text:
1. This letter is not the perfect specimen of proper English grammar and diction. Which of the following would describe the style of this letter and how does that style impact the errors it displays?

A. formal and impersonal

B. informal and personal

C. formal, yet personal

D. informal, yet impersonal

E. none of the above

2. Fill in blank #1 with the most suitable salutation.

A. Ms. (or Mrs.) Allen

B. Legacy Client

C. Dorothy

D. Sir

E. Dorothy Allen

3. Fill in blank #2 with the most suitable close. HINT: Your choice should be consistent with the answer you provided for question #1.

A. Sincerely

B. Yours Truly

C. Thank You

D. Cordially

E. As Always

4. Is it appropriate that the signature lines in this letter include the sender’s title? Why or why not?

A. Yes. It is a standard business practice to always use the sender’s title.

B. No. The sender’s title is optional. For example, it is more appropriate to use it B2B (business-to-business) than B2C (business-to-client).

C. Yes. The title communicates the sender’s relevance with respect to the subject of the letter and its central message.

D. No. The title is optional, so you should consider it with respect to the context and tone of the message.

5. The textbook identifies a number of functions a business letter can fulfill. Which of those is the most accurate way to describe this letter’s function?

A. to introduce the business to a potential client

B. to announce or sell a product or service

C. to communicate feelings and emotions

D. to document an event or decision

E. to deliver important or specific information

6. What rhetorical function would you associate with the phrase, “As the holiday season approaches…”?

A. introductory

B. informative

C. persuasive

D. residual

E. transitional

7. In paragraph #2, “good health and good cheer” is a phrase that uses a common communication technique to reinforce a message. What is that technique?

A. alliteration

B. repetition

C. transition

D. coordination

E. none of the above

8. Where else in this letter does a word or phrase use the technique noted in question 6 to reinforce an important idea? HINT: There are at least four additional applications of the technique.

9. In paragraph #2, “good health and good cheer” is also an example of parallel structure—a technique that helps establish coherence and flow in a letter. List three sets of words that use parallel structure.

10. McLean recommends tying the opening paragraph to the closing paragraph. Identify the two ways this letter accomplishes that.

11. Why is there no return address on this letter?

A. Because it is written in an informal style, so such deviations are acceptable.

B. Because the recipient’s familiarity with this business means she already knows it.

C. Because the letterhead’s graphic takes up too much space.

D. Because the address is printed in the footer of this letterhead.

E. All of the above.

12. Technically, paragraph #2 is flawed because it skips from one idea to another when a well-developed paragraph should focus on one and only one idea. You could correct this weakness by using two paragraphs to expand each idea. Would that be an effective solution? Why or why not?

13. Which of the following summarizes best the central message of this letter?

A. We cherish Legacy clients like you.

B. We are thinking of you.

C. We hope you are enjoying good health and cheer.

D. We are proud of being your home away from home.

E. all of the above


text:
# Strongly isovariant (aka fixed point reflecting / stabiliser preserving) morphisms

I have some questions about what feels like basic topics in quotients of schemes by group actions. Consequently I suspect there are well-known references; I couldn't find them, though.

## First definition: isovariant morphisms

Throughout, I fix a finite group $$G$$. For a $$G$$-scheme $$X$$ we have the stabilizer scheme $$Stab(X) \subset G \times X$$, which may be informally described as the scheme of pairs $$(g, x)$$ with $$gx=x$$. In other words the fibers of $$Stab(X) \to X$$ are given by the scheme-theoretic stabilizers of the points. If $$f: X \to Y$$ is a morphism of $$G$$-schemes, there is a natural map $$Stab(X) \to Stab(Y) \times_Y X =: f^* Stab(Y)$$. We call $$f$$ isovariant (a.k.a. stabilizer preserving, fixed point reflecting) if $$Stab(X) \to f^* Stab(Y)$$ is an isomorphism.

## Preliminary question

If $$f: X \to Y$$ is isovariant, then for every $$x \in X$$ the stabilizer of $$G$$ at $$x$$ is equal to the stabilizer of $$G$$ at $$f(x)$$.

Question: are there conditions for the converse to hold?

If I interpret the stacks project correctly, the converse does hold if $$f$$ is unramified. Can we do better?

## More definitions: strong morphisms

From now on, I fix a $$G$$-scheme $$S$$ and assume that the quotient $$S/G$$ exists (in algebraic spaces, if we want to be fancy). For safety, assume that the quotient is universal (e.g. $$|G|$$ invertible). Let $$P$$ be some property of morphisms of schemes, such as flat, smooth, etc. A morphism $$f: X \to S$$ is called strongly $$P$$ if $$X/G$$ exists, $$f/G$$ is $$P$$, and $$X = X/G \times_{S/G} S$$.

Note that if $$f': X' \to S/G$$ is any morphism, then $$f = f' \times_{S/G} S$$ is isovariant. Morphisms that arise in this way are called strongly isovariant.

## Main questions

Basically I want to know: if $$f: X \to S$$ is $$P$$ and isovariant, then under what additional hypotheses is $$f$$ strongly $$P$$?

I'm particularly interested in the case $$P = \emptyset$$ (which isovariant morphisms are strongly isovariant), $$P$$ means "flat" and $$P$$ means "smooth".

It is easy to see that this is true (possibly under mild assumptions) if $$P$$ means "étale". Also if I'm not mistaken, smooth + strongly flat implies strongly smooth.
• With considerable pain, I seem to have convinced myself if the following. Perhaps let everything be affine over a noetherian base with trivial action, to be safe, but surely this cannot matter too much. Consider $\pi: X \to S$. (1) suppose $\pi$ is smooth and preserves stabilizers in the naive sense (pointwise). Then $\pi$ is isovariant. (2) suppose $\pi$ is flat, isovariant and finite type. Then $\pi$ is strongly flat. (3) suppose $\pi$ is smooth and isovariant. Then $\pi$ is strongly smooth. I'd still be very interested in references for statements like this. – Tom Bachmann Jan 29 at 22:16

format: plain


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126 Double slit interference: photons Photons are fired, one at a time toward a barrier that contains two slits. The

Question:

126 Double slit interference: photons Photons are fired, one at a time toward a barrier that contains two slits. The diagram illustrates that prob- ability with which they arrive at various locations o a screen. The arrows indicate two locations, A and B, on the screen. a) A photon is fired toward the screen and it hits location B. The next photon is then fired to- ward the screen. Which of the following is true about the next photon to arrive on the screen? Explain your answer. i) The photon will definitely not arrive at A and will definitely not arrive at B. ii) The photon will definitely not arrive at A and will definitely arrive at B ii) The photon will definitely not arrive at A and might arrive at B. iv) The photon will definitely not arrive at B and will definitely arrive at A. v) The photon will definitely not arrive at B and might arrive at A. b) Which of the following modifications (select all that could be correct) could cause the next photon to arrive at A? Explain your answer. i) Increasing the intensity of light. ii) Decreasing the intensity of light. ili) Increasing the wavelength of light. iv) Decreasing the wavelength of light. v) Increasing the slit separation. vi) Decreasing the slit separation.

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Consider the eireuit shown below with â‚¬ = 5 uF,R; = 4 Q,and Rz = 2 Q The battery is ideal with aHl emf of â‚¬ = 18 V. Initially, the switeh is open and the capaeitor is uneharged The switeh is then elosed and the capacitor is allowed to charge When the charge On the capacitor reaches 10 uC...


text:
# Mathematical Induction

• May 23rd 2012, 06:59 AM
gurnster76
Mathematical Induction
Use mathematical induction to prove that

n
Σ 2/r(r^2-1) = (1/2) - (1/n(n+1)) n= 2,3,4.......
r=2

i have done the basic proof and got to the point

k+1
Σ = (1/2) - (1/k(k+1)) + 2/(k+1)(k+1^2-1)
r=2

I think i need to get to (1/2) - 1/(k+1)(k+2) but im not sure how to get there, any help would be greatly appreciated!
• May 23rd 2012, 07:35 AM
Plato
Re: Mathematical Induction
Quote:

Originally Posted by gurnster76
Use mathematical induction to prove that
n
Σ 2/r(r^2-1) = (1/2) - (1/n(n+1)) n= 2,3,4.......
r=2

i have done the basic proof and got to the point

k+1
Σ = (1/2) - (1/k(k+1)) + 2/(k+1)(k+1^2-1)
r=2

Work with
$\frac{1}{2}-\frac{1}{k(k+1)}+\frac{2}{(k+1)((k+1)^2-1)}=~?$
• May 23rd 2012, 09:50 AM
gurnster76
Re: Mathematical Induction
That's where im stuck.
• May 23rd 2012, 10:04 AM
Plato
Re: Mathematical Induction
Quote:

Originally Posted by gurnster76
That's where im stuck.

$\frac{2}{(k+1)((k+1)^2-1)}=\frac{2}{(k+1)(k^2+2k)}=\frac{2}{k(k+1)(k+2)}$
• May 24th 2012, 12:35 AM
gurnster76
Re: Mathematical Induction
but how do i get from

(1/2) - (1/k(k+1)) + (2/(k+1)(k+2)

to (1/2) - (1/(k+1)(k+2)


text:
## Basic College Mathematics (9th Edition)

$\frac{5}{6}$, $\frac{1}{1}$
Ratio of Elvis Presley’s song to the sales of song just ahead it. i.e. Elvis Presley’s song to The Beatles song. As per Values given 10 million to 12 million (10 million)/(12 million) = 10/12= (10 ÷ 2)/(12 ÷ 2)= $\frac{5}{6}$ Ratio of Elvis Presley’s song to the sales of song just ahead it is $\frac{5}{6}$ Ratio of Elvis Presley’s song to the sales of song just behind it. i.e. Elvis Presley’s song to The Whitney Houston’s song. As per Values given 10 million to 10 million (10 million)/(10 million) = 10/10= $\frac{1}{1}$ Ratio of Elvis Presley’s song to the sales of song just behind it is $\frac{1}{1}$


text:
# Thread: Remainder and Error of Taylor Series

1. ## Remainder and Error of Taylor Series

Estimate ln(0.3) by using S10 for the Taylor series for lnx. Approximately what does the remainder of the series equal for this value of x? Is the absolute value of this remainder an upper bound for the absolute value of the error in approximating ln(0.3) by S10(0.3)?

2. ## Re: Remainder and Error of Taylor Series

Hey citcat.

As I recall, you can have the error bound of a Taylor series approximation that is based on the nth derivative. Here is the idea (its called the Lagrange Error Bound):

http://clem.mscd.edu/~talmanl/PDFs/A...s/Lagrange.pdf


text:
# Extrude the faces of a cube symmetrically along XYZ

## Sandbox

For the purposes of the current task, a cube of unit length is rendered in oblique projection with ASCII symbols as follows:

  +-----+
/     /|
+-----+ |
|     | +
|     |/
+-----+

• + for the vertices.
• - for the X edges. The unit length along X is represented by five - between two vertices.
• | for the Y edges. The unit length along Y is represented by two | between two vertices.
• / for the Z edges. The unit length along Z is represented by one / between two vertices.
• Vertices are only drawn where all the three planes intersect.
• Edges are only drawn where exactly two planes intersect.

When a unit face is extruded, it is offset by a unit length from its original position and four new edges are created for each direction (positive and negative).

You can think of the extrusion as of drawing the axes of a 3D Cartesian coordinate system where each axis is represented as a cuboid with cross section 1x1 and length n away from (0,0,0)

Extruded by 1 along X:

  +-----------------+
/                 /|
+-----------------+ |
|                 | +
|                 |/
+-----------------+

Given three numbers for the XYZ axes, extrude the faces of a unit cube symmetrically by the indicated amounts and render the result with the ASCII symbols as specified above.

## Input

x, y, z – non-negative numbers - extrusion lengths for the respective axes. 0 means no extrusion. The input can be three numbers, a list of three numbers, a triple, a string or anything that’s convenient for you.

## Output

The ASCII drawing of the cube after extrusion. Leading and trailing wihtespaces are allowed.

## Test cases

X Y Z
0 0 0

+-----+
/     /|
+-----+ |
|     | +
|     |/
+-----+

1 0 0

+-----------------+
/                 /|
+-----------------+ |
|                 | +
|                 |/
+-----------------+

0 0 1
+-----+
/     /|
/     / |
/     /  +
/     /  /
/     /  /
+-----+  /
|     | /
|     |/
+-----+

1 1 0

+-----+
/     /|
+-----+ |
+---|     | +-----+
/    |     |/     /|
+-----+     +-----+ |
|                 | +
|                 |/
+-----+     +-----+
|     | +
|     |/
+-----+

2 0 1

+-----+
/     /|
+-----------+     +-----------+
/                             /|
+-----------+     +-----------+ |
|          /     /|           | +
|         +-----+ |           |/
+---------|     | +-----------+
|     |/
+-----+

1 1 1

+-----+
/     /|-+
+-----+ |/|
+---|     | +-----+
/    |     |/     /|
+-----+-----+-----+ |
|    /     /|     | +
|   +-----+ |     |/
+---|     | +-----+
|     |/| +
+-----+ |/
+-----+

## Winning criteria

The shortest solution in bytes in every language wins. Please add a short description of the method used and your code.

• Is there an upper bound for each of the extrusions? – Embodiment of Ignorance May 12 at 17:17
• @Embodiment of Ignorance - 9 should be enough – Galen Ivanov May 12 at 18:03

# JavaScript (ES6),  525 ... 475 471  459 bytes

Saved 13 bytes thanks to @Neil

Takes input as an array [X,Y,Z]. Returns a matrix of characters.

a=>([X,Y,Z]=a,m=[],W=X*6+Z*2,'1uol9h824lsqpq17de52u1okgnv21dnjaww111qmhx1gxf4m50xg6pb42kzijq21xyh34t0h2gueq0qznnza2rrimsg3bkrxfh3yrh0em').replace(/.{7}/g,s=>[[c,x,A,y,B,w,C,h,D,t]=parseInt(s,36)+s,Y*W,Z,X,Y|!W,Z*X,X*Y*!Z,X*Z*!Y,Y*Z*!X][c]&&(g=H=>V<h&&((m[r=y-(3-B)*p+(t>1?H-V:V)+Y*3+Z*2]=m[r]||Array(W*2+9).fill )[x-2*(3-A)*p+(t>1?V:H-V*t)+W]= \${c>5?'  + ':t>1?'|-':'-|'}+/+[(H%~-w?0:+t?4:2)|!(V%~-h)],g(++H<w?H:!++V)))(V=0,p=a[c%3],w-=-3*C*p,h-=-D*p))&&m

Try it online!

## How?

### Drawing steps

The output consists of 15 sides, drawn in a specific order.

### Implementation

The drawing function is $$\g\$$. It works with the following parameters:

• $$\(x, y)\$$: where to start drawing
• $$\w\$$: width
• $$\h\$$: height
• $$\t\$$: type of side

The vertices are always drawn. Depending on the value of another parameter $$\c\$$, the edges are either drawn or erased.

If $$\t=0\$$, the function draws a front side:

...........      ...........
..+-----+..      ..+     +..
..|     |..  or  ..       ..
..|     |..      ..       ..
..+-----+..      ..+     +..
...........      ...........

If $$\t=1\$$, it draws a top side:

...........      ...........
...+-----+.      ...+     +.
../     /..  or  ..       ..
.+-----+...      .+     +...
...........      ...........

If $$\t=2\$$, it draws a right side:

......+....      ......+....
...../|....      .....  ....
....+ |....  or  ....+  ....
....| +....      ....  +....
....|/.....      ....  .....
....+......      ....+......

The coordinates $$\(x,y)\$$ and the size $$\(w,h)\$$ of each side either have a constant value or depend on exactly one of the input variables: $$\X\$$, $$\Y\$$ or $$\Z\$$. Besides, each side is drawn if and only if a specific condition is met.

Both the input variable to use and the condition to check are identified with the parameter $$\c \in[1..8]\$$.

To summarize, a side is fully described with:

$$\begin{cases} c\\ t\\ x=o_x+m_x\times P\\ y=o_y+m_y\times P\\ w=o_w+m_w\times P\\ h=o_h+m_h\times P \end{cases}$$

where $$\P\$$ is the input variable deduced from $$\c\$$.

Therefore, we need to store the following 10 parameters for each side:

$$[c, o_x, m_x, o_y, m_y, o_w, m_w, o_h, m_h, t]$$

Below are the parameters of the 15 sides that need to be drawn:

$$\begin{array}{|c|c|} \hline \text{side} & c & o_x & m_x & o_y & m_y & o_w & m_w & o_h & m_h & t\\ \hline 0 & 4 & 0 & 0 & 2 & -3 & 7 & 0 & 4 & 6 & 0\\ 1 & 4 & 6 & 0 & 2 & -3 & 4 & 6 & 3 & 0 & 2\\ 2 & 2 & 6 & -2 & 2 & 2 & 4 & 0 & 3 & 4 & 2\\ 3 & 3 & 6 & 6 & 2 & 0 & 4 & 0 & 3 & 0 & 2\\ 4 & 3 & 0 & -6 & 2 & 0 & 7 & 12 & 4 & 0 & 0\\ 5 & 2 & 2 & 2 & 0 & -2 & 7 & 0 & 3 & 4 & 1\\ 6 & 3 & 2 & -6 & 0 & 0 & 7 & 12 & 3 & 0 & 1\\ 7 & 2 & 0 & -2 & 2 & 2 & 7 & 0 & 4 & 0 & 0\\ 8 & 5 & 6 & -2 & 2 & 2 & 4 & 0 & 1 & 2 & 2\\ 9 & 4 & 2 & 0 & 0 & -3 & 7 & 0 & 3 & 0 & 1\\ 10 & 1 & 0 & 0 & 2 & -3 & 7 & 0 & 1 & 3 & 0\\ 11 & 1 & 6 & 0 & 2 & -3 & 1 & 3 & 3 & 0 & 2\\ 12 & 6 & 0 & 0 & 2 & 0 & 7 & 0 & 4 & 0 & 0\\ 13 & 7 & 2 & 0 & 0 & 0 & 7 & 0 & 3 & 0 & 1\\ 14 & 8 & 6 & 0 & 2 & 0 & 4 & 0 & 3 & 0 & 2\\ \hline \end{array}$$

We force all values into the range $$\[0..9]\$$ by applying the following operations:

$$m_x\gets (m_x+6)/2\\ m_y\gets m_y+3\\ m_w\gets m_w/3$$

It results in 15 numbers of exactly 10 decimal digits, which are stored as 15 groups of 7 digits in base 36.

For instance, the first side is encoded as 4032070460 and stored as 1uol9h8.

• I think Array(W*2+9).fill   saves a byte. – Neil May 12 at 11:01
• Oops,sorry, my bad, I must of saw it wrong or something. Disregard. – Embodiment of Ignorance May 12 at 17:11
• @EmbodimentofIgnorance No worries. :) – Arnauld May 12 at 17:14
• Great visualization in addition to your traditionally cool description! Respect! – Galen Ivanov May 13 at 18:12
• @GalenIvanov Thank you! I really enjoyed working on this challenge. – Arnauld May 13 at 18:14

# APL (Dyalog Classic), 162161132 130 bytes

{' |+/+ +-? ??? ++'[⍉(⊃~∘0)⍤1⍣3⍉↑a↑¨⍨↓(--1-⌊/∘,)2-/1⌽↑⍳⍴a←16|29|1+{2⊥+/¨∊¨=⌿¨⍵(⍉⍵)(⍉↓⍵)}⌺2 2 2{⍉⍤2⍉⌽0,⍵}⍣6↑3/↓2/6⌿1<⍵+.=⍨↑⍳1+2×⍵]}

Try it online!

• generate a 3d bool array of unit cubies
• replicate each cubie 6 3 2 times along x y z
• surround with zeroes
• for each 2×2×2 subarray compute a number between 0 and 16 that determines the corresponding output char: (1 + 4*cx + 2*cy + cz) mod 16 where cx,cy,cz are the number of same-value "rods" along axis x,y,z, i.e. vectors along that axis that consist of the same value: 0 0 or 1 1. we make an exception if the subarray is all-zero (or all-one - those don't matter) and we consider its number 0 instead of 28
• for each cell compute where the corresponding char should be drawn on screen
• for each cell build a transparent (0-padded) matrix that contains only one opaque "pixel"
• mix the matrices - at this point we have a 5d array with dimensions inx,iny,inz,outx,outy
• reduce the first three axes, keeping only the first non-transparent (≠0) item along them
• use a lookup table (a string) to convert the numbers into -|/+

thanks Scott Milner for spotting that some +s rendered as ?s

• How do you actually test this? I wanted to try extrusions of 2,3,4 for example, but nothing obvious seemed to work. – Neil May 11 at 17:50
• the algo is very wasteful, so 2 3 4 runs out of memory, but 2 3 3 is within reach – ngn May 11 at 17:54
• @Neil i made a small fix and now 2 3 4 works. lost a byte as a side effect :) – ngn May 11 at 18:13
• An APL solution that's half as long as a Charcoal one on an ascii-art challenge?! What's going on?! – Shaggy May 11 at 18:27
• Not 100% sure, but I'm pretty sure that this isn't desired behavior on 0 1 1 – Scott Milner May 12 at 1:09

# Charcoal, 325 bytes

≔×⁶Ｎθ≔×³Ｎη≔⊗Ｎζ¿‹¹№⟦θηζ⟧⁰«Ｂ⁺⁷⊗θ⁺⁴⊗η↗↗⊕⊗ζ+⁺⁵⊗θＰ↙⊗⊕ζ↓+↓⁺²⊗η↙+↙⊕⊗ζ»«Ｆ‹⁰θ«↗+↗←+←⊖θ↙+/Ｐ→θ↓+↓²+⊖θ+»Ｆ‹⁰η«Ｊ⁶¦³↗+↗↓+↓⊖η↙+/Ｐ↑η←+←⁵↑+↑⊖η+»Ｆ‹⁰ζ«Ｊ⁸±²↓+↓↓↗+↗⊖ζ↑+↑²Ｐ↙ζ←+←⁵↙+↙⊖ζ+»Ｊ⁶¦⁰Ｆ‹⁰ζ«Ｇ↓³↙⊕ζ←⁷↑³↗⊕ζ ↙+↙⊖ζ↓+↓²+→⁵Ｐ↗ζ↑+↑²Ｐ←⁶↗+↗⊖ζ»Ｆ‹⁰η«Ｇ←⁶↑⊕η↗³→⁶↓⊕η ↑+↑⊖η←+←⁵↙+/Ｐ↓η+⁵Ｐ↗²↓+↓⊖η»Ｆ‹⁰θ«Ｇ↗²→⊕θ↓⁴↙²←⊕θ →+⊖θ↗+/↑+↑²Ｐ←θ↙+/Ｐ↓³←+←⊖θ»Ｐ↗∧∧θη²Ｐ↓∧∧ζθ³Ｐ←∧∧ηζ⁶+

Try it online! Link is to verbose version of code. Explanation:

≔×⁶Ｎθ≔×³Ｎη≔⊗Ｎζ

Input the extrusions, but premultiply them to save bytes.

¿‹¹№⟦θηζ⟧⁰«Ｂ⁺⁷⊗θ⁺⁴⊗η↗↗⊕⊗ζ+⁺⁵⊗θＰ↙⊗⊕ζ↓+↓⁺²⊗η↙+↙⊕⊗ζ»«

If at least two of the extrusions are zero, then simply draw a cuboid of dimensions (2x+1, 2y+1, 2z+1). Otherwise:

Ｆ‹⁰θ«↗+↗←+←⊖θ↙+/Ｐ→θ↓+↓²+⊖θ+»

Print the left extrusion, if any.

Ｆ‹⁰η«Ｊ⁶¦³↗+↗↓+↓⊖η↙+/Ｐ↑η←+←⁵↑+↑⊖η+»

Print the down extrusion, if any.

Ｆ‹⁰ζ«Ｊ⁸±²↓+↓↓↗+↗⊖ζ↑+↑²Ｐ↙ζ←+←⁵↙+↙⊖ζ+»

Print the back extrusion, if any.

Ｊ⁶¦⁰

The remaining extrusions all meet at this point (which doesn't get drawn until the very end!)

Ｆ‹⁰ζ«Ｇ↓³↙⊕ζ←⁷↑³↗⊕ζ ↙+↙⊖ζ↓+↓²+→⁵Ｐ↗ζ↑+↑²Ｐ←⁶↗+↗⊖ζ»

Print the front extrusion, if any, taking care to erase parts of the left and down extrusions that may overlap.

Ｆ‹⁰η«Ｇ←⁶↑⊕η↗³→⁶↓⊕η ↑+↑⊖η←+←⁵↙+/Ｐ↓η+⁵Ｐ↗²↓+↓⊖η»

Print the up extrusion, if any, taking care to erase parts of the back and left extrusions that may overlap.

Ｆ‹⁰θ«Ｇ↗²→⊕θ↓⁴↙²←⊕θ →+⊖θ↗+/↑+↑²Ｐ←θ↙+/Ｐ↓³←+←⊖θ»

Print the right extrusion, if any, taking care to erase parts of the down and back extrusions that may overlap.

Ｐ↗∧∧θη²Ｐ↓∧∧ζθ³Ｐ←∧∧ηζ⁶+

Draw the joins between the latter extrusions.

• 325 bytes in Charcoal?! On an ascii-art challenge?! That's enough to make me not even attempt this in Ja(vaScri)pt! – Shaggy May 11 at 18:25
• @Shaggy This might not be an optimal approach, and there might be ways of golfing it that I've overlooked anyway. I do have to say though that ngn's method looks intriguing, it looks as if he draws the shape into an internal array and then performs edge detection to generate his output. – Neil May 11 at 19:45
• @Shaggy I've come up with a 195 byte solution, which I've posted separately as it's a completely different approach. Still well short of APL though. – Neil May 25 at 14:26

# Charcoal, 195164 144 bytes

≔⁺³×⁶Ｎθ≔⁺²×³Ｎη≔⊕⊗ＮζＦ…·±ζζＦ…·±ηηＦ…·±θθ«Ｊ⁻λι⁺κι≔⟦⟧δＦＥ²↔⁻⁺ιμ·⁵ＦＥ²↔⁻κνＦＥ²↔⁻⁺λξ·⁵⊞δ⌊⟦‹μζ‹νη‹ξθ‹¹№Ｅ⟦μνξ⟧‹π⊕ρ¹⟧≡Σδ¹+²§ |-/⁺⌕δ¹⌕⮌δ¹¦³+⁴§ +Σ…δ⁴¦⁶§ |-/⌕δ⁰

Try it online! Link is to verbose version of code. I'm posting this as a separate answer as it uses a completely different approach to drawing the extrusion. Explanation:

≔⁺³×⁶Ｎθ≔⁺²×³Ｎη≔⊕⊗Ｎζ

Input the extrusions and calculate half the size of the enclosing cuboid, but in integer arithmetic because Charcoal's ranges are always integers. The origin of the output maps to the centre of the original unit cube.

Ｆ…·±ζζＦ…·±ηηＦ…·±θθ«

Loop over all coordinates within (including the boundary) the cuboid containing the extrusion.

Ｊ⁻λι⁺κι

≔⟦⟧δＦＥ²↔⁻⁺ιμ·⁵ＦＥ²↔⁻κνＦＥ²↔⁻⁺λξ·⁵⊞δ⌊⟦‹μζ‹νη‹ξθ‹¹№Ｅ⟦μνξ⟧‹π⊕ρ¹⟧

From the given coordinates, peek in all eight diagonal directions to determine whether the extrusion overlaps in that direction. The peeked coordinates are checked that they still lie within the cuboid, and then the number of axes in which the coordinate lies within the original cube must be greater than 1. Note that since the cube has an odd display height, the Y-axis values that are peeked are integers while the other axes use fractional coordinates.

≡Σδ

Consider the number of directions in which the extrusion overlaps. There are five cases of interest where we want to print something, as in the case of zero, that means that this is empty space and we don't want to print anything, while in the case of eight, that means that this is inside the extrusion and anything we did print would be overprinted by a layer nearer the eye point.

¹+

If the extrusion only overlaps in one direction then this is an outer corner and we need to output a +.

²§ |-/⁺⌕δ¹⌕⮌δ¹¦

If the extrusion overlaps in two directions then this is an outer edge. Which sort of edge is determined from the separation between the two overlaps; 6 and 7 are rear-facing edges and will get overwritten, 4 is a diagonal edge, 2 is a vertical edge and 1 is a horizontal edge. (I actually calculate 7 minus the separation as it seems to be easier to do.)

³+

If the extrusion overlaps in three directions then this is an inner corner in the case where one of the extrusions is zero and we need to output a +.

⁴§ +Σ…δ⁴¦

If the extrusion overlaps in four directions then there are two cases: faces (any direction), and inner corners in the case with three positive extrusions. In the latter case there are an odd number of overlaps towards the viewer.

⁶§ |-/⌕δ⁰

If the extrusion overlaps in six directions then this is an inner edge. It works like the complement of an outer edge except that we're only interested when one of the two empty spaces is the direction towards the eye point (the last entry in the array).

# K (ngn/k), 172 bytes

{s:1+|/'i-:&//i:1_--':1_4#!#'2*:\a:16!29!1+2/(!3){+'+x}/'{2+':''1+':'0=':x}'{++'x}\6{+'+0,|x}/{6}#{3}#'{2}#''s#1<+/x=!s:1+2*x;" |+/+ +-? ??? ++"s#@[&*/s;s/i;{x+y*~x};,//a]}

Try it online!

obligatory k rewrite of my apl solution

same algorithm, except that 3d->2d rendering is done with (the k equivalent of) scatter index assignment instead of creating 2d matrices for each 3d element and mixing them

• How would your ngn/apl perfrom in comparison to your Dyalog APL solution? – Galen Ivanov May 13 at 18:19
• @GalenIvanov it wouldn't be a fair comparison because in my apl solution i waste a lot of memory to save a few bytes, and in k shorter happens to be faster in this case – ngn May 13 at 18:36
• I asked about comparison between two APL solutions - yours APL/Dyalog one and a hypothetical solution in ngn/apl – Galen Ivanov May 13 at 18:42
• oops, i don't know why i read that as "ngn/k"... it's again unfair - ngn/apl is hobbyist javascript, dyalog is professional c – ngn May 13 at 18:50
• @GalenIvanov probably not. ngn/apl is missing recent additions to the language like the rank operator (⍤) and stencil (⌺) – ngn May 13 at 18:58


text:
We’re going to create a summary of BBC News Articles and place them at the top using a Firefox extension. This article is about the gnarly algorithm Term Frequency-Inverse Document Frequency (TF-IDF). We’re going to create a real-world usage in the form of a Firefox extension. I know what you’re thinking. “TF-IDF? Yawn 😴” but bare with me, it’s quite interesting!

When we’re done, it’ll look like this:

News article found here.  Red box highlights the most important sentences. The red box is not in  the final product, is only used for illustration purposes.

### Term frequency * Inverse Document Frequency

Don’t worry, the name of the algorithm makes me fall asleep every time I hear it said out loud too. This algorithm is 2 algorithms multiplied together. Let’s see how both of these work:

#### Term Frequency

Term frequency (TF) is how often a word appears in a document, divided by how many words there are.

Let’s say you’re reading a news article on Brexit. The word “Brexit” will appear a lot, so the term frequency of the word “Brexit” is high.

Quite often, we would want to build a dictionary (hashmap) of term frequencies alongside the term. Like {word: term frequency of that word} and then iterate through this dictionary to find out which word appears the most times.

Now, what if I told you that the term frequency dictionary would look a little like this:

{"and": 0.87, "the": 0.73}

You can see how these common English words aren’t useful to us. Of course, most English texts have these words in them, but we call English words like these stopwords. Stopwords usually refer to the most common words in a language, although there isn’t one singular definition. You have to choose stopwords per usage. You have to decide on what words to use. Before processing some text you’ll usually want to remove stopwords to better process the text.

Words with capital letters in them differ from words without capitals. In programming, “Africa” and “africa” are two different things. Because of this, we want to turn everything into lowercase or uppercase to better process our text. We’re going to turn all words into lowercase.

Given a string, we want to remove stop words and turn it into lowercase. Our extension will give us a string of all text on a BBC news article. Don’t worry about where we get the text from yet, that’s done later in the Firefox extension section. For now, assume we have text that looks like this:

... These are external links and will open in a new windowA neat feature of many modern laptops is the ability to power them up through the USB port. Unlike the rectangular USB ports of old, the newer type - USB-C - can carry enough power to charge your machine.That’s great news: it means ...

The above text is shortened to prevent the reader from falling asleep. GitHub freaks out on this page when I try to insert the stopwords_array, so I had to split it up like this.

This is the function which will “prettify” our documents. Line 3 is an array of stopwords I found on StackOverflow. I added “share” and “linkthese” since these are commons words in the news article we don’t want.

Line 5 is Regex. The square brackets mean or. [,.] means “activate on a comma or a full stop ”. /g means global. Once you find one ‘,’ or ‘.’ don’t stop, continue searching the string. The empty string is what we’re replacing it with. If we find a full stop or a comma, replace it with nothing — delete it. This is because the words “Africa.” and “Africa” would be classified as two different words without this.

Line 4 splits the document up into separate words. The map function applies a function to every element in an array. Once the string is split up into an array of words we apply the toLowerCase() method to every element. It makes every word lowercase.

From my Instagram, @Brandon.codes. The arrows are Arrow Functions (Lambda / Anonymous functions).

We then return the lowercase words once we’ve filtered out stop words. Filter() creates a new array with only the elements for which the function inside returns True.

If a word is a stop word, it’ll result in True which means we’ll get a new array of only the stopwords in the document. We use the negation operator “!” to get the opposite, which is what we want. To return a list of words with no stopwords in it.

Now we want to count how many times each word appears in the document. This will be useful for both Term Frequency and Inverse Document Frequency. First, we want to get all the unique words from an array of words.

We convert the array into a set because sets have no repetitions. This lets us get only the unique words in the array. Sets also don’t have an order, so we can’t use array indices to access elements. We need to turn it straight back into an array. For more about Set Theory, check out this article I wrote.

Okay, now time to count how many times a word appears in the words array.

This function goes through every single unique word and counts how many times that word appears in the array of words. The Term frequency function is quite long, so I’m going to break it down.

Line 6 divides the document up into sentences. Sometimes sentences have white space before them. “Brandon. Dogs.” Has whitespace before “Dogs.” we apply the trim() method to each item to get rid of these trailing whitespace.

Regarding line 7, The first 146 characters of the first word is social media links. The rest of that word is a title or sub-title. Here, look:

Share this withEmailFacebookMessengerMessengerTwitterPinterestWhatsAppLinkedInCopy this linkThese are external links and will open in a new windowRyanair is tightening the rules on what passengers pay to take luggage onto the plane, in order to "speed up boarding".

This is annoying, as the title is an essential part of the story and needs to be taken into account. So we remove the first 146 characters of the first word to get:

Ryanair is tightening the rules on what passengers pay to take luggage onto the plane, in order to "speed up boarding"

Remember this formula?

The variable “TFVals” is calculating this formula. If we run the sentence “Hello, my name is Brandon. Brandon Brandon. The elephant jumps over the moon” through the term frequency function, we’ll get something that looks like this:

We have the term frequencies of words, but we want to calculate the most important sentences, not words. To do that, we go through every single sentence and see what words come up in that sentence which are in TFVals.

TF only returns 1 “Brandon”, so not every Brandon has a TF score (like in the image above). For  this example, bare in mind that TF only returns 1 Brandon because we are  calculating the singular term (word) frequency.

We just need to add them all up and divide by how many words we have. Since we’re only adding up the TF values of non stop words, it’s only fair if we divide by how many non stop words there are, instead of how many words there are in a sentence. If we don’t divide by how many words we have, long sentences have an advantage over shorter ones.

This is what line 20 onwards does below. We go through every single sentence and calculate the TF values of each sentence, just as we did above. New code starts at line 20.

And that’s it. But we have a problem with using only Term Frequency. As you may have seen earlier, “Brandon Brandon” was the highest scoring TF out of all 3 sentences we looked at.

Popularity isn’t enough. We don’t want sentences that have the most keywords in them as they may not make sense, or they may be repetitions of each other. Such as in the “Brandon” Brandon” sentence. It has a high TF value but doesn’t hold much content.

It doesn’t contain much information and isn’t useful. We want a sentence that is both rare, unique and contains keywords common in the article. This is where inverse document frequency comes in.

#### Inverse document frequency

Term frequency is how common a word is, inverse document frequency (IDF) is how unique or rare a word is. The formula for IDF is:

t is the term and d is the documents. We have multiple documents, we’re treating each sentence as its own document.

IDF used over many documents, whereas TF is built for one document. You can decide what a document is. In this article, each sentence is its own document.

The first few steps of IDF is the same as TF. We prettify the document, count the words in the document and get all the unique words.

Lines 1–6 is nothing new. The for loop on line 17 loops through every sentence in the document. Since each sentence is a new ‘document’, we need to count the words of each sentence individually. We have to prettify them to get rid of the stopwords and turn them into an array of words. We push the wordcount object of each new sentence into wordCountSentences.

There are 3 sentences here. We don’t care how many times Brandon appears  over the whole document. I tried to pick the words to give you  something interesting without making the maths too large, but in the  process I developed narcissism.

We’re now going to go through every single word and count how many times that word appears in every sentence and calculate the IDF score using the below formula.

I use log 10 here. Note: all logarithms are constantly scaled, so it doesn’t matter much on what logarithm you use.

Now we just do this for every non stop word.

New code starts on line 25.

Now we want to get the IDF Values of all the sentences, we use the same code from TF here but replace some things to make it work.

If I’m being truthful with you, I did a simple “find and replace” the variables. Instead of “TF” in the comments, I replaced them with IDF. Instead of “TFVals”, I replaced it with “IDFVals”. Nothing important has happened here, so feel free to skip this part. New code starts on line 38.

Now that we have written code to calculate the IDF value of sentences;  we have completed the formula from earlier. “is a dog” has the highest  IDF value.

We now know how unique or rare a sentence is. This isn’t that useful since we want the sentence to be information-rich too. We want some way to combine the popularity of TF with the uniqueness of IDF. This leads us to our next section…

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### TF-IDF revisited

via GIPHY

“Are we done yet? Can I wake up now?” Yes, very nearly done!

We now have TF and IDF functions implemented. The only thing left to do is to multiply them together.

The objects TF and IDF both stem from the same data, so TF isn’t going to contain something that isn’t in IDF. Because of this, we can iterate through one object and use the same key. We multiply the value in TFVals by the value from in IDFVals.

Our next step is to calculate the 3 most important sentences in our TF-IDF Object. Iterating over the [key, value] of the object with a couple of if statements works perfectly.

You’ll see at the bottom we return the formatted string. We format it so it looks nice when we insert it into the webpage. Each <br> is a line break, a space in the text. The black dots are bullet points. We’re now going to implement this algorithm into a Firefox extension. 🔥🦊

### Getting & changing text in a BBC news article

Go to any BBC news article, right-click and press “inspect element”. You’ll see a nice box at the bottom of the screen. Use the element selector tool in the top left-hand corner and hover over the article. We can see that the whole article is encompassed within a CSS class of ‘story-body’.

If we go further in, we can see that all the actual text in the article is encompassed by paragraph tags, inside this CSS class.

We’re going to use JQuery to select the text.

This line selects all the <p> tags within the story-body class. Now we want to get the text, we do this by applying the method .text().

We want to add our text to the top of the article. JQuery has a method called prepend which lets us prepend data to the top of an object.

And we’re done! We can now identify the most important sentences in a BBC News article and display them right at the top. Just time to turn it into an extension.

### Firefox extension basics

Firefox extensions have 2 main parts. The Javascript you wrote and the manifest.json file which tells Mozilla what your extension does. We’re going over manifest.json now.

manifest_version tells Firefox what version of manifest you are using. Name tells Firefox what the name of your extension is. Version tells Firefox what version number your extension is. These 3 are mandatory.

description tells Firefox what your extension does.

content_scripts tells Firefox what scripts to load when the URL matches what you have inputted. In order for the scripts you have specified to run, the current URL must match at least one of the URLs you have specified. You can use 2 special characters here:

1. *” Matches zero or more characters. In this instance, I don’t know whether the user will load HTTP or HTTPS so I have it step to load both. I also don’t know what exact article the user will look at, so I have it set to activate on any article.
2. ?” matches exactly one character.

For example: "*na?i" would match "illuminati" and "annunaki", but not "sagnarelli".

Since  we’re going to be using jQuery, we’ll import the jQuery JS file as well  into the website, before our script executes. You can grab the jQuery  file from here. Copy and paste into a file named “jquery.js”.

From here, click “Load Temporarary Add-on…” and then click any of the files in the extension. Once you do, you should see this:

Mozilla have a nice article on the basics of Firefox extensions, here.

Now load any BBC news article to play with it!

### Conclusion

You’ve now seen the awesome power of TF-IDF and a real-world application for it. This idea came to me because I have email anxiety. I get so nervous about reading emails that I wanted a quick summary of them to calm my thoughts. Alas, this is my first time ever writing Javascript. I started out on something easier like BBC news articles.

Here are some ways you can improve upon this code, if you so wish:

• Dynamically select how many sentences in a summary you want. You can find out the average TF*IDF value in the whole article and anything over X you can include in the summary. This makes it so long articles are treated as equally as shorter articles.
• Extending this to work on any other websites you wish.

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# [Maxima] texput and extension of the tex function

Zach elzacho at gmail.com
Wed Dec 3 11:52:22 CST 2008

Hey list,

I am looking for more control over the tex function's output.  I have been
looking at the texput function today and am wondering how much it can do.
For instance, can I use it to make the tex function do similar stuff with my
functions as the tex function does with, say, the integrate function.  When
the tex function sees integrate it decides to output:

\int_<<third argument>>^<<fourth argument>> <<first argument>> d<<second
argument>>,

at least in the 4 argument case.  As far as I can tell, something like this
cannot be done using texput (or set_tex_environment).  Am I wrong or is
there another facility that I am missing?

This would be very useful and cool when using imaxima/imath as I can match
any kind of notation in maxima's output.  It almost makes
emacs+imaxima+latex an ideal environment developing papers.

Zach
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# Highlights

• Novel automatic quality evaluation module inspired by the relationship between labels and inputs
• Overfitting control module to ensure enough trainable samples and avoid overfitting
• Method can retain both high accuracy and good generalization at different noise levels

# Introduction

The motivation behind the paper is that it is still a struggle to get a massive amount of clean and precise labels to train a good model, and that low-accuracy segmentation model trained on noisy labels will severely impact the result. The authors also claim that up until now, most approaches for low-quality annotations are designed for the classification tasks.

# Methods

The overall proposed end-to-end architecure is detailed in Figure 2. The authors used the dice loss as segmentation loss and U-Net as their segmentation network.

## Quality Awareness Module (QAM)

The QAM runs parallel with the segmentation module to calculate weight values to apply to the loss from each sample. During training, the re-weighted loss is used for both the segmentation module and the QAM.

The network used for the QAM is a VGG network, where the number of input channels is changed to $$n + 1$$, where $$n$$ is the number of segmentation classes, to take both image and labels as input. The final layer is also replaced by a one-channel average pooling layer to assess the weight for each sample in the mini-batch.

The resulting loss used by the segmentation model then becomes:

The authors mention that they apply a softmax (which does not seem to be pictured in the equation) to rescale all the weights between 0 and 1.

## Overfitting Control Module (OCM)

The motivation behind the OCM is that overfitting problems can arise from the use of the QAM, if the QAM has a larger weight on a small subset with clear-labeled samples. The authors also state that if the QAM makes mistakes during the training process, the relative weight of this sample can be too small or too big for the network to correct.

To counter this, the authors add an extra restriction between the quality awareness module and the softmax layer. They simply use a weighted hyperbolic tangent (as shown in the equation below), to limit the relative ratio of the output quality score from ($$-\infty$$, $$\infty$$) to ($$-\lambda$$, $$\lambda$$). The authors state they empirically set $$\lambda$$ to be 2.

## Data

The authors tested their method on the public JSRT dataset, which contains 247 X-rays images of the chest. The labeled structures are the heart, clavicles, and lungs.

To generate noisy-labeled images, the authors used different types and levels of noises. They randomly selected 0%, 25%, 50% and 75% samples from the training set and eroded or dilated them with $$1 \leq n_i \leq 8$$ and $$5 \leq n_i \leq 13$$pixels. Examples of the resulting noisy-labeled segmentations are shown in Figure 1.

# Results

Table 1 and Figure 3 compare the performance of the baseline method and their proposed architecture when faced with more or less noisy labels, and with varying degrees of noise.

Figure 4 gives some insight into how the QAM affects the final loss during the training.

# Remarks

• The unclear wording in the paper, along with the lack of details in the network architecture figure, make it unclear whether the QAM uses the original segmentation loss or the re-weighted loss
• No clear explanation (e.g. ablation study) as to why they apply both a softmax and a tanh to the output of the QAM, and whether only one of them would have the same effect (which intuitively would seem to be the case)


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# Differences

This shows you the differences between two versions of the page.

 vpde_lecture32 [2020/04/16 13:33]trinh [Section 19.1: Uniqueness for zero Dirichlet heat equation] vpde_lecture32 [2020/04/16 13:34]trinh [Section 19.1: Uniqueness for zero Dirichlet heat equation] Both sides previous revision Previous revision 2020/04/16 13:34 trinh [Section 19.1: Uniqueness for zero Dirichlet heat equation] 2020/04/16 13:33 trinh [Section 19.1: Uniqueness for zero Dirichlet heat equation] 2020/04/16 13:33 trinh [Section 19.1: Uniqueness for zero Dirichlet heat equation] 2020/04/16 13:15 trinh 2020/04/16 11:23 trinh 2020/04/16 09:45 trinh created Next revision Previous revision 2020/04/16 13:34 trinh [Section 19.1: Uniqueness for zero Dirichlet heat equation] 2020/04/16 13:33 trinh [Section 19.1: Uniqueness for zero Dirichlet heat equation] 2020/04/16 13:33 trinh [Section 19.1: Uniqueness for zero Dirichlet heat equation] 2020/04/16 13:15 trinh 2020/04/16 11:23 trinh 2020/04/16 09:45 trinh created Line 82: Line 82: E(t) \equiv 0 E(t) \equiv 0  - for all time. Looking at the form of the integrand, you would conclude that + for all time. Looking at the form of the integrand, you would conclude that the only way this occurs is if the integrand is itself zero, or  w^2(x, t) = 0 w^2(x, t) = 0  - for all $x\in[0, L]$ and for all $t \geq 0$. So $u(x, t) \equiv v(x, t)$ and the solutions must be the same. + for all $x\in[0, L]$ and for all $t \geq 0$. So $w = u - v \equiv 0$ and thus u(x, t) \equiv v(x, t)\$ and the solutions must be the same. ===== Section 19.2: Uniqueness for other BCs of the heat equation  ===== ===== Section 19.2: Uniqueness for other BCs of the heat equation  =====


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# The default alignment of images, that are inserted in Web page, is: a. left b. right c. inline with text d. middle

The default alignment of images, that are inserted in Web page, is:
a. left
b. right
c. inline with text
d. middle


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# Advantage of having more measurements

• I
Hello! I have some points in the plane, with errors on both x and y coordinates. The goal of the experiment is to check if the points are consistent with a straight line or not i.e. if they can be described by a function of the form ##y = f(x)=a+bx## or if there is some nonlinearity involved (e.g. ##y = f(x)=a+bx+cx^2##). Assume first we have only 3 points measured. In this case, the approach is to calculate the area of the triangle formed and the associated error, so we get something of the form ##A\pm dA##. If ##dA>A##, then we are consistent with non-linearity and we can set a constraint (to some given confidence level) on the magnitude of a possible non-linearity (e.g. ##c<c_0##). If we have 4 points, we can do something similar and we can for example calculate the area of the triangle formed by the first 3 points (in order of the x coordinate), ##A_1\pm dA_1## and the area of the last 3 points ##A_2\pm dA_2## and then sum them add and do error propagation to get ##A\pm dA## then proceed as above (in the case of this experiment we expect to not see a non-linearity so we just aim for upper bounds). My question is, what is the advantage of having more points? Intuitively, I expect that the more points you have, the more information you gain and hence the better you can constrain the non-linearity. But it seems like the error gets bigger and bigger, simply because we have more points and error propagation (you can assume that the errors on x and y are the same, or at least very similar for different measurements). So, assuming the points are actually on the line, for 3 points we get ##0\pm dA_3## and for, say 10 points we get ##0\pm dA_{10}## with ##dA_{10}>dA_3##, so the upper bounds we can set on the non-linearity are better (smaller) in the case of 3 points. But intuitively that doesn't make sense. Can someone help me understand what I am doing wrong. Why is it better to have more points? Thank you!

anuttarasammyak
Gold Member
My question is, what is the advantage of having more points?
I believe the more observations or trials we do, the more information we get to know the physical system including its proper disturbance, noise or probabilistic behaviors.

Last edited:
I believe the more observations or trials we do, the more information we get to know the physical system including its proper disturbance, noise or probabilistic behaviors.
Well yeah, this is what I believe intuitively, but I am not sure how to show it mathematically.

anuttarasammyak
Gold Member
Law of large numbers and central limit theorem would be of your interest.

Dale
Mentor
2020 Award
I don’t understand the point of the areas. Why not just estimate c directly using least squares. Or even a Bayesian estimation

jedishrfu
I don’t understand the point of the areas. Why not just estimate c directly using least squares. Or even a Bayesian estimation
But in order to estimate c, I would need to know the functional form of the non-linearity. However the actual form is very model dependent so in our case we don't want to set constraints on a given model we just want to set a constraint on any deviation from linearity, regardless of its actual form. Am I miss understanding your point?

Basically I want to quantify how far the points are from being on a straight line. I decided to use this area as a quantifier, but I am totally open to suggestions for better ways to do it.

Law of large numbers and central limit theorem would be of your interest.
I know about these in general, I am just not sure how they apply to my particular case. For example, in general the error would go as ##1/\sqrt{N}##, where N is the number of measurements, but I don't see that in my expressions above explicitly, so I am probably doing something wrong.

I do not know what exactly is your system but I expect many (x,y) data plots show some dense and sparse pattern and it becomes clearer for larger N as the below linked experiment video shows as an example. link https://www.hitachi.com/rd/research/materials/quantum/doubleslit/index.html
I don't have many points, tho. Here is a paper that might explain it better (the physics of it is involved, but the details are not important for my question), in figure S2. In the experiments so far, people used to measure 3 points and get something like in figure S2. What it is usually done in literature is to calculate the area created by these 3 points and the error associated to it (by propagating the error from each of the 3 points), and from there set a constraint on the non-linearity (so far all the areas are smaller than the uncertainties, so we were able to just set upper limits). My question is simply, if I am able to measure a 4th point on that plot, how would that help me (I am sure it would, as I would gain more data, but I am not sure mathematically how is the error on the area reduced by adding one more point)?

jedishrfu
Mentor
What you’re trying to do is what a linear regression does. It finds the best line through a set of points. If it looks to be a poor line after a lot of points then you must consider that there’s a different relationship.

Sometimes folks will apply a linear regression to the log values of x or y or both. This scheme can discover polynomial functions like ##y = x^2 ## because a log plot would show a straight line for ##log(y) = 2 log(x)##

Here’s more on linear regression:

https://en.wikipedia.org/wiki/Linear_regression

and this video

What you’re trying to do is what a linear regression does. It finds the best line through a set of points. If it looks to be a poor line after a lot of points then you must consider that there’s a different relationship.

Sometimes folks will apply a linear regression to the log values of x or y or both. This scheme can discover polynomial functions like ##y = x^2 ## because a log plot would show a straight line for ##log(y) = 2 log(x)##

Here’s more on linear regression:

https://en.wikipedia.org/wiki/Linear_regression

and this video

I know what linear regression is, that is not what I am trying to do... as I said in the previous reply, the paper I linked to might explain better what I want to do, especially figure S2. There they measure 3 points, calculate the area of the triangle created by them and quantify the deviation from linearity based on the value of that area. I don't see how doing a linear regression to these 3 points would help me quantify that non-linearity.

jedishrfu
anuttarasammyak
Gold Member
There they measure 3 points, calculate the area of created by them and quantify the deviation from linearity based on the value of that area.
I observe in S2 they set a half of volume of hexagonal with axis of three momentum vectors as NL, right ? Do these three vectors come from one time experiment data ? I would like to understand how you want to add data or vectors to it in your question.

I observe in S2 they set a half of volume of hexagonal with axis of three momentum vectors as NL, right ? Do these three vectors come from one time experiment data ? I would like to understand how you want to add data or vectors to it in your question.
I am not sure what you mean. What hexagonal volume are you referring to?

anuttarasammyak
Gold Member
Equation (6) and its explanation by S2.

Equation (6) and its explanation by S2.
Equation (6) is just the area of that triangle in figure S2. In the experiment they measure the 6 points ##m\nu_i^{AA_j}## from the x and y axis in figure S2, and from there they calculate the area created.

anuttarasammyak
Gold Member
Equation (6) seems to have dimension of volume p^3 in momentum space
$$|(A \times B)\cdot C|$$
not area for me.

But equation (6) seems to have dimension of volume p^3 in momentum space
$$(A \times B)\cdot C$$
If you look just before equation (5), ##m_\mu## is just a constant, without units.

anuttarasammyak
Gold Member
I see. And the paper saying "Equivalently, in our geometrical picture it is the volume of the parallelepiped defined by −→mν1,2 and −→mµ." assures my view.

Going back to your point what would you like to do more than this triplet vectors ? Making a quartet by incorporating another vector ? Getting a set of the triplet by many experiments?

Last edited:
jim mcnamara
Mentor
You mentioned area. Area==zero. That is how to test for collinearity of points:
https://www.geeksforgeeks.org/program-check-three-points-collinear/

You can also use the distance test, if that makes any difference to you.

Now we are on the same page I hope.

The above is the best way to test when you want yes/no answers. Or. Use some kind of Minimum area test, if you are okay with a not "perfect" result. What you do in this case is up to you. This is arbitrary you realize. Regression seems okay here. As others mentioned.

This is an example for "not perfect", which you already know:
https://cran.r-project.org/web/packages/olsrr/vignettes/regression_diagnostics.html

Tolerance test of multi-collinearity -- what you are asking about i.e., "more points":
https://www.statisticshowto.com/tolerance-level-statistics/

Dale
Mentor
2020 Award
But in order to estimate c, I would need to know the functional form of the non-linearity.
Not really. You can always do a series expansion and approximate your nonlinearity as a polynomial. You only need to know the functional form if you want to make accurate predictions. But if you only want to detect nonlinearity a polynomial is fine.

Basically I want to quantify how far the points are from being on a straight line. I decided to use this area as a quantifier, but I am totally open to suggestions for better ways to do it.
I suggest least squares regression to a polynomial.

My question is simply, if I am able to measure a 4th point on that plot, how would that help me (I am sure it would, as I would gain more data, but I am not sure mathematically how is the error on the area reduced by adding one more point)?
With one more point you could fit a third order polynomial.

Last edited:
Stephen Tashi
in figure S2. In the experiments so far, people used to measure 3 points and get something like in figure S2.
On the page following that figure, the paper says:
Our procedure above applies to cases with enough experimental data. For systems lacking (sufficiently precise)measurements, we can still derive projections provided that an acceptable estimation of the F21 constant is availablefrom either theory calculation or hyperfine splitting data (whenever available).

So I think the three points in figure S2 are themselves are not necessarily 3 single measurements, but instead , each of those points may be the mean value of many measurements.

Not really. You can always do a series expansion and approximate your nonlinearity as a polynomial. You only need to know the functional form if you want to make accurate predictions. But if you only want to detect nonlinearity a polynomial is fine.

I suggest least squares regression to a polynomial.

With one more point you could fit a third order polynomial.
I am not sure I understand, I do want to set very accurate bounds on the non-linearity. Basically I want to describe my points by ##y=ax+b+g(x)##, with ##g(x) << ax,b##. From there I want to set constraints as tight as possible on the ##g(x)##. If I use a polynomial won't that influence how tight the constraints are? On a more practical aspect, in all the paper on this topic they use this area method, so I assume that if polynomial were to work they would have used them. But given that they use areas in literature, I would still like to find out the answer to my question in the case of using areas to define non-linearity.

You mentioned area. Area==zero. That is how to test for collinearity of points:
https://www.geeksforgeeks.org/program-check-three-points-collinear/

You can also use the distance test, if that makes any difference to you.

Now we are on the same page I hope.

The above is the best way to test when you want yes/no answers. Or. Use some kind of Minimum area test, if you are okay with a not "perfect" result. What you do in this case is up to you. This is arbitrary you realize. Regression seems okay here. As others mentioned.

This is an example for "not perfect", which you already know:
https://cran.r-project.org/web/packages/olsrr/vignettes/regression_diagnostics.html

Tolerance test of multi-collinearity -- what you are asking about i.e., "more points":
https://www.statisticshowto.com/tolerance-level-statistics/
I am not sure I understand what you mean. Of course area=0 for collinear points. But in practice they won't be on a straight line, as we have experimental errors. So the area will be of the form ##3 \pm 5##, which is not zero, but it is consistent with zero within the error. My question is, if I add one more point, and I calculate the area formed by these 4 points, what to I gain compared to the case of having only 3 points. Sending me link to statistic webpages doesn't help me. I know the basics, I just don't know how to apply it to my problem.

On the page following that figure, the paper says:

So I think the three points in figure S2 are themselves are not necessarily 3 single measurements, but instead , each of those points may be the mean value of many measurements.
On yes, the points in figure S2 are the results of many measurements. In the experiment one measures the x and y for a given point several times, then places it on that plot in S2. After measuring 3 such points we quantify the non-linearity by calculating that area. My question is, if I measure a 4th point, with the same uncertainty as the other 3 points. Do I get anything in terms of better constraining the non-linearity of the 3 points.

I see. And the paper saying "Equivalently, in our geometrical picture it is the volume of the parallelepiped defined by −→mν1,2 and −→mµ." assures my view.

Going back to your point what would you like to do more than this triplet vectors ? Making a quartet by incorporating another vector ? Getting a set of the triplet by many experiments?
I would like to add another point to figure S2. In terms of the mathematical description of the problem, the vector will be become 4D (not they are 3D).


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# If a 220/440 V, 50 Hz, single phase transformer is operated on 220 V, 40 Hz supply then

This question was previously asked in
MPSC Assistant Engineer EE Mains 2018 - Paper 1
View all MPSC AE Papers >
1. the eddy current loss and hysteresis loss of the transformer will decrease
2. the eddy current loss and hysteresis loss of the transformer will increase
3. the hysteresis loss of the transformer increases while eddy current loss remains the same
4. the hysteresis loss remains the same, whereas eddy current loss decreases

Option 3 : the hysteresis loss of the transformer increases while eddy current loss remains the same
Free
CT 1: Basic Concepts
18600
10 Questions 10 Marks 6 Mins

## Detailed Solution

Concept:

Hysteresis losses: These are due to the reversal of magnetization in the transformer core whenever it is subjected to alternating nature of magnetizing force.

$${W_h} = \eta B_{max}^xfv$$

$${B_{max}} \propto \frac{V}{f}$$

Where

x is the Steinmetz constant

Bm = maximum flux density

f = frequency of magnetization or supply frequency

v = volume of the core

At a constant V/f ratio, hysteresis losses are directly proportional to the frequency.

Wh  f

Eddy current losses: Eddy current loss in the transformer is I2R loss present in the core due to the production of eddy current.

$${W_e} = K{f^2}B_m^2{t^2}V$$

$${B_{max}} \propto \frac{V}{f}$$

Where,

K - coefficient of eddy current. Its value depends upon the nature of magnetic material

Bm - Maximum value of flux density in Wb/m2

t - Thickness of lamination in meters

f - Frequency of reversal of the magnetic field in Hz

V - Volume of magnetic material in m3

At a constant V/f ratio, eddy current losses are directly proportional to the square of the frequency.

We  f2

Iron losses or core losses or constant losses are the sum of both hysteresis and eddy current losses.

Wi = W + We

At constant V/f ratio, Wi = Af + Bf2

Calculation:

In the given question, voltage kept constant and frequency is varied.

The V/f ratio is not constant.

Hysteresis losses, Wh ∝ V1.6 f-0.6

As the frequency is decreased to 40 Hz from 50 Hz, the hysteresis losses will increase.

Eddy current losses, We ∝ V2

As eddy current losses are independent of frequency, the eddy current losses remain the same.


text:
# Help me!

Please do help me with this question, Thanks!

$\quad If\quad a+b+c=0,\quad then\quad find\quad \frac { { a }^{ 2 } }{ bc } +\frac { { b }^{ 2 } }{ ca } +\frac { { c }^{ 2 } }{ ab }$

Note by Swapnil Das
4 years, 5 months ago

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## Comments

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The given expression can be converted to this,$(a^3+b^3+c^3)/abc$.Now,because $a+b+c=0,a^3+b^3+c^3=3abc$,hence answer is 3.Proof:$a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)$.

- 4 years, 5 months ago

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Thank You!

- 4 years, 5 months ago

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Most welcome!!

- 4 years, 5 months ago

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The same question was in my FIITJEE package. :p

Anyway, Here's the solution.

I cannot type in LaTeX, because I am operating from my iPad.

Taking the LCM of the 3 terms to be added, we get,

(abc)(a^3+b^3+c^3)/(abc)^2

=(a^3+b^3+c^3)./(abc)

Now, Since a+b+c=0, a^3+b^3+c^3=3abc.

=3abc/abc=3. QED.

Method 2

Jugaad method. Try 1+1-2 :P

- 4 years, 5 months ago

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:3 :3 :3 :3 :3 Cheers!

- 4 years, 5 months ago

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Cheers! I hope you like the 2nd Method. :P

- 4 years, 5 months ago

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Let $a+b+c= \sigma_1 \\ ab+bc+ac=\sigma_2 \\ abc = \sigma_3 \\ \varphi_3 = a^3+b^3+c^3$

By Newton's sums we have:

$\varphi_3=\sigma_1^3-3\sigma_1\sigma_2 + 3\sigma_3 = (0)^3-3(0)\sigma_2+3\sigma_3 = 3\sigma_3$

The given expression can be written as : $\dfrac{\varphi_3}{\sigma_3} = \dfrac{3\sigma_3}{\sigma_3} = \boxed{3}$

- 4 years, 5 months ago

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So Mr.Mathematician, @Nihar Mahajan , Thank U, but Can you tell me what are Newton's sum or provide a link to it so that I learn about them?

- 4 years, 5 months ago

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Click here for Newton's identities and click here too :).

- 4 years, 5 months ago

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Hey, I am not an Idiot!

- 4 years, 5 months ago

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Don't take it seriously. That was just for trolling :P :P

- 4 years, 5 months ago

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I didn't take it seriously brother!

- 4 years, 5 months ago

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It's quite funny how Nihar got to know about that site ;P

- 4 years, 5 months ago

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I knew that you would comment something here. :P

- 4 years, 5 months ago

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+1 for ur note & +10 for the other link. How did u get to that link?? Did anyone share the link with u or u just figured it out??

- 4 years, 5 months ago

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hehe , actually once Pi Han Goh had mentioned that link in some note (I dont remember).Thats why I know about it.

- 4 years, 5 months ago

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Lol :P. @Pi Han Goh where did u get the link. How many hours do u stay online on brilliant @Nihar Mahajan ??

- 4 years, 5 months ago

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I may "appear" to be attentive on brilliant , but behind the screen , I am actually doing some other work :)

- 4 years, 5 months ago

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Multitasker!

- 4 years, 5 months ago

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And is the Newton's Sum method easier than the usual method?

- 4 years, 5 months ago

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It depends from person to person whether its easier or not.

- 4 years, 5 months ago

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I really wanna learn about Newton's Sum, well which method was easier to you?

- 4 years, 5 months ago

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Both are convenient for me. I just added this for variety of methods.

- 4 years, 5 months ago

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You need to specify that $a,b,c \in \mathbb R - \{0\}$ otherwise it would be undefined.

- 4 years, 5 months ago

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Correct.I was also going to say that.

- 4 years, 5 months ago

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If a+b+c=0 Then,a^3+b^3+c^3-3abc=0 Therefore,a^3+b^3+c^3=3abc

Now in the question ,LCM of denominators will be "abc''

Thus ,on the numerator we get,a^3+b^3+c^3

Which is equal to 3abc

Therefore substituting it,we get the answer as 3...

Answer::3

- 4 years, 5 months ago

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Thank You!

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Such questions r in class 9 RD Sharma book.

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# Curved Beam Analysis on a Hand-Forged, Non-Rectangular Cross Section

In a recent post I discussed curved beam analysis with a rectangular cross section, and as a follow-up a comparison of that type of analysis with finite element results. Many curved beams, such as the lifting hook shown above, do not have rectangular cross-sections. This is an interesting topic, not only because of the engineering involved, but also because it is an interesting specimen of late nineteenth and early twentieth century craftsmanship, as the piece was (as indicated on the drawing) hand forged. Thus we combine analysis with craftsmanship.

## Overview of the Application

The hook shown above was intended to be used with a pile pulling chain. The complete assembly is shown below, taken from the 1929 catalogue.

The hook shown in the drawing is the largest of the series. The eye on the back side of the hook–and the shackle to go with it–was for the tugger line. The chain (a curved beam problem in its own right) was wrapped around the pile (usually wood) and the hook connected back into the chain. The top of the chain was attached to the crane line and the pile pulled out of the ground (hopefully.)

## Loads and Moments on the Hook

As was the case before, the main axis of loading is the axis between the centre of the lifting eye of the hook and the centre of the radius of curvature of the inside of the hook. The moment is thus the product of the main force $P$ and the distance from the centre of the radius of curvature to the centroid of the section. It’s harder to see with this hook, but that’s just about where the lozenge shaped shaded section is on the hook, which suggests that Vulcan applied some kind of analysis in some way to the design of this hook.

The load on the hook is ostensibly the load given in the literature above, i.e. 61 kips. However, that’s not the way it was used, because the hook was wrapped around the pile in a “choker” arrangement and hooked back to the chain. Let’s say for analysis purposes that the basic load on the hook is 61/2 = 30.5 kips. We still haven’t determined the moment because we still haven’t determined the centroid of the section.

## Developing the Equations

The problem here is obvious: with rectangular cross-sections, the centroid is easy to find, and the integration is likewise simple. Here we have a composite, non-rectangular cross-section. Belyaev gives us values for several non-rectangular cross-sections but doesn’t give us a way to put them together. For that we will turn to Boresi et.al. (1993).

Let’s start by stating the basic equation of tangential stress, from the previous post:

$\sigma_{\theta} = \frac {P}{A}+\frac {M}{\rho}\frac{z}{S}$ (1)

where

• $P =$ basic load on hook
• $A =$ cross-sectional area of hook
• $M = PR_0 =$ moment on curved beam
• $R_0 =$ distance from centre of curvature to centroid of section
• $\rho$ = distance of a particular point of interest from the centre of curvature
• $z$ = distance from centroid of section to neutral axis of bending stress
• $S$ = result of integration from the inner to the outer radius of the section

From Boresi et.al. (1993), let us rewrite this as follows:

$\sigma_{{\theta}}=\frac{P}{A}+\frac{P{\it R_{0}}\,\left(A-\rho\,{\it A_{m}}\right)}{A\rho\,\left({\it R_{0}}\,{\it A_{m}}-A\right)}$ (2)

where

$A_{m}=\int_{A}\frac{dA}{\rho}$ (3)

This integral appears often in Belyaev and we will have recourse to it.

The advantage of using Equation (2) is the following:

• We can take several sections and compute their values of $A$, adding them up for the total area.
• We can take several sections and compute their values of $A_m$, likewise adding them up for the total integral
• We can compute the centroid $R_0$ of the entire section by computing the areas and centroids of each part and them compute $R_0$ by summing the products of the centroids and sections of each area and then dividing them by the total area.

Now let’s look at a CAD-drawn diagram of the critical area shown in the original drawing.

The geometry of this section is defined by the radii, which are a little tricky to pull together. This is not unusual in Vulcan designs. It’s also worth noting that, although the drawing calls for the section to be 2 3/4″ at its thickest, the section calls for a 3″ thickest point. For our analysis we’ll use the latter. That lack of precision–and some of the other vague dimensioning on the drawing–may be frustrating to the modern engineer, but it reflects a feature on Vulcan drawings of the era: many decisions were left to the craftsmen at the shop floor.

In any case the three regions we’ll use for analysis are obvious. Also obvious are the centroid location and cross-sectional area, both of which came from the CAD software. It’s possible to compute $A_m$ by a) dividing the section into thin vertical strips, b) computing the area for each strip by multiplying the width by the length, c) measuring the distance from the centroid of the strip to the centre of curvature, d) dividing the area by the centroidal distance for each strip, and e) summing up the results.

From the drawing the distance from the centre of curvature to the centroid is 3 3/4″ + 2.994 ~ 6 3/4″, and the area is 15.3 sq.in. Let’s see how this compares with using the method of Boresi.

There are three zones shown above:

1. The “ellipsoid” closest to the centre of curvature;
2. The trapezoid in the tapered section; and
3. The semicircle at the far side.

The variables and their values are as follows:

With all of that, the formulae for these three sections are as follows:

#### Section 1 (ellipsoid):

$A = 1/2\,\pi \,ab_4$ (4a)
$R = R_2-4/3\,{\frac {a}{\pi }}$ (4b)
$A_m = 2\,b_4+{\frac {\pi \,b_4\left (R_2-\sqrt {{R_2}^{2}-{a}^{2}}\right )}{a}}-2\,b_4\sqrt {{R_2}^{2}-{a}^{2}}\arcsin({\frac {a}{R_2}}){h}^{-1}$ (4c)

#### Section 2 (Trapezoid):

Note the diagram to the right, which will helpful to understand the variables.

$A = 1/2\,\left ({\it b_2}+{\it b_1}\right )\left (h\right )$ (5a)
$R = {\frac {R_2\left (2\,{\it b_2}+{\it b_1}\right )+R_1\left ({\it b_2}+2\,{\it b_1}\right )}{3\,{\it b_2}+3\,{\it b_1}}}$ (5b)
$A_{m}=\left(b_{1}+R_{1}\frac{b_{2}-b_{1}}{h}\right)-\left(b_{2}-b_{1}\right)$ (5c)

#### Section 3 (Semicircle)

These equations assume the semicircle is “pointing outward,” i.e., the semicircle is outside of the diameter that bisects the circle relative to the radius of curvature. Equation (6c) also assumes that $R_1 > b_3$.

$A = 1/2\,{b_3}^{2}\pi$ (6a)
$R = R_1+4/3\,{\frac {b_3}{\pi }}$ (6b)
$A_m = R_1\pi -2\,b_3-\pi \,\sqrt {{R_1}^{2}-{b_3}^{2}}+2\,\sqrt {{R_1}^{2}-{b_3}^{2}}\arcsin({\frac {b_3}{R_1}})$ (6c)

#### Putting Them Together

Making all of the appropriate substitutions yields the following:

The results for $A$ and $R_0$ are very close.

Substituting these values and the load into Equation (2) and plotting the results yields the following:

The stress at the inside radius is 22.8 ksi and at the outside radius -7.7 ksi.

## Some Notes About the “Blacksmith”

The “note to blacksmith” on the drawing reveals more than the draughtsman probably meant to, because it means that these were basically forged by hand. Vulcan was started as a foundry and had one until after World War II, but several parts on the hammers were consistently forged:

The last is on the same size scale as the hook. Although I haven’t investigated it, it’s likely that Vulcan had some kind of forging hammer to assist the blacksmith in making these parts. That said, the basic geometry of the part came into being through the interaction of the blacksmith and the forging hammer. Such shows the level of craftsmanship that existed in the manufacture of the Vulcan product line, a significant part of its success. It’s also an interesting glimpse into a manufacturing base long gone; its loss has and will prove expensive to our economy.

## Other Reference

• Boresi, A.P., Schmidt, R.J. and Sidebottom, O.M. (1993) Advanced Mechanics of Materials. Fifth Edition. New York: John Wiley and Sons, Inc.


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The 95% daily VaR corresponds to the result on the: