is greater than the angle opposite to the shorter side.

Experimental verification:

Step 1: Three triangles ABC with different positions and size in different orientations are drawn in such a way that BC is the longest and CA is the shortest side in each triangle.

Step 2: The angle opposite to the greater side BC (i.e.∠B) are measured and tabulate in the table:

 Figure Angle opposite to BC:∠A Angle opposite to CA:∠B Result i) ii) iii)

Conclusion: In any triangle, the angle opposite to the longer side is greater than the angle opposite to the shorter side.

#### Experiment no.3: In any triangle, the side opposite to the greater angle is longer than the side opposite to the smaller angle.

Experimental verification:

Step 1: Three triangle ABC with different positions and sizes in different orientations are drawn in such a way that∠A is the greatest and∠B is the smallest angles in each triangle.

Step 2: Measure the side BC opposite to the greatest single angle A and the side CA opposite to the smallest angle B and fill up the table below:

 Figure Angle opposite to ∠A (BC) Angle opposite to ∠B (CA) Result i) ii) iii)

Conclusion: In any triangle, the side opposite to the greater angle is longer than the side opposite to the smaller angle.

#### Experimental no.4: Among all straight line segments down to a given line from a point outside it, the perpendicular is the shortest line.

Experimental verification:

Step 1: Three straight line segments XY of different length in different orientations are drawn.A point P is taken outside of each line segment. Three line segments PA, PB, PC and a perpendicular line PM are drawn from P to XY.

Step 2: Measure the lengths of each line segment PA, PB, PC and PM. Then tabulate them below:

 Figure PA PB PC PM i ii iii

Conclusion: Among all straight line segments down to a given line form a point outside it, the perpendicular is the shortest line.

#### Experiment no.5: The square of the hypotenuse in a right-angled triangle is equal to the sum of squares of its base and perpendicular.

Step 1: Draw three right-angled triangles of different dimensions.

Step 2: Fill up the given table.

 Figure AB BC CA AB2 BC2 CA2 AB2 + BC2 Results i) ii) iii)

Conclusion:The square of hypotenuse in right-angled triangle is equal to the sum of squares of it's base and perpendicular.

Experiment no.5 is known as Pythagoras Theorem. Around 2500 years ago, Pythagoras was a Greek Mathematician. He invented a fact on the right-angled triangle which is named as Pythagoras Theorem. This theorem is used in all other fields of mathematics not only used in geometry.

##### Things to remember

The angle is mainly divided into the following types:

• Acute angle: less than 90
• Right angle: exactly 90
• Obtuse angle: more than 90 but less than 180
• Straight angle: exactly 180
• It includes every relationship which established among the people.
• There can be more than one community in a society. Community smaller than society.
• It is a network of social relationships which cannot see or touched.
• common interests and common objectives are not necessary for society.
##### Triangle angle example 1 | Angles and intersecting lines | Geometry | Khan Academy

Here , y = 56o + 4o
$\therefore$ y = 105o [exterior angle in a traiangle is equal to sum of two no - adjacent interior angles]

Again , x + 56o+49o = 180o [sum of three interior angle of a triangle is 180o]

$\therefore$ x = 180o - 56o = -49o
= 180o - 105o = 75o
Hence , x = 75oand y = 105o

Here , a + 75o+ 60o= 180o [sum of three angles of traingles]
or , a + 135o = 180o
or , a = 180o - 135 = 45o

Here , b + 60o = 180 [straight angle]
$\therefore$ b = 180o - 60o = 120
Hence , a = 45o and b = 120o

Here , a + 51o + 60o = 180o [ sum of three angle of a triangle is 180o]
or , a = 180o - 51o - 60o = 1800 - 111o = 69o
Again , b = 60o + 51o = 111o [being exterior angle and sum of opposite interior angles of a triangle]
Hence , a = 69o and b = 111o

Here , AOB = DOC = 75o[∴AOB and DOC are vertically opposite angles]
Again , p+ 68o+75o =180o
or , p + 143o = 180o
or , p = 180o - 143o = 37oAns.
Now , in ABO
or , q+ 75o + P = 180o
or , q + 75o + 37o = 108o
$\therefore$ q = 180o - 112o = 68o

Here , x = 61o + 82o = 143o [being exterior angle and sum of two opposite interior angle]
Again , y + 82o = 180o
$\therefore$ y = 180o - 82o = 98o
Hence , x = 143oand y = 98o

InΔ BCE
p = 85o + 30o = 180o [sum of three interior angles of a triangle]
or , p + 115o = 180o
or , p = 180o - 115o = 65o
Here , ABE = 85o + 30o = 115o [In , BCe exterior angle = sum of two non-adjacent interior angle]

Now , inΔABO ,
q + ABO + 46o = 180 [The sum of three interior angle of a triangle]
or , q + 115o + 46o = 180o[ABO = ABE = 115]
or , q + 161o = 180o
or , q = 180o - 1610 = 19o
Hence , p = 65o and q = 19o

Here , AED = ACB [corresponding angles DE || BC]
or , x = 35
Now , in ΔABC
y+80+x= 180 [sum of all interior angle of a triangle]
or , y + 35 + 80 = 180
or , y = 180 - 80 - 35 = 180 - 115 = 65
Hence , x = 35 and y = 65

Here , ODC = OAB [alternate angles and AB||CD]
or , OCD = 38
Now , in DCO
x + 81+ODC = 180 ]sum of all three angles of a triangle]
or , x + 81 + 38 = 180
or , x + 119 = 180
or , x = 180 - 119 = 61 Ans.

In DOC ,
70 + DOC = DCE [ being exterior angle and sum of two opposite interior angle]
or , DOC = 150 - 70 = 80
Now , AOB = DOC = 80 [Vertically oppostie angle]
Again , in ABO ,
x + 63 + AOB = 180[sum of three angles of a triangle]
or , x + 63 + 80 = 180
x + 143 = 180
or , x = 180 - 143 = 37 Ans.

Here , EAB = AEC [in aec , exterior angle = sum of two non-adjacent interior angle]
or , x = 37
Again , DCA = AEC + CAE [ In? AEC , exterior angle = sum of two non-adjacent interior angle]
or , 63 = 37 + y
or , y = 63 - 37 = 26.
Here , ACB = 90 [from given]
Now , In?ACB
CAB+ACB+CBA = 180 [sum of three interior angle of triangle]
or , x + y + 90 + z = 180
or , 37 + 26 + 90 + z = 180,
or , 153 + z = 180
or , z = 180 - 153 = 27
Hence , x = 37 , y = 26 and z = 27

Given : AP = AQ and PB = QC
To prove ABC is an isoceles triangle
i.e. AB = AC

 Statements Reasons In APB and AQC1. PB = QC2 . APB = AQC3. AP = AQ4. APB≅AQC5. AB = AC6. ABC is an isoceles traingle. 1. Given2. Base anglen of an isoceles angle APQ3.Given4. S.A.S Axiom5. Corresponding sides of congruent triangles are equal.6. From statement (5)

Given: In isoceles triangle ABC , BO and CO bisect ABC and ACB respectively.
To prove: BOC is an isoceles triangle i.e. BO = CO
Proof :

 Statements Reasons 1. ABC = ACB2. OBC = $\frac{1}{2}$ ABC and OCB = $\frac{1}{2}$ACB3. OBC = OCB4. OB = OC $\therefore$ BOC is an isoceles triangle. 1. Base angle of an isoceles triangle.2. OB and OC bisect ABC and ACB respectively.3. From (1) and (2) , i.e. half of the equal angles4. From statement (3) , i.e. base angles are equal

Given :
In PQR ,PS⊥QR , QPS = 45 and QPR = 90
To prove: PQR is an isoceles triangle.

 Statements Reasons 1. PQS + QPS = 90or , PQS + 45 = 90or , PQS = 90 - 45 = 452. QPR = 903. QRP +PQR =90or , QRP + 45 =90QRP = 90 - 45 = 454. PQ = PR5. PQR is an isoceles triangle. 1. In right angled triangle PSQ , sum of acute angle is one right angle. 2. Given. 3. In PQR , sum of acute angles is one right angle 4. From statement (1) and (3) , i.e. base angles of PQR are equal. 5. From statement (4)

 Statements Reasons 1. a = 60 Each angles of a equilateral triangle is 60. 2. b + a + 53 = 180or , b + 60 + 52 = 180or , b = 180 - 60 - 53 = 67 2. Sum of the all angles made at a point E of st.line AD and lying on one side or AD. 3 . EBA = b = 67or , B = 76 3. Being base angles of an isoceles triangle EAB. 4. In EAB , c + b + B = 180 or , c + 67 + 67 = 180or , c = 180 \ 67 - 67 = 46 4. Sum of interior angles of a triangle. 5. C = DCE = 53 5. Base angle of all of a isoceles triangle i.e DE = DC 6. d + 53 = 180or ,d + 53 + 53 =180d = 180 - 53 - 53 = 74a = 60 , b = 67 , c = 46 and d = 74 6. The sum of all three interior angles of a triagle is 180

 Statements Reasons 1. BEH + 140 = 180BEH = 180 - 140 = 140 1.Sum of linear pairs. 2. f = BEH = 40 2. Alternate angles and AC||GI 3. d = BHE = 56 3. ALternate angles an AC||EI 4. f + e + d = 180 4. From the reason 1
 5. 40 + e + 56 = 180e = 180 - 96 = 84e = 84 , f = 40 and d = 56 5. From the statement 2 and 3

Given : In ABC , CD and BE are perpendicular drawn from the point B an C on Ab and Ac respectively as well as BE = CD
To proveABC is an isoceles triangle , i.e , AB= AC.

 Statements Reasons 1. BE = CD 1. Given 2. AEB = ADC 2. Both are right angles 3. EAB = DAc 3. Common angles 4. AEB≅ ADC 4. By S. A. S facts 5. Ab = AC 5. Correspondind sides of congruent triangles AEB an ADC

Given :D is the mid point of a line segment AB , EF is the line passing through the mid-point D. AX and BY are perpendiculars drawn from A and B on EF respectively i.e. AX ⊥ EF and BY ⊥ EF.

 Statements Reason 1. In ΔAXD and ΔBYD(i). AD = BD(ii). ADX = BDY(iii). AXD = BYD 1(i) being D as mid point (ii). vertically opposite angle(iii). Both are right angles 2. ΔAXD≅ΔBYD 2. BY S.A.A fact 3. AX = BY 3. Corresponding sides of congruent triangle.

Given , InΔABC , D is midpoint of BC. From the midpoint D , DF⊥AB and DE⊥AC are drawn where Df = DE.
To prove , Δ ABC is an isoceles triangle.
or ,AB = AC

 Statements Reasons DFB = DEC both are right angle BD = DC From given , being D is mid point. BF = DE Given ΔBFD≅CED By RHS axiom DBF = DCE Corresponding angles of congruent triangles AB = AC Fom statement (5) , base angles of ΔABC are equal.

 Statements Reasons 1. InΔAOB and ΔOCD(i) OAB = OCD(ii) AO = OD(iii) BOA = COD 1. (i). Both are right angle(ii) Given(iii) Vertivally Opposite angles are equal. 2. ΔABO≅ΔOCD 2. A.S.A fact 3. AB = CD $\therefore$ Breadth of river = CD 3. Corresponding sides of congruent.


text:
# Long pathnames truncated in results window

0 votes

When matches in the results window have very long pathnames, it is not easy to identify which files are in which folder, as the path gets truncated. Adjusting the Location column width is not possible as there is not enough room on the screen.
Is there any way to view the whole path (aside from viewing the Reports tab, which is not ideal as those paths are not hyperlinked to the actual file)?
If not, please raise a feature request to (optionally) display the full path, maybe over two lines or some other wrapping technique.
Thanks.

asked

## 1 Answer

0 votes

There are various options for dealing with long path names in the File path display preference settings, although no multi-line wrapping option.

• Don't truncate/Truncate end - End of path is truncated, e.g. C:\folder\sub\... Although the display might still be truncated if you hover over the field it should display the full value of the location.
• Truncate start - First part of path is truncated, e.g. ...\sub\end
• Truncate middle - Middle part of path is truncated, e.g. C:\folder\...\end
answered by (62k points)
Sort of.  It would be more useful to be able to see the whole path somewhere.
Also, although "truncate start" and "truncate middle" do what they're supposed in the result pane, unlike "truncate end" and "do not truncate", the hover over doesn't display anything.
As a compromise, is there any chance the hover over could be changed to show the whole path, possibly as multiple lines, regardless of which truncate option is used? The current hover over is limited by the screen width.


text:
How to obtain data for Benford's Law analysis?

First off, let me be specific as to what I mean by "Benford's Law analysis": I'm looking to test, given some data, if the set of first significant digits are independent, identically distributed Benford random variables. As such, I would prefer the following:

• The order in which the data is presented matters e.g. the data is in chronological order.
• Reasonable expectation that the first significant digits might be IID.

Thus, for instance, doing this particular analysis on some sequence of population growth (1000 one year, 2000 the next, 4000 the year after that, etc.) would not be helpful, as certainly we would not expect those first significant digits to be independent of each other.

The only thing that comes to mind that meets the parameters I have set would be a chronologically ordered sequence of withdrawals/transactions from an account. But then how to obtain such data? I sure hope this question doesn't come across as lazy.

Below is a short list of data sites that may be useful for this purpose. This list is not a comprehensive list by any means. Not all of the data on these sites will be chronological, but I think there are many datasets in these few sites that could be useful for testing Benford's.

http://testingbenfordslaw.com/

https://www.gapminder.org/data/

https://www.theguardian.com/data

https://www.data.gov/

https://www.kaggle.com/datasets

https://www.ncdc.noaa.gov/data-access

Also, using Google's new Dataset Search yield's a bunch of interesting tests with links to the datasets.

$P(d) = log_b(d+1) - log_b(d) = log_b(1 + \frac1d)$
For other number bases $b$ where $b\geq1$ in case you wanted to test it on a dataset that is not decimal. The number set satisfies Benford's if the leading digit $d$ $(d \in \{1, ..., b-1\})$ occurs according to the general form.


text:
# The number of states for fermions, bosons, and Boltzman in statistical mechanics

This is related with Equation 8.58 in Kerson Huang's 2nd edition of Statistical Mechanics.

The partition functions for the ideal gases are given as $Q_N (V,T) =\sum_{\{ n_p \}} g\{n_p \}e^{-\beta E\{n_p \}}$ where $E\{n_p \} =\sum_p \epsilon_p n_p$ and the occupation numbers are subject to the $\sum_p n_p =N$

The textbook says, for a bose gas and a boltzmann gas $n_p=0,1,2, \cdots$ and for the fermi gas $n_p=0,1$. I understand this is due to the pauli-exclusion principles. But what i couldn't understand is eq 8.58 which is following

$g\{n_p\} = 1$ for both bose and fermi gas

$g\{n_p\} = \frac{1}{N!} \left( \frac{N!}{\prod_p n_p !} \right)$ for boltzaman gas

Please explain why we have this results.

Bose particles cannot be identified as different in a given state, whereas boltzmann particles can (even though both types can occupy a given energy state with more than one particle). Thus boltzmann statistics need to take into account the permutations ($n!$) of the $n$ particles into a given state, in contrast to the bose particles (which are not identified as separate).

For example for Bose-Einstein statistics the configurations:

(ike, mike) and (mike, ike) are considered the same.

Whereas for Maxwell-Boltzmann statistics the configurations:

(ike, mike) and (mike, ike) are considered different (each particle "ike" or "mike" can be identified as "ike" or "mike").

wikipedia Identical particles

• How about fermi gas? still the $g\{n_p\}$ are given as same in bose gas. Also in bose gas you mean that $\left( \frac{N!}{\prod_p n_p !}\right) =1$? Oct 23 '14 at 14:30
• for fermi-dirac particles, you already stated the answer, in each given state there can be either $1$ or $0$ due to pauli-exclusion principle. It all breaks down to particle identification in quantum mechanics (see the link) (plus it is not clear what exactly the factors mean in your question, but in any case this is the core of the answer) Oct 23 '14 at 14:33
• Here i am considering the degeneracy $g\{n_p \}$ of given particles. I understand your points, but i still got confused why both boson and fermion have degeneracy 1. Oct 23 '14 at 14:41
• @phy_math, great made a small update to elucidate further Oct 23 '14 at 14:50


text:
## Introduction

Unlike other climate change drivers, the temperature and cloud response—and in some cases, the precipitation response1—to absorbing aerosol is dominated by rapid adjustments.2 Recent modeling studies suggest that these adjustments, which include the aerosol–cloud semi-direct effect (SDE), lead to a negative radiative perturbation, thereby offsetting some of the positive direct forcing of absorbing aerosol, resulting in a relatively weak surface temperature response.2,3,4 Climate models from the Precipitation Driver Response Model Intercomparison Project (PDRMIP) show that a tenfold increase in BC leads to a robust increase in globally averaged low-level cloud, but a reduction in mid-level and in particular, high-level cloud.2 These responses are consistent with the simulated vertical co-location of absorbing aerosol and cloud, including relative humidity reductions and cloud burn-off when BC and cloud are co-located,5,6 but also enhanced low-level cloud due to increases in stability when BC is located above the cloud.7,8,9,10,11 Qualitatively similar cloud responses are obtained in several other recent studies.12,13,14 Consistent with these cloud changes, and in particular the large high-cloud reduction and enhanced longwave cooling, the annual global mean rapid adjustment in these simulations ranges from −0.39 to −1.44 W m−2 for a tenfold increase in BC;2,12 −2.8 W m−2 for 25 × BC;14 and −0.25 W m−2 for present-day BC emissions.13 Thus, model simulations indicate that rapid adjustments—primarily due to cloud responses—offset up to 60% of the direct radiative forcing associated with BC.2

The climate impact of absorbing aerosols, however, is associated with significant uncertainty.15,16 This is related to several factors, including underestimation of BC emission inventories and absorption aerosol optical depth, which implies too little aerosol solar absorption in the atmosphere and therefore, underestimation of BC direct radiative forcing.15,17,18,19,20,21 The vertical profile of absorbing aerosol is also important, as it impacts the direct radiative forcing, as well as the rapid adjustments and precipitation response.9,15,18,22,23,24,25 In the context of the aerosol–cloud SDE, different cloud responses are obtained depending on the vertical co-location of absorbing aerosol and cloud.9,15,26

Large inter-model diversity exists in simulated BC vertical profiles, with models likely overestimating the amount of BC aloft.15,18,27,28 This could be related to a variety of factors, including vertical resolution,29 convective processes,28,30 scavenging,31 or aging.32 Thus, given these uncertainties in aerosol simulations—particularly related to the vertical profile of absorbing aerosol—the corresponding aerosol–cloud SDE may not be properly constrained by the aforementioned studies.

Aerosol observations, including the Cloud–Aerosol Lidar with Orthogonal Polarization (CALIOP)33 allow the the aerosol vertical profile to be constrained. Here, we estimate the aerosol–cloud SDE using three different climate models (Methods), including the Community Atmosphere Model version 4 (CAM4)34 and CAM5,35 as well as the Geophysical Fluid Dynamics Laboratory (GFDL) Atmospheric Model version 2.1 (AM2.1),36 forced with monthly observationally constrained fine-mode aerosols without dust and sea salt37 (Methods; STANDARD simulations). Unlike the anthropogenic fraction, the fine-mode fraction of aerosol amount—which is largely anthropogenic—can be constrained by observations. We also conduct analogous observationally constrained aerosol simulations, but with [a.] vertically uniform aerosol atmospheric heating (VERTUNIF simulations) and [b.] half the forcing (i.e., monthly surface reduction in solar radiation and atmospheric solar heating reduced by 50%; HALF simulations). These simulations are compared with PDRMIP 10xBC aerosol simulations (Methods). We do not attempt to quantify interannnual variability, instead focusing on the climatological mean signal. However, there is a considerable year-to-year variability in biomass burning, which implies similarly large internannual variability in the aerosol–cloud semi-direct effect.

## Results

### Observationally constrained fine-mode aerosols

Figure 1 shows the annual mean atmospheric solar absoprtion (FATM) and reduction in surface solar radiation (FSFC) for observationally constrained fine-mode aerosols, without dust and sea salt. The maximum atmospheric heating occurs over central Africa (due to biomass burning), India (due to biofuel burning and fossil fuel combustion), and China (due to fossil fuel combustion). The sunlight absorption by aerosols comes largely from BC, but brown carbon (or organic aerosols) also contributes to the absorption. Figure 1 also shows the vertical profile of the atmospheric solar heating rate response in CAM4 STANDARD (other models are similar). Most of the heating occurs near the surface, peaking at ~925 hPa, and then rapidly decaying to zero near ~500 hPa. Although aerosol simulations generally reproduce a similar spatial distribution of FSFC and FATM, their vertical aerosol heating profile is more uniform, with relatively large heating that extends through the upper troposphere.2 Based on PDRMIP 10xBC simulations, the atmospheric solar heating response remains relatively large up to 100 hPa, with most models exhibiting a secondary peak near 200 hPa (Fig. 1c). The percent of average shortwave heating above 500 hPa, relative to the column average, ranges from 70% in CAM5-PDRMIP to 110% in HadGEM2, with a multi-model PDRMIP mean of 86%. This is considerably larger than that based on the observationally constrained heating profile, at 2% (by design, VERTUNIF yields 100%). Moreover, the lower-tropospheric heating maximum is elevated relative to observationally constrained fine-mode aerosols (~850 vs. 925 hPa). These large differences in simulated versus observed vertical aerosol heating profile imply corresponding differences in the aerosol–cloud SDE (i.e., cloud adjustment).

### Cloud responses

Figure 2a shows global annual mean vertical profiles of the cloud response to observationally constrained fine-mode aerosol forcing without dust and sea salt (STANDARD). Weak increases occur near the surface (especially in CAM4), which then give way to much larger decreases that span the entire troposphere, extending from ~950 to 250 hPa. Consistently, all models show global annual mean cloud reductions for low-, mid-, and high-level clouds (Methods), with the largest decreases in low- and mid-level clouds (Table 1). The annual mean low cloud response varies between −0.43% in GFDL to −0.14% in CAM4. The annual mean mid-level (high) cloud response varies from −0.26% in CAM4 to −0.21% in GFDL (−0.04% in GFDL to −0.12 % in CAM5). Figure 2b shows that these cloud responses are consistent with the corresponding change in relative humidity (RH). All three models show weak increases in RH near the surface, which then transitions into larger decreases in RH throughout most of the troposphere.

Across the three models, similar spatial cloud responses exist in several regions. Figure 3 shows low clouds generally increases over areas with large atmospheric heating, including Africa and parts of southeast Asia (Supplementary Fig. 1 shows model agreement). Low-cloud reductions occur over Indonesia, the Indian Ocean, and parts of the Northern Hemisphere mid- and high latitudes. Generally, similar signed cloud responses occur over both land and ocean, except for high-level cloud, which increases over land, but decreases over ocean in all models. The bulk of this high cloud increase occurs over central Africa, and is associated with an increase in vertical motion and convective mass flux, particularly in the upper troposphere (Supplementary Fig. 2). Although convective mass flux is generally reduced in response to the stabilizing effects of aerosol atmospheric heating—particularly in the lower troposphere—the large amount of atmospheric heating over Africa (Fig. 1) acts to destabilize the mid- and upper troposphere, resulting in rising air and an increase in high cloud. To a lesser extent, this is also the case over parts of southeast Asia.

### Mechanisms for cloud changes

All three models yield an increase in lower-tropospheric stability (S), primarily due to warming at 700 hPa (Supplementary Table 1; Supplementary Fig. 2). This increase in stability traps moisture near the surface, leading to a corresponding increase in near-surface RH and cloud, including significant increases in cloud and relative humidity at the model’s bottom level. CAM4 yields the largest increase in cloud at the model’s bottom level (Table 1), which is consistent with its S-based marine stratocumulus parameterization (Methods). However, above the near-surface, decreases in lower-tropospheric RH are consistent with an overall decrease in low cloud. For example, RH925–700 decreases between −0.19 to −0.37% in GFDL and CAM4, respectively (Supplementary Table 1). Similarly, decreases in mid- and upper-tropospheric RH are consistent with decreases in mid and high cloud. RH600–500 decreases between −0.09 to −0.25% in GFDL and CAM4; RH300–250 decreases between −0.01 and −0.22% in GFDL and CAM5, respectively. A more detailed regression analysis (Supplementary Discussion; Supplementary Table 2) shows that RH is generally the best predictor, with S (vertical motion) also important for the low cloud (mid and high cloud) response (Supplementary Fig. 3). Thus, observationally constrained aerosol forcing yields overall cloud reductions, most of which is driven by decreases in RH and subsequent cloud burnoff.

The vertical profile of simulated climatological clouds differs in our models (Supplementary Fig. 4). For example, CAM5 shows the largest amount of cloud throughout the atmospheric column; GFDL shows the least amount of low-level cloud; and CAM4 shows the least amount of mid-level cloud. However, there does not seem to be a clear connection between the cloud response, and the amount of climatological cloud, implying parameterization differences (Methods) likely drive the bulk of any contrasting cloud responses in these models.

### Semi-direct cloud responses and rapid adjustments

Figure 4a–c shows the aerosol–cloud SDE (Methods) for all three STANDARD simulations. All models yield a positive global annual mean SDE. The largest SDE occurs in GFDL at 0.52 W m−2, followed by CAM5 and CAM4 at 0.30 and 0.29 W m−2, respectively (Table 1). Similar (but smaller in magnitude) results exist for HALF (simulations with 50% less aerosol forcing). For example, GFDL yields an SDE of 0.30 W m−2, followed by CAM5 and CAM4 at 0.17 and 0.14 W m−2, respectively (Table 1). Thus, HALF simulations generally yield an SDE about 50% as large as in STANDARD. Several regions of SDE model agreement exist, including the Indian Ocean, most of south America, and parts of the NH mid-latitudes (Fig. 5a). Moreover, the SDE is dominated by shortwave effects, which drive the positive SDE; longwave effects act to mute the positive SDE. These changes are consistent with the corresponding changes in clouds. Low clouds tend to have a net cooling effect on the planet, due to high albedo; high clouds tend to have a net warming effect, due to their greenhouse effect. Mid-level clouds are a combination of both, but we find they tend to affect the SDE similarly to low-level clouds. Thus, the decrease in low- and mid-level cloud drives the positive SDE (in the shortwave), and the decrease in high-level clouds weakens this through enhanced longwave cooling. We also note that the rapid adjustments are positive in all models (Table 1), and dominated by the aerosol–cloud SDE. Although non-cloud adjustments (e.g., changes in temperature and moisture) also contribute to the rapid adjustments, this contribution is smaller (−0.09 to 0.03 W m−2 in STANDARD) than that due to cloud adjustments.

In contrast, several recent studies have shown the rapid adjustments (RAPADJ) to absorbing aerosol are negative.2,4 PDRMIP 10xBC aerosol simulations yield a multi-model mean rapid adjustment of −0.94 W m−2, ranging from −0.55 to −1.48 W m−2, similar to previously published values.2 Additional analyses reveals that the rapid adjustment is dominated by the longwave component, with a multi-model mean of −1.05 W m−2, ranging from −0.45 to −2.24 W m−2 (Supplementary Table 3). The corresponding multi-model mean shortwave component is 0.12 (−0.22 to 0.76) W m−2. The importance of longwave radiation to the rapid adjustment is consistent with the corresponding cloud changes (Fig. 2c). PDRMIP models yield increases in low cloud (0.07; 0.03 to 0.11%), decreases in mid-level cloud (−0.17; −0.50 to 0.02%), and most importantly, relatively large decreases in high cloud (−0.37; −0.79 to −0.09%), which is consistent with the negative RAPADJ–LW. Furthermore, PDRMIP cloud responses are consistent with the corresponding RH change (Fig. 2d). Although PDRMIP models show smaller decreases in high cloud over land (including increases over Africa and parts of India/southeast Asia), as compared with ocean, high cloud decreases over both ocean and land.

Unfortunately, most PDRMIP models did not archive the relevant diagnostics to quantify the importance of the aerosol–cloud SDE to the rapid adjustment. Calculation of the shortwave cloud forcing due to absorbing aerosol using standard techniques leads to a positive bias (in the shortwave), by increasing the radiative warming when the absorbing aerosol lies above clouds.38 Thus, it is necessary to use either the clean sky cloud radiative forcing or the clear-sky instantaneous radiative forcing (IRF), and only four models archived these diagnostics (Methods). These four PDRMIP 10xBC simulations yield an aerosol–cloud SDE ranging from −0.41 to −1.26 W m−2, with a multi-model mean of −0.97 W m−2 (Table 1). Moreover, the SDE is dominated by longwave radiation, with SDELW ranging from −0.35 to −1.57 W m−2, with a multi-model mean of −0.82 W m−2. Thus, the aerosol–cloud SDE is a relatively large negative number in PDRMIP 10xBC simulations, dominated by longwave radiation. This agrees with the aforementioned cloud changes, and in particular, the large decrease in high cloud (Supplementary Table 3). Although this conclusion is consistent with a recent PDRMIP 10xBC study using alternative methodologies,4 some discrepancies related to the magnitude of the cloud (and non-cloud) adjustment exist (Methods).

### Sensitivity simulations

To test the hypothesis that the cause of the SDE sign difference using observationally constrained aerosols, relative to PDRMIP and other aerosol simulations, is related to the vertical aerosol heating profile (Fig. 1), we perform sensitivity simulations with observationally constrained aerosol forcing, but we apply it uniformly in the vertical (VERTUNIF simulations). Figure 2 shows that VERTUNIF yields very different results relative to STANDARD, and moreover, VERTUNIF yields cloud changes very similar to PDRMIP. Relatively large increases in cloud and RH occur near the surface, extending up to ~800 hPa. Above ~800 hPa, decreases in both cloud and RH exist, with relatively large reductions above ~300 hPa. Consistently, VERTUNIF yields increases in low cloud, decreases in mid-level cloud, and large decreases in high cloud (Table 1; Supplementary Fig. 6). Several regions of cloud response model agreement exist, particularly for high cloud (Supplementary Fig. 1). These changes are again consistent with ΔRH (Supplementary Discussion). Relative to STANDARD, VERTUNIF RH925–700 now increases and RH300–250 yields larger and more consistent decreases (Supplementary Table 1). In turn, the effective radiative forcing (ERF) swaps sign from positive to negative, as does the rapid adjustment and the aerosol cloud SDE (which still dominates the rapid adjustment). GFDL VERTUNIF yields an aerosol–cloud SDE of −0.41 m−2, which increases to −0.67 m−2 in CAM4 (Fig. 4d–f). SDE model agreement (Fig. 5b) also not only shows the dominance of a negative SDE, most notably over India and Southeast Asia, but also parts of Africa, Canada, and the North Atlantic. Moreover, the SDE is now dominated by longwave effects (cooling), consistent with the large decrease in high-level cloud. These results are also qualitatively consistent with other studies13,25 that found weaker decreases in high cloud and a less negative SDE when BC above 200 and 500 hPa was restricted.

## Discussion

Using observationally constrained fine-mode aerosol forcing, we find a robust positive aerosol–cloud SDE, that acts to warm the climate system. This response is consistent with the corresponding cloud changes, including relatively large decreases in low- and mid-level cloud. Aerosol simulations, however, yield the opposite response—a negative SDE, primarily due to large decreases in high cloud and enhanced longwave cooling. These differences are related to the vertical profile of aerosol atmospheric heating, with aerosol simulations yielding a relatively vertically uniform profile, with strong heating that extends up to 100 hPa. The negative SDE and the corresponding cloud changes are qualitatively reproduced with a simple idealized simulation forced with observationally constrained aerosols, but with vertically uniform atmospheric heating.

We note that our approach has uncertainties of its own (Methods), including lack of consistency between the aerosol forcing and simulated meteorology, as well as the simulated cloud fields. The fine-mode aerosol radiative forcing is also subject to observational errors, including aerosol optical depth and aerosol optical properties, such as the single-scatter albedo.39,40 Sensitivity tests were performed to quantify this uncertainty in observationally constrained fine-mode aerosol forcing,37 resulting in an atmospheric heating uncertainty range of ±0.25 W m−2. Uncertainty in CALIOP’s aerosol vertical profile also exists, including attenuation of the lidar signal. For example, optically thick aerosol layers above African biomass burning can completely attenuate the signal, leading to no retrievals below these levels.41 Our use of CALIOP’s total aerosol vertical distribution, regardless of aerosol type, also leads to additional uncertainty in the absorbing aerosol vertical profile.

Our simulations show a large cloud response over southern Africa and southeast Atlantic (Fig. 3). Some studies have argued that over these regions (and other regions where absorbing aerosols are present), there are potential meteorological influences that confound aerosol–cloud interactions, such as the SDE.42,43,44,45,46 For example, the presence of mid-tropospheric moisture and the southern Africa easterly jet correlate with the increase in the absorbing aerosols, and these meteorological changes appear to impact the cloud development over the southeast Atlantic, independent of the aerosol effect.45,46 Although we have not attempted to formally separate dynamical versus thermodynamical impacts, we do show that cloud changes are generally consistent with the traditionally defined SDE and changes in relative humidity and stability (e.g., Fig. 2; Supplementary Table 2). Nonetheless, a better understanding of how aerosol-induced changes in meteorology affect cloud responses is warranted.

Our focus has been to quantify the aerosol–cloud semi-direct effect based on anthropogenic aerosols, which is why the emphasis is on fine-mode aerosols without dust and sea salt. Although sea salt is a non-absorbing aerosol, dust is a weakly absorbing aerosol and thus has the potential to affect clouds via the semi-direct effect.10,26,47 Moreover, some dust is anthropogenic in nature due to agricultural activities and land-use change.48 Furthermore, climate change itself has likely perturbed dust emissions,49,50 which also implies an anthropogenic contribution to dust. Our analysis does not account for possible dust-related SDE effects.

Although our observationally constrained aerosol forcing contains more atmospheric warming than model estimates, we find similar conclusions—including a positive SDE—with smaller atmospheric heating (HALF simulations). This is consistent with the importance of the vertical aerosol atmospheric heating profile, as opposed to its magnitude. Furthermore, global models may not be able to simulate sub-grid- scale features (e.g., cloud-layer thickness, cloud-top entrainment) that may influence the SDE. However, they should adequately simulate the large-scale features important for the SDE, including changes in stability and relative humidity, and modification of the large-scale atmospheric circulation. Thus, despite the caveats of our approach, we conclude that aerosol simulations lead to a negatively biased SDE. We also suggest that this bias may translate into other aspects of the climate response, including surface temperature and precipitation.

## Methods

### Global climate models

This study uses the National Center for Atmospheric Research (NCAR) Community Atmosphere Model version 4 (CAM4)34 and CAM5.35 Both models have a horizontal resolution of 1.9° × 2.5°; CAM4 has 25 vertical layers and CAM5 has 30. Their main shared physical parameterization is the Zhang-McFarlane deep convection scheme (bulk mass flux with CAPE closure).51 CAM4 uses a shallow convection scheme that involves three-level adjustment of moist static energy52 and a prognostic single-moment microphysics scheme, including diagnostic cloud fraction.53 Cloud fraction depends on several factors, including RH, S, water vapor, and convective mass fluxes. Three types of cloud are diagnosed: low-level marine stratus, convective cloud, and layered cloud. Layered clouds form when RH exceeds a pressure-dependent threshold. Marine stratocumulus clouds are diagnosed using an empirical relationship based on S. Convective cloud fraction is related to updraft mass flux in the deep- and shallow-cumulus schemes. The remaining cloud types are diagnosed on the basis of relative humidity.34

CAM5 uses a mass flux scheme with convective inhibition closure for shallow convection54 and a prognostic double-moment microphysics scheme55 with ice supersaturation56 and a diagnostic cloud fraction scheme for cloud microphysics and macrophysics. Although deep-cumulus cloud fraction is diagnosed as in CAM4, shallow cumulus fraction in CAM5 is directly computed using the definition of convective updraft mass flux from the new shallow convection scheme. Liquid stratus fraction is derived from the assumed triangular distribution of the total relative humidity. The sensitivity of liquid stratus fraction to the changes of grid-mean RH differs between the two models. Ice stratus fraction is also diagnosed, using a modified version of Slingo’s formula.

The rapid radiative transfer model (RRTMG) provides the radiative transfer calculations in CAM5, which is an accelerated and modified version of the correlated k-distribution model, RRTM.57,58,59 RRTMG divides the solar spectrum into 14 shortwave bands that span 0.2 μm to 12.2 μm. A two-stream δ-Eddington approximation is used and assumes homogeneously mixed layers while accounting for absorption and scattering in the calculation of reflectance and transmittance. Distinction is made between the direct solar beam and scattered radiation. The calculation of shortwave radiation in CAM4 is based on a δ-Eddington approximation.60,61,62 The solar spectrum is divided into 19 intervals, and the δ-Eddington solution for the reflectivity and transmissivity for each atmospheric layer under clear and cloud skies is calculated. The layers are then combined together, accounting for multiple scattering between layers, allowing upward and downward spectral fluxes to be estimated.

We also use the Geophysical Fluid Dynamics Laboratory (GFDL) Atmospheric Model version 2.1 (AM2.1),36 with a horizontal resolution of 2° × 2.5° and 24 vertical layers. Moist convection is represented by the Relaxed Arakawa–Schubert formulation.63 In this parameterization, convection is represented by a spectrum of entraining plumes which produce precipitation. Closure is determined by relaxing the cloud work function for each cloud in the spectrum back to a critical value over a fixed time scale. Large-scale clouds are parameterized with separate prognostic variables for specific humidity of cloud liquid and ice, with an updated treatment of mixed-phase clouds.64,65 Stratocumulus cloud cover is based on large-scale subsidence, diabatic cooling by radiation, and turbulent entrainment of warm and dry air from above the inversion.66 Shortwave radiation is based on an exponential sum fit with 18 spectral bands.67 The δ-Eddington method is used to solve for the layer reflection and transmission, while the thick-averaging method is used to combine layers.

Low-, mid-, and high-level clouds are directly calculated by each model. CAM4/5 assumes a maximum-random overlap, with low clouds based on pressure levels from the surface pressure to 700 hPa; mid-level clouds are based on 700–400 hPa; and high-level clouds are based on 400 hPa to the model top. For GFDL AM2.1, clouds are assumed to randomly overlap. Low clouds are calculated over 1000–680 hPa; mid-level clouds from 680 to440 hPa and high clouds from 440 to 10 hPa. For PDRMIP models, these cloud diagnostics were not available for all models. Thus, low-, mid-, and high-level clouds are approximated as simple vertical averages of the cloud fractions over given pressure levels.2 Low clouds are based on 1000–680 hPa; mid-level clouds are based on 680–440 hPa; and high clouds are based on 440–50 hPa. These are the same pressure levels used by ISCCP to define low-, mid-, and high-level clouds.

### PDRMIP simulations

Ten global climate models simulated a control experiment using year 2000 greenhouse gas and aerosol concentrations or emissions, and an experiment in which anthropogenic BC concentration or emissions were increased by a factor of ten (10xBC).2 Five models fixed their control BC concentrations to the monthly multi-model mean present-day concentration of AeroCom Phase II; the 10xBC experiment used a tenfold increase of this anthropogenic BC concentration. In the remaining five models, the control simulation was performed using present-day BC emissions, and the 10xBC experiment used a tenfold increase in the anthropogenic BC emissions. CESM-CAM4, GISS-E2-R, IPSL-CM5A, and NorESM1 are concentration driven; CanAM4, CESM-CAM5, HadGEM2-ES, and MIROC-SPRINTARS are emissions driven. Generally, present-day emissions refer to the year 2000; however, in the case of HadGEM2-ES, the year 1860 was used (the 10xBC experiment used a tenfold increase in anthropogenic BC emissions from the year 2000). All PDRMIP simulations analyzed here are based on an atmosphere-only setup, where sea surface temperatures are prescribed. Most simulations were performed for 15 years. The fast response of a tenfold increase in BC is obtained by taking the difference between the 10xBC experiment and the corresponding control simulation.

CESM1-CAM5 10xBC responses shown here are based on new simulations, analogous to PDRMIP. These new integrations include the double-radiation calls to estimate the instantaneous radiative forcing (which were missing from the original CESM1-CAM5-PDRMIP 10xBC simulation). The HadGEM2-ES instantaneous radiative forcing of 3.55 W m−2 also comes from a second ensemble member of the PDRMIP 10xBC experiment, and is somewhat different than that previously published (3.29 W m−2).2 This difference is due to internal variability and associated meteorological differences and aerosol transport between the two HadGEM2-ES ensemble members.

### Experimental design

We conduct experiments with fixed sea surface temperature (fSST), which uses a repeating cycle of monthly climatological SSTs. The fSST setup prohibits significant global mean temperature change while keeping fast atmospheric responses, like the SDE. Experiments are run for 20 years, and the last 15 years are used in this study. Monthly observationally constrained fine-mode aerosol radiative effects (atmospheric heating rate and surface solar radiation reduction) are interpolated to each model’s horizontal resolution and incorporated into their radiation modules. The atmospheric heating rate is vertically interpolated to each model’s hybrid pressure levels. Although aerosol forcing is almost independent of solar zenith angle (θ) when the angle is small, aerosol forcing approaches zero as θ approaches 90°. Thus, the added aerosol radiative effect is multiplied by a scaling factor that depends on zenith angle.8,68 The climate response is estimated as the difference between the simulation with observationally constrained fine-mode aerosol (without dust and sea salt), and a corresponding control runs without observationally constrained fine-mode aerosol (STANDARD signal). The idealized vertically uniform heating simulations are analogous, but the monthly mean aerosol atmospheric heating is vertically averaged at each grid box, and this vertically averaged value is prescribed to all pressure levels >100 hPa. The climate response of vertically uniform heating (VERTUNIF signal) is obtained by taking a difference with the same control simulation. We also perform simulations analogous to STANDARD, but with half the aerosol forcing (HALF). At each grid box, the (monthly) reduction in surface solar radiation and atmospheric solar heating are reduced by 50%. Significance of all climate responses is based on a Student’s t test for the difference of means, using the pooled variance.

In the case of the CAM4 and GFDL models, which only include aerosol-direct effects, the radiative effects of the default aerosols are neglected (in both observationally constrained aerosol and control simulations) by removing them from the list of radiatively active species. With CAM5, which includes both direct and indirect aerosol effects, the radiative effects of the default aerosols cannot be simply neglected. Thus, CAM5 simulations (observationally constrained and control) use prescribed (as opposed to prognostic) modal aerosols. Although this represents a double counting of aerosol effects in the observationally constrained CAM5 simulation, the response (observationally constrained minus control) will remove the bulk of the radiative effects due to the default prescribed aerosols (which will be similar in observationally constrained and control simulations).

### Observationally constrained aerosol forcing

Aerosol optical depth (AOD) is a popular measure of aerosol amount, and known to be the most reliable global-scale aerosol product. However, the ground-based AErosol RObotic NETwork (AERONET) locations are heterogeneously distributed over the planet, while less reliable satellite AODs have nearly global coverage. Satellite AOD from the Moderate Resolution Imaging Spectroradiometer (MODIS) and Multi-angle Imaging SpectroRadiometer (MISR) is therefore nudged toward AERONET AOD to obtain a globally reliable AOD from 2001 to 2010.69,70 The AOD Angstrom exponent is also derived by adjusting the satellite data toward AERONET data. Fine-mode aerosol optical depth (fAOD) at 500 nm is obtained by using AERONET fAOD and the total AOD to derive the fine-mode fraction (FMF). The AOD Angstrom exponent data are converted into the FMF data, which are then nudged toward AERONET FMF data to derive reliable FMF and fAOD over the globe. Observational data gaps—which are primarily confined to polar regions—are filled by the Goddard Chemistry Aerosol Radiation and Transport (GOCART) model. GOCART accurately simulates most of the prominent AOD features in the satellite observations, within a factor of two for aerosol source and outflow areas.71 However, several GOCART biases have been identified, including an underestimation of aerosol extinction over India, overestimation of aerosol extinction in dust source regions, and overestimation of aerosol aloft over mid-latitude transport regions.72 Aerosol optical properties, such as the single-scatter albedo (SSA) and asymmetry parameter (ASY), are obtained by nudging GOCART SSA and ASY toward AERONET data.

Aerosol vertical profiles are obtained from the space-borne Cloud–Aerosol Lidar with Orthogonal Polarization (CALIOP).33 CALIOP is able to distinguish between dust and other type of aerosols, which generally do not depolarize light, resulting in six aerosol categories, including dust, marine, smoke, polluted dust, polluted continental, and clean continental.73 However, these aerosol categories were not used—the total aerosol vertical distribution was applied regardless of aerosol type. Daytime CALIPSO Level 2.0 data from June 2006 to October 2011 were processed to obtain a clear-sky aerosol extinction coefficient at 532 nm at the T42 spatial resolution and 500-m vertical resolution of the Monte Carlo Aerosol Cloud Radiation (MACR) model.74,75 The data gaps are filled using available neighboring data through linear interpolation. A climatological seasonal cycle for the entire data period is then constructed. The aerosol vertical profile from CALIPSO is scaled to match the AOD observations obtained by integrating AERONET, MODIS, and MISR data.37 To adjust the magnitude of AOD over cloud by CALIPSO data, 5° × 2° June 2006 to January 2012 globally gridded monthly mean daytime CALIPSO lidar Level 3.0 data76 is used, specifically, the ratio of clear-sky AOD to above-cloud AOD. This is used to modify the aerosol amount over cloud at each grid cell in the MACR model. For coarse-mode aerosols (which are not used here), the same vertical profiles are used. Because coarse-mode aerosols are not very absorbing, the effect on the vertical profile is very small.75

The direct aerosol effect is obtained by incorporating the integrated global aerosol data into an updated version of the MACR model.37 MACR is based on the Monte Carlo Independent Column Approximation (McICA) and uses satellite observations to describe multilayer cloud, surface albedo, and stratospheric column ozone, as well as ERA-Interim Reanalyzes for precipitable water. The 2001–2010 surface albedo and stratosphere column ozone come from the Clouds and the Earth’s Radiant Energy System (CERES) Synoptic Radiative Fluxes and Clouds (SYN) monthly product. Clouds come from the 2001–2010 averages of the merged daytime monthly mean CERES ISCCP-D2-like product, which is based on combined geostationary cloud retrievals and cloud retrievals from Terra/Aqua MODIS.75 Clouds were combined into four types: low, mid, high, and convective clouds. Between the four types, a random overlap scheme was applied. The key element of this updated version is the Monte Carlo radiative transfer solver combined with the Monte Carlo integration in the multi-dimensional parameter space, including solar zenith angles, aerosol–cloud configurations, and spectral bands. Despite monthly resolution of the three-dimensional model inputs, MACR was run for all 365 days of the year to address variations in the solar zenith angle, declination, and eccentricity of Earth’s orbit. For each day, 40,000 photons were deployed, and monthly inputs were interpolated in time and the model output was averaged over each month. The benefits of the Monte Carlo approach for radiative transfer is that it provides more accurate flux values relative to two stream approximations for both clear and cloudy skies.77

The global average atmospheric heating (FATM) and reduction in surface solar radiation (FSFC) for fine-mode aerosols without dust and sea salt, which include both natural and anthropogenic aerosols, is +3.64 W m−2 and −3.75 W m−2, respectively (−0.11 W m−2 TOA forcing). These estimates are several times larger than anthropogenic forcings estimated from models,78 which are 0.75 W m−2 for FATM and −1.02 W m−2 for FSFC. This large difference is consistent with model underestimation of absorbing aerosol, including black carbon optical properties and emissions, as well as omission of absorbing brown carbon.15,17,18,19,20,21 Furthermore, the observationally constrained FATM and FSFC are larger over land than oceans. FATM is 2.67 (5.82) W m−2 over ocean (land) and FSFC is −2.99 (−5.47) over ocean (land). Despite the relatively large aerosol forcing in STANDARD, we obtain similar conclusions—including cloud responses and a positive SDE—in our reduced aerosol-forcing simulation (HALF). This is consistent with the importance of the vertical aerosol atmospheric heating profile, as opposed to its magnitude.

We note that although this approach uses observations to constrain the fine-mode aerosol radiative effect, the results are subject to observation errors. AERONET SSA, in particular, is subject to uncertainties due to various assumptions used in the retrieval algorithms. Climatological AERONET SSA differs from in situ measured SSA by up to 0.02, with underestimation in low Ångström exponent areas and overestimation in high Ångström exponent areas.40 The maximum error of AERONET AOD is between 0.01 and 0.015 for Level 2.0 data.39 A direct validation of AERONET absorption AOD (AAOD) with unmanned aerial vehicle observations shows that AERONET AAOD errors are less than 20%.79

Sensitivity tests were performed to quantify the uncertainty (primarily due to AOD and SSA) in the observationally constrained fine-mode aerosol forcing used here.37 Fine-mode aerosols yield atmospheric absorption of 3.88 W m−2. This varies from 3.63 W m−2 (least absorbing case) to 4.08 W m−2 (most absorbing case), yielding an uncertainty range of about ±0.25 W m−2. These uncertainty estimates are based on modifying BC AOD and BC/organic aerosol (OA) SSA. Three different sets of simulated SSA are used:

$$\begin{array}{c}{\mathrm{SSA1}} = \left( {0.19{\mathrm{xBC}}\,{\mathrm{AOD}} + 0.85{\mathrm{xOA}}\,{\mathrm{AOD}} + {\mathrm{sulfate}}\,{\mathrm{AOD}}} \right.\\ \left. { +\; {\mathrm{sea - salt}}\,{\mathrm{AOD}} + {\mathrm{0}}{\mathrm{.96xdust}}\,{\mathrm{AOD}}} \right){\mathrm{/total}}\,{\mathrm{AOD}};\end{array}$$
(1)
$$\begin{array}{c}{\mathrm{SSA2}} = \left( {0.14{\mathrm{xBC}}\,{\mathrm{AOD}} + {\mathrm{0}}{\mathrm{.80xOA}}\,{\mathrm{AOD}} + {\mathrm{sulfate}}\,{\mathrm{AOD}}} \right.\\ \left. { +\; {\mathrm{sea - salt}}\,{\mathrm{AOD}} + {\mathrm{0}}{\mathrm{.96xdust}}\,{\mathrm{AOD}}} \right){\mathrm{/total}}\,{\mathrm{AOD;and}}\end{array}$$
(2)
$$\begin{array}{c}{\mathrm{SSA3}} = \left( {{\mathrm{0}}{\mathrm{.19xBC}}\,{\mathrm{AOD}} + {\mathrm{0}}{\mathrm{.98xOA}}\,{\mathrm{AOD}} + {\mathrm{sulfate}}\,{\mathrm{AOD}}} \right.\\ \left. { +\; {\mathrm{sea - salt}}\,{\mathrm{AOD}} + {\mathrm{0}}{\mathrm{.96xdust}}\,{\mathrm{AOD}}} \right){\mathrm{/total}}\,{\mathrm{AOD}}{\mathrm{.}}\end{array}$$
(3)

BC AOD above refers to the GOCART BC AOD at 550 nm. Parameters (e.g., 0.19 for BC SSA) in the above equations come from observational studies.80,81 In addition, in SSA2 (more absorbing case), the magnitude of BC AOD was doubled, given that simulated BC may be significantly underestimated.15 These are then nudged toward the AERONET SSA, giving three sets of semi-observational SSA.37 In terms of fine-mode dust and sea salt, with a baseline atmospheric heating estimate is 0.23 W m−2, sensitivity tests yield a corresponding uncertainty range from 0.16 to 0.26 W m−2. These sensitivity tests are based on using GOCART versus GISS ModelE2 fine-mode dust and sea salt.37

Uncertainty in the absorbing aerosol profile as derived from CALIPSO also exists, including attenuation of the lidar signal. For example, optically thick aerosol layers above African biomass burning can completely attenuate the signal, leading to no retrievals below these levels.41 Although our observationally constrained aerosol vertical profile is not based on individual CALIPSO aerosol categories, such attenuation can also lead to misclassification of aerosol type by the classification algorithm.82,83 However, our use of CALIPSO’s total aerosol vertical distribution, regardless of aerosol type, also leads to additional uncertainty in the absorbing aerosol vertical profile. This uncertainty is much smaller than the overall differences between the CALIPSO profile and simulated BC profiles.


text:
#### Solution for 12125 is what percent of 3:

12125:3*100 =

(12125*100):3 =

1212500:3 = 404166.67

Now we have: 12125 is what percent of 3 = 404166.67

Question: 12125 is what percent of 3?

Percentage solution with steps:

Step 1: We make the assumption that 3 is 100% since it is our output value.

Step 2: We next represent the value we seek with {x}.

Step 3: From step 1, it follows that {100\%}={3}.

Step 4: In the same vein, {x\%}={12125}.

Step 5: This gives us a pair of simple equations:

{100\%}={3}(1).

{x\%}={12125}(2).

Step 6: By simply dividing equation 1 by equation 2 and taking note of the fact that both the LHS
(left hand side) of both equations have the same unit (%); we have

\frac{100\%}{x\%}=\frac{3}{12125}

Step 7: Taking the inverse (or reciprocal) of both sides yields

\frac{x\%}{100\%}=\frac{12125}{3}

\Rightarrow{x} = {404166.67\%}

Therefore, {12125} is {404166.67\%} of {3}.

#### Solution for 3 is what percent of 12125:

3:12125*100 =

(3*100):12125 =

300:12125 = 0.02

Now we have: 3 is what percent of 12125 = 0.02

Question: 3 is what percent of 12125?

Percentage solution with steps:

Step 1: We make the assumption that 12125 is 100% since it is our output value.

Step 2: We next represent the value we seek with {x}.

Step 3: From step 1, it follows that {100\%}={12125}.

Step 4: In the same vein, {x\%}={3}.

Step 5: This gives us a pair of simple equations:

{100\%}={12125}(1).

{x\%}={3}(2).

Step 6: By simply dividing equation 1 by equation 2 and taking note of the fact that both the LHS
(left hand side) of both equations have the same unit (%); we have

\frac{100\%}{x\%}=\frac{12125}{3}

Step 7: Taking the inverse (or reciprocal) of both sides yields

\frac{x\%}{100\%}=\frac{3}{12125}

\Rightarrow{x} = {0.02\%}

Therefore, {3} is {0.02\%} of {12125}.

Calculation Samples


text:
# zbMATH — the first resource for mathematics

Necessary and sufficient conditions for a variety of Leibniz algebras to have polynomial growth. (English. Russian original) Zbl 1184.17003
J. Math. Sci., New York 152, No. 2, 282-287 (2008); translation from Fundam. Prikl. Mat. 12, No. 8, 207-215 (2006).
Summary: We study the behavior of the codimension sequence of polynomial identities of Leibniz algebras over a field of characteristic 0. We prove that a variety V has polynomial growth if and only if the condition $N_2 A,\widetilde{V_1 } \not\subset V \subset \widetilde{N_c A}$ holds, where $$N_2A$$ is the variety of Lie algebras defined by the identity $$(x_1 x_2)(x_3 x_4)(x_5 x_6)\equiv 0,\widetilde{V_1}$$ is the variety of Leibniz algebras defined by the identity $$x_1(x_2 x_3)(x_4 x_5)\equiv 0$$, and $$\widetilde{N_cA}$$ is the variety of Leibniz algebras defined by the identity $$(x_1 x_2)(x_{2c+1}x_{2c+2})\equiv 0$$.

##### MSC:
 17A32 Leibniz algebras
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##### References:
 [1] I. I. Benediktovic and A. E. Zalesskii, ”T-ideals of free Lie algebras with polynomial growth of the sequence of codimensions,” Izv. Akad. Nauk BSSR. Ser. Fiz.-Mat. Nauk, 3, 5–10 (1980). [2] A. Blokh, ”A generalization of the concept of a Lie algebra,” Sov. Math. Dokl., 6, 1450–1452 (1966). [3] S. P. Mishchenko, ”Varieties of polynomial growth of Lie algebras over a field of characteristic zero,” Math. Notes, 40, 901–905 (1986). · Zbl 0624.17007 [4] S. P. Mishchenko, ”Varieties of Lie algebras with two-step nilpotent commutant,” Izv. Akad. Nauk BSSR. Ser. Fiz.-Mat. Nauk, 6, 39–43 (1987). · Zbl 0646.17013 [5] S. P. Mishchenko, ”Growth in varieties of Lie algebras,” Russ. Math. Surv., 45, No. 6, 27–52 (1990). · Zbl 0718.17004 · doi:10.1070/RM1990v045n06ABEH002710 [6] S. P. Mishchenko and O. Cherevatenco, ”A Leibniz variety with weak growth,” Kazanskoe Mat. Obshch. Tr. Mat. Centra im. N. I. Lobachevskogo, 31, 103–104 (2005). [7] S. P. Mishchenko and O. Cherevatenco, ”A Leibniz variety with weak growth,” Vestn. SamGU (2006). [8] S. Mishchenko and A. Valenti, ”A Leibniz variety with almost polynomial growth,” J. Pure Appl. Algebra, 202, No. 1–3, 82–101 (2005). · Zbl 1179.17002 · doi:10.1016/j.jpaa.2005.01.013
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# Mean Absolute Deviation for a Stable Distribution as a Function of the Tail Exponent

Consider the standard Lévy-Stable (or Alpha Stable) distribution $S(\alpha,\beta, \mu, \sigma)$ where $\alpha$ is the tail exponent, $1 \leq \alpha \leq 2$. Picking the symmetric case with $0$ mean (that is, $\beta =0$, $\mu =0$ the characteristic function becomes $C(t)= e^{-\left| t \sigma \right| ^{\alpha }}$.

We can numerically extract the mean absolute deviation by inverting the characteristic function. From the shape of the distribution shown below in the graph, and, where $X$ is the r.v., from the property that $E(|X|)_{\alpha=2}= \frac{2 }{\sqrt{\pi }} \sigma$ (the only explicit solution we get) and $E(|X|)_{\alpha=1}= \infty$ ( the Cauchy case), a good guess is: $$E(|X|)= \left(\frac{a}{\log(\alpha)}\right)^{f(\alpha)} \sigma,$$ where $a=\frac{2 \log (2)}{\sqrt{\pi }}$ and $f ( \alpha)= 1$ when $\alpha =2$:

I wonder if someone knows of an explicit expression.

Added Comment: By tinkering I figured out we could track the function with $E(|X|)=\frac{a_1 \sqrt{\alpha }}{(a_2+\alpha)\, B\left(a_3 \alpha, a_4\right)}$,where $B$ is the Beta function, but the number of parameters is high.

Answer: Thanks to help it looks like $E(|X|)= \frac{2 \Gamma \left(1-\frac{1}{\alpha }\right)}{\pi } \sigma$. Zolotarev: where $m(X)$ is the median and $\lambda=\sigma^\alpha$, $$E|X- m(X)|^s= \frac{1}{\pi} 2^{1-s/2} (2 \lambda)^{s/\alpha} \Gamma(1-\frac{s}{\alpha}) \Gamma(s) \sin\frac{\pi}{2} s$$

Answer: Thanks to help it looks like $E(|X|)= \frac{2 \Gamma \left(1-\frac{1}{\alpha }\right)}{\pi } \sigma$. Zolotarev: where $m(X)$ is the median and $\lambda=\sigma^\alpha$, $$E|X- m(X)|^s= \frac{1}{\pi} 2^{1-s/2} (2 \lambda)^{s/\alpha} \Gamma(1-\frac{s}{\alpha}) \Gamma(s) \sin\frac{\pi}{2} s$$


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# How to do arithmetic with a basis element from modular forms space

Hi, I'm new to Sage and coding in general. I want to look at the 1-dim'l space CuspForms(Gamma0(7), 3) but with character. Here's my code for that

S = G = DirichletGroup(7, QQ)
list(G)
G.gens()
e = G.0
S = CuspForms(e, 3, prec=20)
print(S)
print(Gamma0(7).cusps())
print(S.basis())
g = S.0

Here, g is the only basis element for S. I want to do F = g*(L+4) where L is some eta-quotient of level 7. However, I get a long error with 'NoneType' object is not iterable at the end. So I'm thinking I need to define g in a different way, since it seems to be that g is not a q-series type object the way I've defined it?

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It would be helpful to have the complete code you used to produce L as well...

( 2020-03-24 14:07:11 -0600 )edit

Yes, sorry. It's L = EtaProduct(7, {1:4, 7:-4}). I also mixed up the level on accident but edited to fix. It's 7.

( 2020-03-25 10:19:19 -0600 )edit

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You can get the associated formal power series as follows:

sage: g
q - 3*q^2 + 5*q^4 - 7*q^7 - 3*q^8 + 9*q^9 - 6*q^11 + 21*q^14 - 11*q^16 - 27*q^18 + O(q^20)
sage: parent(g)
Cuspidal subspace of dimension 1 of Modular Forms space of dimension 3, character [-1] and weight 3 over Rational Field
sage: g.q_expansion()
q - 3*q^2 + 5*q^4 - 7*q^7 - 3*q^8 + 9*q^9 - 6*q^11 + 21*q^14 - 11*q^16 - 27*q^18 + O(q^20)
sage: parent(_)
Power Series Ring in q over Rational Field
more

Yes, but if I try something like g*L.q_expansion() or g*L.qexp(20) or things like that I get that error.

( 2020-03-25 10:21:06 -0600 )edit
1

Try g.q_expansion() * L.q_expansion()

( 2020-03-25 12:21:18 -0600 )edit

It worked!!! So when I do print(g.q_expansion()*L.q_expansion) something goes wrong, but leaving out "print" seems to fix this... weird.

( 2020-03-26 18:52:59 -0600 )edit

missing pair of parentheses after expansion

( 2020-03-27 01:38:34 -0600 )edit

The answer of FrédéricC is already wonderful, here is maybe a way to make your own way in the next similar situation. First of all, let us consider the code:

G = DirichletGroup(7, QQ)
S = CuspForms(e, 3, prec=20)
g = S.gens()[0]
print( f"g = {g}" )
R = g.qexp().parent()
print( f"R = {R}\nIts default precision is {R.default_prec()}" )
q = R.gen()

L = EtaProduct(7, {1:4, 7:-4})
print( f"L = {L}" )

The above gives

g = q - 3*q^2 + 5*q^4 - 7*q^7 - 3*q^8 + 9*q^9 - 6*q^11 + 21*q^14 - 11*q^16 - 27*q^18 + O(q^20)
R = Power Series Ring in q over Rational Field
Its default precision is 20
L = Eta product of level 7 : (eta_1)^4 (eta_7)^-4

Now i noted that the method L.qexp needs an argument. At least for me in my version. This can be seen / asked in the ipython interpreter for sage as follows...

sage: L.qexp?
Signature:      L.qexp(n)
Docstring:
Alias for "self.q_expansion()".

EXAMPLES:

sage: e = EtaProduct(36, {6:8, 3:-8})
sage: e.qexp(10)
q + 8*q^4 + 36*q^7 + O(q^10)
sage: e.qexp(30) == e.q_expansion(30)
True
Init docstring: Initialize self.  See help(type(self)) for accurate signature.
File:           /usr/lib/python3.8/site-packages/sage/modular/etaproducts.py
Type:           method
sage:

and the above L.qexp goes to the doc string of the method qexp in the class L is an instance for:

sage: L.__class__
<class 'sage.modular.etaproducts.EtaGroupElement'>

So we really need an argument for it in my version.

sage: version()
'SageMath version 9.0, Release Date: 2020-01-01'

In my case L.qexp() produces an error, since one argument is needed, but this is easily corrected, L.exp(20) would be ok. Since the precision of the series for $g$ is $20$, we should use the same precision... Well:

sage: L.qexp(20)
q^-1 - 4 + 2*q + 8*q^2 - 5*q^3 - 4*q^4 - 10*q^5 + 12*q^6 - 7*q^7 + 8*q^8 + 46*q^9 - 36*q^10 - 26*q^11 - 44*q^12 + 46*q^13 - 28*q^14 + 42*q^15 + 188*q^16 - 132*q^17 - 96*q^18 + O(q^19)
sage: g.qexp()
q - 3*q^2 + 5*q^4 - 7*q^7 - 3*q^8 + 9*q^9 - 6*q^11 + 21*q^14 - 11*q^16 - 27*q^18 + O(q^20)

and the series are stopping at the "multiplicative precision" equal to $20$. For $L$ we have an expression modulo $O(q^{19})$ (since it starts with $1/q$), for $g$ we have an expression modulo $O(q^{20})$ . This situation may lead to losing the initial precision.

Now i tried:

• g + O(q^6), but it fails. (Although fL + O(q^6) works.)
• g * fL, but it fails. (Although g(q) * fL works.)

To make things work, replace g by g(q)...

So finally, the following code does the job for me. Framework:

PREC = 20
G = DirichletGroup(7, QQ)
S = CuspForms(e, 3, prec=PREC)
g = S.gens()[0]
R = g.qexp().parent()
q = R.gen()

L = EtaProduct(7, {1:4, 7:-4})
fL = L.qexp(PREC)

g(q) * (fL + 4)(q) + O(q^6)

(Addition, multiplication, and truncation work.) Final result:

1 - 3*q + 2*q^2 + 7*q^3 - 29*q^4 + 21*q^5 + O(q^6)
more