text:
# A 4-pole dc motor takes a 50 A armature current. The armature has lap connected 480 conductors. The flux per pole is 20 mWb. Calculate the gross torque developed by the armature of the motor.

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UPRVUNL JE EE 2014 Official Paper
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1. 52.6 N-m
2. 39.5 N-m
3. 76.4 N-m
4. 15.2 N-m

Option 3 : 76.4 N-m
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ST 1: Logical Reasoning
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20 Questions 20 Marks 20 Mins

## Detailed Solution

Concept:

Speed of the motor is given as

$${ω} = \frac{{{N} \times 2\pi }}{{60}}$$

Torque developed by the motor is

T = P / ω

where,

Power developed P = E Ia

$$E = \frac{ϕ ZNp}{60A}$$

∴ $$T= \frac{\frac{ϕ ZNp}{60A}I_a}{\frac{{{N_s} \times 2\pi }}{{60}}}=\frac{ϕ ZpI_a}{2\pi A}$$

Calculation:

Given,

Armature current Ia = 50 A

Number of poles p = 4

Lap wound A = p = 4

Number of conductors Z = 480

The flux per pole ϕ = 20 mWb

$$T=\frac{ϕ ZpI_a}{2\pi A}=\frac{20\times 10^{-3}\times 480\times 50 \times4}{4\times 2\pi }$$

T = 76.4 N-m

format: plain


text:
1. Oct 25, 2012

### Artusartos

Proposition: Let V be an A-invariant subspace of $F^n$ $A \in M_(F)$. Let $B= v_1, ... , v_k$ be a basis of V.

Extend it to a basis $\bar{B} = v_1, ... , v_n$ of F^n.

Then $[T_A]_{\bar{B}} = \begin{bmatrix} [T_A|_v:V \rightarrow V]_B & X \\zeros & C \end{bmatrix}$

Where C is the matrix of $\mu_x : F_A^n/V \rightarrow F_A^n/V$ with respect to $\bar{v_{k+1}} , ... , \bar{v_n}$, which is a basis of $F_A^n /V$

EXAMPLE:

For $\begin{bmatrix} 2 & 1 \\0 & 3 \end{bmatrix}$, V=span(e_1) (already invariant)

$\mu _x : F^2/V \rightarrow F^2/V$has matrix [3] with respect ot hte basis e_2.

Why does it have matrix [3]? Where did that come from?

A-3I = $\begin{bmatrix} -1 & 1 \\0 & 0 \end{bmatrix}$

N(A-3I) = span(e_1 + e_2)

$A(e_1 + e_2) = \begin{bmatrix} 3 \\3 \end{bmatrix} = 3(e_1 + e_2)$

Why did they find the null space? And why is A being multiplied by the vector that spans the null space?

The extended basis is $\bar{B} = e_1, e_1 + e_2$

$[T_A]_{\bar{B}} = \begin{bmatrix} 2 & 0 \\0 & 3 \end{bmatrix}$


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# Style Guide¶

This document provides some general guidelines for writing documentation, both code and user documents. It is by no means inclusive.

For Python coding conventions, please refer to the Python Style Guide.

## Standard for Coding Docstrings¶

OpenMDAO uses documentation strings, or docstrings, written in reStructuredText (reST) to document code. Please follow the docstring standard described below.

We would like to acknowledge the NumPy/SciPy standard, which we referred to frequently when developing our own standard.

### Docstring¶

A docstring describes a module, function, class, or method definition. A docstring is a special attribute of an object (object.__doc__) and, for consistency, is surrounded by triple double-quotes, for example:

"""A simple iteration driver. Basically runs a workflow, passing the output
to the input for the next iteration. Relative change and number of
iterations are used as termination criteria.
"""

A docstring may be one line or spread over several lines. Line length should not exceed 80 characters.

### Sections of a Docstring¶

The following sections can be included in a docstring, but in many cases, are optional.

1. Summary (required)

You need at least a one-line summary that does not use variable names or the function name, for example:

class ExternalCode(Component):
""" Run an external code as a component. """

You may want a longer summary (a few sentences) to clarify functionality. Do not discuss implementation details or background theory in the summary.

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*Parameters*

value: ndarray
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row_start: integer
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to the anchor, and can be negative.

row_end: integer
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field_start: integer
Starting field in the given row_start as denoted by
delimiter(s).

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We need this to figure out if the template is too small or large.

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In the above example, we italicized Parameters (indicated by the single asterisk on either side of the word) and used it as a heading for the parameters section. We did this merely to show how it can be done. Parameters and returns should be clear to users, so section headings aren’t required. On the other hand, you might want them for the optional sections.

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scipy.random.norm: Random variates, PDFs, etc.
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In general, use underline and overline for the title of document (in the index file for the document). In OpenMDAO, we typically use a double line for titles as shown in the following example:

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In general it is preferred that the heading levels go no more than four levels deep. If there are more than four levels, readers may forget where they are in the hierarchy. Of course, there are always exceptions, and the subject matter should be your guide.

When creating your document, be sure each heading level has a different underline style, and be consistent. The following example shows how heading levels may be set up:

Heading 1     (Title of the file)
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Notice that the italics designation (a set of asterisks) is used for Heading 3. Heading levels 2 and 3 are similar in size, so asterisks are used to better designate the hierarchy.

• If you are adding text to an existing file, you must use the same underlining scheme as the file’s creator.
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The tech writer will review new documentation to make sure you are consistent. Although, if you are not consistent, Sphinx will complain when you try to build.

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#### Class Names¶

Always capitalize class names (e.g., Component, Assembly, Driver, Engine, etc.). Sometimes, a class represents a concept having the same name. In that case, the name of the concept would generally not be capitalized.

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Capitalize only the first letter of Fortran unless you are referring to a version earlier than Fortran 90, when it was known as FORTRAN (e.g., FORTRAN 77).

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Capitalize Python when referring to the programming language, for example, a Python module. However, python should be lower case when it refers to an OpenMDAO path name, script name, command, part of a URL, etc., that is lower case.

This rule also applies to other programming languages or software programs; for example, Enthought, Inc. is capitalized, but enthoughts.traits.api is not.

#### reStructuredText¶

Please capitalize the appropriate letters and type it as one word: reStructuredText.

#### website¶

In our OpenMDAO documents, we will not capitalize website but use all lower case letters (one word). Please do not use any of the following variants: Web site, web site, or Website. Not only is website easier to type, but in 2003, in a survey of over 150 WordBiz Report subscribers, 65% of those surveyed preferred website – even those who thought Web site was correct. And as noted in The American Heritage Dictionary, 4th ed., the use of website reflects the trend of other technological expressions which have moved to unhyphenated forms as they become more familiar (e.g., email, online). The main thing is consistency, so please be consistent and use website.

### Numbers¶

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Inline literal test is designated by back quotes (the same computer key as the tilde) enclosing the specified text. Inline literal text can be used in many situations; in OpenMDAO documentation please use it to designate the following:

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directory names:      docs/dev-guide
api names:            openmdao.main.api
Python packages:      openmdao.lib
file names            example.rst

which will result in text that looks like this:

• /OpenMDAO/dev/shared/working_main
• docs/dev-guide
• openmdao.main.api
• openmdao.lib

Additionally, if a word or phrase contains an underscore (_) or a dot (.), use literal text, so it is easier to read, e.g., _init_ function, self.driver, and optimization_constrained.py.

### Abbreviations and Acronyms¶

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Abbreviation:       mph,for miles per hour; MB/s, for megabytes per second

Acronyms and abbreviations should go in the glossary. We have one glossary and one index for all OpenMDAO user documents.

• When to spell out: In general, spell out the term when you introduce it. You may also spell out an abbreviation or acronym if you think your audience may not be familiar with it.

• How to spell out: Generally put the spelled-out version first, with the abbreviation or acronym in parentheses, for example:

Internet service provider (ISP)

If the abbreviation or acronym is much more familiar than the spelled-out version, you can put the abbreviation or acronym first, followed by the spelled-out version in parentheses, or you can explain that the abbreviation is short for the spelled-out version and place the spelled-out version in italics.

You can share your personal URL (Uniform Resource Locator).
An Internet address is sometimes called a URL, short for Uniform Resource Locator.
• File types: Use all caps for abbreviations of file types.

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Filename extensions, which indicate the file type, should be in lowercase.

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• Plural: Don’t add an apostrophe before the “s” when you form the plural of an abbreviation.

CDs, URLs

Abbreviations:

2D - Abbreviation for two-dimensional. No hyphen in the abbreviation.

3D - Abbreviation for three-dimensional. No hyphen in the abbreviation.

### Hyphens and Dashes¶

- Hyphen:

Use a hyphen (-) for the following:

• Simple compound modifiers (twentieth-century writers) – Do not use a hyphen with adverb-adjective modifiers (wholly owned subsidiary)
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• Two-word numbers less than a hundred (twenty-nine)

Note

Do not put a hyphen in the word plugin. While plug-in is a correct variation, we must be consistent, and we chose to omit the hyphen. (This follows the trend of omitting hyphens in technological terminology.)

- En dash:

Use an en dash (–) for the following:

• To indicate a range (pp. 25–36, June–July 2006, 1:00–2:00 p.m., etc. Note that there are no spaces on either side of the dash.
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In reST an en dash is formed by typing two hyphens (minus signs).

### Comma (in a Compound Sentence)¶

• Use a comma before “and” when you have a compound sentence, for example:

Many analysis components will require some representation of geometry, and that representation could vary in detail from simple parameters, e.g., length, up to a full 3D mesh.

A comma is required before the “and” because the sentence has two independent clauses, i.e., each clause has a subject and a verb, making the sentence compound.
• Do not use a comma before “and” when the construction is merely a compound verb, as in:

Some of these effects were derived from empirical data and are essentially valid over an engine speed ranging from 1000 RPM to 6000 RPM.

In this case the sentence has one subject (some) but two verbs (were derived and are). It is not a compound sentence.


text:
# Question f639b

Aug 19, 2017

Use the Ratio Test (it's usually a good idea whenever factorials are involved) to show that this series converges.

#### Explanation:

Let a_{n}=(2/n)^{n}*n!.

We want to know whether the series ${\sum}_{n = 1}^{\infty} {a}_{n}$ converges or not.

To apply the Ratio Test, we consider the sequence $| {a}_{n + 1} / {a}_{n} |$. In this example, all the terms in the original sum are positive, so we may dispense with the absolute value signs and note that a_{n+1}/a_{n}=(2/(n+1))^(n+1) * (n+1)!*(n/2)^(n)*1/(n!).

This expression simplifies to:

${a}_{n + 1} / {a}_{n} = 2 {\left(\frac{n}{n + 1}\right)}^{n}$.

It is well known that ${\left(\frac{n + 1}{n}\right)}^{n} = {\left(1 + \frac{1}{n}\right)}^{n} \to e$ as $n \to \infty$. Therefore, ${a}_{n + 1} / {a}_{n} \to \frac{2}{e} \approx 0.73576 < 1$ as $n \to \infty$.

By the Ratio Test, this is enough to imply that the series sum_{n=1}^{infty}(2/n)^{n}*n! converges.

Aug 19, 2017

See below.

#### Explanation:

Using the Stirling asymptotic formula

n! approx sqrt(2pin)(n/e)^n we have

sum_(n=1)^oo(2/n)^2n! approx sum_(n=1)^oo(2/n)^2sqrt(2pin)(n/e)^n = sum_(n=1)^oo sqrt(2pin)(2/e)^n and

${\sum}_{n = 1}^{\infty} \sqrt{2 \pi n} {\left(\frac{2}{e}\right)}^{n} \le {\sum}_{n = 1}^{\infty} \left(n + 1\right) {\left(\frac{2}{e}\right)}^{n}$

and from

$\sum \left(k + 1\right) {x}^{k} = \frac{d}{\mathrm{dx}} \left(\sum {x}^{k + 1}\right)$ the convergence for $\sum \left(k + 1\right) {x}^{k}$ is assured if $\left\mid x \right\mid < 1$ or in our case

$\frac{2}{e} < 1$ and as a consequence

sum_(n=1)^oo(2/n)^2n! # converges.


text:
Arc Length

1. Jul 20, 2006

suspenc3

$$y^2 = 4(x+4)^3 0<=x<=2 ......y>0$$

Ok..so do I just simplify it to y=etc... and then find the derivative and solve..I tried this and I get a kinda long derivative..just seems like it should be easier?

Thanks

2. Jul 20, 2006

HallsofIvy

Not all that "long" is it? the derivative of y2, with respect to x, is $\2y \frac{dy}{dt}$ and the derivative of 4(x+4)3 is 12(x+ 4)2 so differentiating both sides you get
$$2y \frac{dy}{dx}= 12(x+4)^2$$
So what is $\frac{dy}{dx}$?

3. Jul 20, 2006

suspenc3

oh, i did it completely wrong..i took the root of both sides and hen tried to get the derivative

4. Jul 20, 2006

benorin

Whether you solve it implicitly (HallsofIvy's post) or explicitly (taking the rott, etc.) the result is the same: the implicit method gives

$$2y \frac{dy}{dx}= 12(x+4)^2\Rightarrow \frac{dy}{dx}= \frac{6(x+4)^2}{y} = \frac{6(x+4)^2}{2(x+4)^{\frac{3}{2}}}=3(x+4)^{\frac{1}{2}}$$

since solving the given equation for y gives $$y=2(x+4)^{\frac{3}{2}$$ and the positive root was chosen since y>0. This is the starting point for the explicit method, just differentiate the last expression

$$y=\frac{d}{dx}\left[ 2(x+4)^{\frac{3}{2}}\right] = 3(x+4)^{\frac{1}{2}}$$

which is actually easier.

5. Jul 20, 2006

suspenc3

oohhh..Ok Thanks, I get it

6. Jul 20, 2006

benorin

7. Jul 20, 2006

suspenc3

cant i just do..$$\int_0^2 \sqrt{1+(3(x+4)^1^/^2)^2}$$

$$\int_0^2 \sqrt{37+9x}$$

8. Jul 20, 2006

benorin

Yes, absolutely: I was in error.

9. Jul 20, 2006

suspenc3

but wha tyou have there..is $$\sqrt{3(x+4)}^2 = 9(x+4)^2$$..

shouldnt it be: $$\sqrt{3(x+4)}^2 = 8(x+4)$$..which can be simplified and then its an easy substitution.

10. Jul 20, 2006

suspenc3

oh,,ok..Thanks!

11. Jul 20, 2006

benorin

use rather u=37+9x so that du=9dx to get

$$\int_0^2 \sqrt{37+9x} dx = \frac{1}{9}\int_{37}^{55} \sqrt{u}du$$

Last edited: Jul 20, 2006
12. Jul 20, 2006

suspenc3

$$y= \frac{x^5}{5} + \frac{1}{10x^3} 1>=x>=2$$

I found the derivative to be $$\frac{5x^4}{6} - \frac{3}{10x^4}$$..im kinda stuck here..should I just keep going and will I eventually see a substituion?

Last edited: Jul 20, 2006
13. Jul 20, 2006

suspenc3

yes that is what i got..except du=9dx

14. Jul 20, 2006

benorin

yep, it was

15. Jul 20, 2006

suspenc3

if I have $$\int_0^\pi^/^3 sqrt{sec^2x}$$..should I substitute u = sec^2x..or can it just be secx..and find the integral?

Last edited: Jul 20, 2006
16. Jul 21, 2006

Do you mean $$\int_0^\frac{\pi}{3}\sqrt{sec^2x} dx$$?

If yes, note that since secx is positive from 0 to $$\pi$$/3, then $$\sqrt{sec^2x}$$ is equivalent to secx.

There is a standard formula for evaulating $$\int_0^\frac{\pi}{3}secxdx$$. Do you know what it is?

Last edited: Jul 21, 2006
17. Jul 21, 2006

benorin

The derivative of $$y= \frac{x^5}{5} + \frac{1}{10x^3}$$ is $$y^{\prime}= x^4 - \frac{3}{10x^4}$$

18. Jul 21, 2006

suspenc3

ohh..sorry its suppose to be $$y= \frac{x^5}{6} + \frac{1}{10x^3} 1>=x>=2$$

19. Jul 21, 2006

suspenc3

and for pizzasky..no I dont know how..I found some stuff on the internet concerning the integral of secx..I never knew this integral up until now..so im guessing theres another way

20. Jul 21, 2006


text:
# How to stop the result?

My program allows me to solve the equation Ht. The aim is to show the probability for each value of d value in my dlist table (dmin and dmax). But, for certain values of d, I have a first negative probability then two other positive ones. What i want, is that my program show no others p (prabability) results when the first probability is negative. that's means, if for a d value the first probability is negative, i need the program show only this negatif probability value and no others probability values.

Clear["Global*"]

c = 0.058;
rc = -0.06;
th = 0.80;

g1[ε_] = 2*(ε + 0.5)*(ε + 0.5);

g2[ε_] = 1 - 2*(-ε + 0.5)*(-ε + 0.5);

dmin = 1;
dmax = 3;
pas = 0.01;
dlist = Table[d, {d, dmin, dmax, pas}]
Length[dlist]

f[d_] = Min[{-0.02 + 0.0231397 d, 0.038524}];

rt = Simplify[((1 + rc) (1 - p th ))/(1 - p) - 1]

Ht = (rt - c) d - f[d]

k[d_, p_] = g1[Ht] - p

For[i = 1, i < Length[dlist] + 1,
i++, {Print[i "ème d fixé = ", dlist[[i]]],
Print[NSolve[k[dlist[[i]], p] == 0, p, Reals]]}]

k[d_, p_] = g2 [Ht] - p

For[i = 1, i < Length[dlist] + 1,
i++, {Print[i "ème d fixé = ", dlist[[i]]],
Print[NSolve[k[dlist[[i]], p] == 0, p, Reals]]}]

• Welcome to Mathematica.SE! I hope you will become a regular contributor. To get started, 1) take the introductory tour now, 2) when you see good questions and answers, vote them up by clicking the gray triangles, because the credibility of the system is based on the reputation gained by users sharing their knowledge, 3) remember to accept the answer, if any, that solves your problem, by clicking the checkmark sign, and 4) give help too, by answering questions in your areas of expertise. – bbgodfrey May 21 at 16:13

This can be accomplished by replacing

NSolve[k[dlist[[i]], p] == 0, p, Reals]

by

If[t = NSolve[k[dlist[[i]], p] == 0, p, Reals]; (p /. First[t]) < 0, First[t], Flatten[t]]

in the Print statements. This can be tested by

NSolve[k[dlist[[61]], p] == 0, p, Reals]
If[t = NSolve[k[dlist[[61]], p] == 0, p, Reals]; (p /. First[t]) < 0, First[t], Flatten[t]]
(* {{p -> 0.0274673}, {p -> 0.170184}, {p -> 0.775767}} *)
(* {p -> 0.0274673, p -> 0.170184, p -> 0.775767} *)

NSolve[k[dlist[[62]], p] == 0, p, Reals]
If[t = NSolve[k[dlist[[62]], p] == 0, p, Reals]; (p /. First[t]) < 0, First[t], Flatten[t]]
(* {{p -> -0.00249263}, {p -> 0.187624}, {p -> 0.775012}} *)
(* {p -> -0.00249263} *)

Note that I have used Flatten to eliminate an unnecessary set of curly brackets. Tor keep them, simply remove Flatten. I hope this helps.

• Hi Frey! Sorry for my late answer, iam a little bit ill this days. I've tried your code but could not found the same result. how did you do from the loop? did you keep the same structure as mine? – EmilioDas May 23 at 15:29
• I used For[i = 1, i < Length[dlist] + 1, i++, {Print[i "ème d fixé = ", dlist[[i]]], Print[If[t = NSolve[k[dlist[[i]], p] == 0, p, Reals]; (p /. First[t]) < 0, First[t], Flatten[t]]]}]. The values returned for i = 61 and i = 62 of the g2` computation are those in my answer. – bbgodfrey May 23 at 18:55
• Okay Frey, Thnaks a lot for the hand. Have a nice Day! – EmilioDas May 24 at 8:45


text:
## Friday, June 1, 2018

### 5/29 Phi Day

This week I wanted to extend some of our talk about the golden ratio. For the last reference see:  http://mymathclub.blogspot.com/2018/05/522-chaos-mod-arithmetic.html   I'm also not quite done testing out the weaving activity so this was easier to schedule right after a busy Memorial Day weekend.

I had two videos in mind that I recently saw:

I also had several group demos that I wanted to start off with. But I started by asking the room what they already knew about the golden ratio. As expected, Fibonacci numbers were mentioned and facts about famous art examples i.e. the Mona Lisa.

That was a good bridge to start with a precise definition of the ratio and from there we covered:
• Phi is not the same as the Fibonacci numbers and in fact for all such sequences defined by $F_n = F_{n-1} + F_{n-2}$ the ratio $\frac{F_n}{F_{n-1}}$  tend to approach the golden ratio.  This one you can test with your own generator numbers.
• Phi in the Pentagon. We derived the basic ratio  of the diagonal (upper center diagram) as a group.

• The general properties of Phi based on its root equation $x^2 -x -1 = 0$ i.e.
$$\Phi^2 = \Phi + 1$$
$$\Phi = 1 + \frac{1}{\Phi}$$

• The idea of finding spirals in a square.   I gave some demo with finding facts about the 10x10 grid. For example divide diagonally and you get its the sum of 2 triangle numbers. The challenge I gave out was for the kids to find a way to break it into a spiral.  Interestingly we found both ways of counting 10 + 9 + 9 + 8 + 8 vs. 10  +  8 + 10 .... which led to an interesting discussion ("You're both right - how can that be?")
I worked these on the board with breaks for the kids to try things out at their tables.  We then watched the videos.  Overall while everyone watched attentively, the first spiral simulation was particularly appreciated,  I think doing both was a mistake and I should have only used the numberphile one. On reflection, there was  a little too much passive viewing  and I would build a final Phi investigation/activity in at the end instead if I repeat.  The general idea of breaking a rectangle into its spiral/continued fraction is probably enough for an entire day on its own.

#### POTW adapted from Mike Lawler

1.  Can you find a polynomial with all integer coefficients and one root equal to ?
2. Can you find a polynomial with all integer coefficients and one root equal to ?
3.  Can you find a polynomial with all integer coefficients and one root equal to ?

[Notes: this was a bit too involved for a take home problem or perhaps its just the end of the year setting in. Next time it could probably be combined with the polynomial deltas to make a complete polynomial day.   In practice, I ended up carefully working part 1 as a group the next week and talked alot about conjugates and the relationship between roots and factors.]


text:
# The distributed winding in alternator -

1. Reduces harmonics from armature emf
2. Increases harmonics in armature emf
3. No impact on armature conductor
4. Provides voltage balance

Option 1 : Reduces harmonics from armature emf
Free
CT 1: Network Theory 1
11667
10 Questions 10 Marks 10 Mins

## Detailed Solution

Distributed winding:

For obtaining smooth sinusoidal emf waveform, conductors are placed is several slots under a single pole in a winding. This type of winding is known as distributed winding.

Although distributed armature winding of alternator reduces emf but still it very useful due to the following reasons

1. It reduces harmonic emf and so waveform is improved.

2. It also diminishes the armature reaction.

3. Even distribution of conductors, helps for better cooling.

4. The core is fully utilized as the conductors are distributed over the slots on the armature periphery.

Note:

Short pitch winding:

• It improves the waveform of generated emf i.e. generated emf can be made to approximate a sine wave more easily and the distorting harmonics can be reduced or totally eliminated.
• The disadvantage of using short-pitch winding is that the total voltage around the coils is somewhat reduced. In order to compensate for this reduction in Generated EMF, more number of turns, and therefore more copper is required.


text:
We included HMH Into Math Grade 7 Answer Key PDF Module 4 Add and Subtract Rational Numbers to make students experts in learning maths.

Which Sum Does Not Belong?

At Fun Times Pizza, customers can order a pizza in the shapes shown below.

Each diagram shows a sum of two rational numbers that represent combinations of pizza slices. Write an addition equation for the sum that is shown in the shaded part of each diagram.

Turn and Talk
Which of the sums do you think does not belong with the others? Justify your answer.

Complete these problems to review prior concepts and skills you will need for this module.

Add and Subtract Fractions and Decimals

Find the sum or difference.

Question 1.
8.04 + 4.15
________
The sum of two decimal numbers
8.04 + 4.15 = 12.19

Question 2.
9.1 + 12.98
________
The sum of two decimal numbers
9.1 + 12.98 = 22.08

Question 3.
88.4 – 39.75
________
The difference of two decimal numbers
88.4 – 39.75 = 48.65

Opposites and Absolute Value

Find the absolute value.

Question 4.
|-48|
________
The absolute value of a negative number will be always a non-negative number.
|-48| = 48

Question 5.
|15|
________
The absolute value of a negative or positive number will be always a non-negative number.
|15| = 15

Question 6.
|-150|
________
The absolute value of a negative number will be always a non-negative number.
|-150| = 150

Add and Subtract on a Number Line

Use the number line to help find each sum or difference.

Question 7.
-1.2 + 2.0
________
-1.2 + 2.0 = 0.8

Question 8.
1.0 – 1.2
________
1.0 – 1.2 = -0.2

Question 9.
-0.3 – 1.7
________
-0.3 – 1.7 = -2.0 = -2

Add and Subtract on a Number Line

Use the number line to help find each sum or difference.

Question 10.
–$$\frac{5}{6}$$ + $$\frac{2}{6}$$
__________
–$$\frac{5}{6}$$ + $$\frac{2}{6}$$
The denominator of both the fractions are the same so we have to subtract the numerators.
–$$\frac{5}{6}$$ + $$\frac{2}{6}$$ = –$$\frac{3}{6}$$ = –$$\frac{1}{2}$$

Question 11.
1$$\frac{1}{6}$$ – $$\frac{2}{3}$$
__________
1$$\frac{1}{6}$$ – $$\frac{2}{3}$$
Rewrite the equation
1 + $$\frac{1}{6}$$ – $$\frac{2}{3}$$
$$\frac{1}{6}$$ – $$\frac{2}{3}$$ = –$$\frac{3}{6}$$
1 – $$\frac{3}{6}$$ = 1 – $$\frac{1}{2}$$ = $$\frac{1}{2}$$

Question 12.
$$\frac{5}{3}$$ – 3$$\frac{1}{6}$$
__________
$$\frac{5}{3}$$ – 3$$\frac{1}{6}$$
$$\frac{5}{3}$$ – 3 – $$\frac{1}{6}$$
$$\frac{5}{3}$$ – $$\frac{1}{6}$$
$$\frac{5}{3}$$ – $$\frac{1}{6}$$ = $$\frac{10}{6}$$ – $$\frac{1}{6}$$ = $$\frac{9}{6}$$ = $$\frac{3}{2}$$ = 1$$\frac{1}{2}$$
-3 + 1 + $$\frac{1}{2}$$ = -2 + $$\frac{1}{2}$$ = -1$$\frac{1}{2}$$


text:
# What constitutes a “context” in Pose mode?

I have an armature associated with a mesh. In Pose Mode I have oriented the armature the way I want, so I press A to select the entire armature and then I press I to insert keyframes. I've done this successfully on simple models while following tutorials, but back in my real model I'm getting the following error and I don't understand what context is being looked for.

No suitable context info for active keying set

What constitutes a "context" in Pose mode? What should I be looking for in my model structure that is leading to such an error?


text:
Contests Virtual Contests Problems Submit Runs Status Rank List Forum

3021.   CIVIC DILL MIX
Time Limit: 1.0 Seconds   Memory Limit: 65536K
Total Runs: 498   Accepted Runs: 251

Roman numerals are an ancient numbering system used extensively throughout Europe through the 13th century (where it was eventually replaced by our current positional system). Vestiges of this system still exist today on clock faces, building cornerstones, Super Bowls and Star Wars episodes. The system uses the following 7 symbols:

Symbols I, X, C and M can be repeated as needed (though never more than three times for I, X and C), so that 3 is represented as III, 27 as XXVII and 4865 as MMMMDCCCLXV. The symbols are always written from the highest value to the lowest, but for one exception: if a lower symbol precedes a higher one, it is subtracted from the higher. Thus 4 is written not as IIII but as IV, and 900 is written as CM. The rules for this subtractive behavior are the following:

1. Only I, X and C can be subtracted.

2. These numbers can only appear once in their subtractive versions (e.g., you can't write 8 as IIX).

3. Each can only come before symbols that are no larger than 10 times their value. Thus we can not write IC for 99 or XD for 490 (these would be XCIX and CDXC, respectively). Note that the first two words in this problem title are invalid Roman numerals, but the third is fine.

Your task for this problem is simple: read in a set of Roman numeral values and output their sum as a Roman numeral.

### Input

Input will consist of multiple test cases. Each test case starts with a positive integer n indicating the number of values to add. After this will come n values (potentially several on a line), all valid Roman numerals with whitespace only coming between values. A value of n = 0 will indicate end of input. All sums will be less than 5000.

### Output

For each test case, output the case number and the sum, both as Roman numerals, using the format shown below. Case numbers should start at I.

### Sample Input

2
XII MDL
4
I I I
I
0

### Sample Output

Case I: MDLXII
Case II: IV

Source: East Central North America 2007
Submit   List    Runs   Forum   Statistics

Tianjin University Online Judge v1.3.0
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text:
## You are here

### Iterated Elliptic And Hypergeometric Integrals For Feynman Diagrams

 Title: Name(s): Iterated Elliptic And Hypergeometric Integrals For Feynman Diagrams. 21 views 3 downloads Ablinger, J., authorBluemlein, J., authorDe Freitas, A., authorvan Hoeij, M., authorImamoglu, E., authorRaab, C. G., authorRadu, C.-S., authorSchneider, C., author text Journal ArticleTextJournal Article 2018-06-01 computeronline resource 1 online resource English We calculate 3-loop master integrals for heavy quark correlators and the 3-loop quantum chromodynamics corrections to the rho-parameter. They obey non-factorizing differential equations of second order with more than three singularities, which cannot be factorized in Mellin-N space either. The solution of the homogeneous equations is possible in terms of F-2(1) Gau beta hypergeometric functions at rational argument. In some cases, integrals of this type can be mapped to complete elliptic integrals at rational argument. This class of functions appears to be the next one arising in the calculation of more complicated Feynman integrals following the harmonic polylogarithms, generalized polylogarithms, cyclotomic harmonic polylogarithms, square-root valued iterated integrals, and combinations thereof, which appear in simpler cases. The inhomogeneous solution of the corresponding differential equations can be given in terms of iterative integrals, where the new innermost letter itself is not an iterative integral. A new class of iterative integrals is introduced containing letters in which (multiple) definite integrals appear as factors. For the elliptic case, we also derive the solution in terms of integrals over modular functions and also modular forms, using q-product and series representations implied by Jacobi's nu(i) functions and Dedekind's eta-function. The corresponding representations can be traced back to polynomials out of Lambert-Eisenstein series, having representations also as elliptic polylogarithms, a q-factorial 1/eta(k) (tau), logarithms, and polylogarithms of q and their q-integrals. Due to the specific form of the physical variable x(q) for different processes, different representations do usually appear. Numerical results are also presented. Published by AIP Publishing. FSU_libsubv1_wos_000437094100030 (IID), 10.1063/1.4986417 (DOI) deep-inelastic scattering, differential-equations, dedekinds eta-function, heavy flavor contributions, inverse binomial sums, master integrals, operator matrix-elements, quantum-field theory, structure-function f-2(x, sunrise graph The publisher’s version of record is available at https://doi.org/10.1063/1.4986417 http://purl.flvc.org/fsu/fd/FSU_libsubv1_wos_000437094100030 FSU Journal of Mathematical Physics.0022-2488 iss. 6, vol. 59

Ablinger, J., Bluemlein, J., De Freitas, A., Van Hoeij, M., Imamoglu, E., Raab, C. G., … Schneider, C. (2018). Iterated Elliptic And Hypergeometric Integrals For Feynman Diagrams. Journal Of Mathematical Physics. Retrieved from http://purl.flvc.org/fsu/fd/FSU_libsubv1_wos_000437094100030


text:
<meta http-equiv="refresh" content="1; url=/nojavascript/"> Triangles and Angles | CK-12 Foundation

# 6.2: Triangles and Angles

Created by: CK-12
0  0  0

## Introduction

Tipi Decoration

Jaime has been working hard on her tipi. She has decided to use the following pattern as part of the design on the material. She thinks that if she uses cloth triangles, that she can sew them onto the cloth of the tipi to make a pattern. This red, black, red pattern is going to stretch along the outside bottom edge of the tipi.

“That is very cool,” her sister Lily said admiring Jaime’s sketch.

“I think so too. It should look very pretty on the brown material,” Jaime added.

“Yes. How do you know that the triangles will be the exact same size?”

“That’s easy. I can measure the top angle of the triangle using a protractor and then create each one by tracing. This one will have an angle of $120^{\circ}$. when I am done,” Jaime explained to Lily.

“What kind of triangle is that?” Lily asked.

Do you know? This lesson is all about different types of triangles. As you learn all about the different types of triangles, think about this problem. What type of triangle is in the pattern? Can you justify your answer?

What You Will Learn

In this lesson, you will learn how to complete the following skills and tasks.

• Classify triangles by angle measures (acute, right, obtuse) and by side lengths (equilateral, isosceles, scalene)
• Confirm that the sum of any angles of a triangle is equal to $180^{\circ}$.
• Describe and analyze triangles and associated angle measures using known information, variable expressions and including triangles formed by two lines intersecting parallel lines.
• Describe and classify triangles found in real-world objects and architecture.

Teaching Time

I. Classify Triangles by Angle Measures (Acute, Right, Obtuse) and by Side Lengths (Equilateral, Isosceles and Scalene)

In this lesson we will examine different kinds of triangles. As we know, triangles are geometric figures that have three sides and three angles. There are different types of triangles too. We can classify or identify them in different ways. One way is by their angles measures and one way is by their sides lengths.

We all know that triangles have three angles. The corners where each of the line segments connects form an angle. When we know the measures of these angles, we can use this information to name and identify the triangles. Let’s look at some of the different types of triangles according to angle measure.

Acute Triangles are triangles that have three angles that are less than $90^{\circ}$. The word “acute” when applied to angles means less than $90^{\circ}$, so an acute triangle has three angles where all three angles are less than $90^{\circ}$.

Right Triangle a right triangle is a triangle with one $90^{\circ}$ angle. The other two angles will be acute, but the key to identifying a right triangle is that it has one right angle.

Obtuse Triangle is a triangle with one angle that is obtuse or greater than $90^{\circ}$.

Now let’s apply this information in an example.

Example

Identify whether each of the following triangles is acute, obtuse or right.

Now let’s break each one down.

With the first triangle, triangle $a$, we can see that one of the angles is greater than 90 degrees. This is an obtuse triangle.

The second triangle has three angles that are less than 90 degrees, this is an acute triangle.

The third triangle also has three angles that are less than 90 degrees. This is also an acute triangle.

The fourth triangle has a right angle. You can see that because it forms a nice neat corner so perfectly. This is a right triangle.

The fifth triangle has an angle greater than 90 degrees. This is an obtuse triangle.

We can also classify or identify triangles by the length of their sides. This means that we look at the line segments that create the triangle.

Equilateral Triangle is a triangle with all three sides equal.

Isosceles Triangle has two sides that are equal in length. Often an isosceles triangle is the trickiest one to identify.

Scalene Triangle is a triangle where none of the sides are the same length. All three sides are different lengths.

Now let’s apply what we have learned and identify some triangles.

Example

Classify each triangle as equilateral, isosceles, or scalene.

We need to examine the lengths of the sides in each triangle to see if any sides are congruent.

In the first triangle, two sides are 7 meters long, but the third side is shorter. Which kind of triangle has two congruent sides? This is an isosceles triangle.

Now let’s look at the second triangle. All three sides are the same length, so this must be an equilateral triangle.

The last triangle has sides of 5.5 cm, 4.1 cm, and 8 cm. None of the sides are congruent, so this is a scalene triangle.

Sometimes, we can classify a triangle by both its sides and its angles.

Example

Identify each triangle by both its sides and angles.

The first triangle is a right isosceles triangle. It has one right angle and two sides that are the same length.

The second triangle is an acute scalene triangle. All three sides are different lengths and all three angle measures are acute.

The third triangle is an obtuse scalene triangle. It has one obtuse angle and three different side lengths.

The last triangle is an obtuse isosceles. It has one obtuse angle and two side lengths that are the same.

Equilateral triangles do not quite fit this pattern. They are always acute. This is because the three angles in an equilateral triangle always measure $60^{\circ}$.

There is one more thing to know about classifying triangles by their angles and sides. We can also tell whether a triangle is isosceles, scalene, or equilateral by its angles. Every angle is related to the side opposite it. Imagine a book opening. The wider you open it, the greater the distance between the two flaps. In other words, the wider an angle, the longer the side opposite it is. Therefore we can say that if a triangle has two congruent angles, it must have two congruent sides, and thus it must be isosceles. If it has three angles of different measures, then its sides are also all of different lengths, so it is scalene. Finally, an equilateral triangle, as we have seen, always has angles of $60^{\circ}$, and these angles are opposite congruent sides.

Notice that the word “congruent” means exactly the same.

II. Confirm that the Sum of any Angles of a Triangle is $\underline{180^{\circ}}$

If you look at all of the angles in a triangle, you will notice something consistent about each one of them. If we add up the number of degrees in each angle of a triangle, you will see that the sum of the angle measures is equal to $180^{\circ}$.

That is a great question. That short answer is, yes, it is always true. But let’s look at an example to understand this a little further.

We can start by looking at an equilateral triangle. The three angles of an equilateral triangle are all equal. We know from the last section, that each of these angles is 60 degrees. Here is an example of an equilateral triangle.

Now let’s look at what happens when we cut out the $60^{\circ}$ angles. Three $60^{\circ}$ angles are equal to $180^{\circ}$ and there are $180^{\circ}$ in a straight line. The sum of the angles of a triangle is $180^{\circ}$, and this will happen no matter what the angle measures are. The angles of a triangle will always form a straight line and be equal to $180^{\circ}$.

Now let’s look at how we can use this information to find missing angle measures.

III. Describe and Analyze Triangles and Associated Angle Measures Using Known Information, Variable Expressions and Triangles Formed by Two Lines Intersecting Parallel Lines

Now that we have looked at how to identify triangles by angle measures and side lengths, we can look at how we can work with missing angle measures. We identify missing angle measures using what we know about triangles and by writing variable expressions. Let’s take a look at how we can problem solve in an example.

Example

What is the missing angle measure of this triangle?

Now we have a triangle here. First, we can use our known information to figure out what kind of triangle it is. Let’s begin by looking at the angles in this triangle. There are two small angles. These are 25 degree angles and they are acute. We can see by looking at this third unknown angle, that is an obtuse angle. This is an obtuse triangle.

Next, we can write an expression to help us to figure out the missing measure.

$25+25+x$

Now, we can write this expression into an equation with a sum of 180 degrees. Then we can solve it for the value of $x$.

$25+25+x &= 180\\50+x &= 180\\x &= 180-50\\x &= 130^{\circ}$

The measure of the missing angle is $130^{\circ}$.

Now let’s look at another example.

Example

What is the measure of the two missing angles if this is an isosceles triangle?

Here we have two missing angles. We know from the problem that this is an isosceles triangle. That means that side lengths are the same and we can see that the two base angles are also congruent. Our given angle is $50^{\circ}$, so we can write a variable expression to help us figure out the measure of the missing base angles.

$x+x+50$

We can expand this expression to an equation that is equal to 180 degrees.

$x+x+50=180$

Next, we combine the like terms before solving this.

$2x+50 &= 180\\2x &= 130\\x &= 65^{\circ}$

Each of the base angles is equal to $65^{\circ}$.

Sometimes, we will have triangles that are created by intersecting lines. When this happens, we can use the information that we know about intersecting lines to figure out the missing angle measures. Let’s look at an example.

Example

Find the value of the missing angles $x$ and $y$

Now to work through this problem, you will need to apply all of the things that you have learned to problem solve the measures of the missing angles.

Let’s start by looking at angle $x$.

You can see that angle $x$ is an acute angle. It is also an adjacent angle with the 140 degree angle already labeled. We know that the sum of adjacent angles is $180^{\circ}$. Now we can write an equation and solve for the missing angle measure.

$140 + x &= 180\\x &= 40^{\circ}$

There are two ways to find the measure of angle $y$. One is to use the sum of the angle measures given that we know the measure of $x$. Let’s do that one first.

$y+40+85 &= 180\\y+125 &= 180\\y &= 55^{\circ}$

The second way is to use vertical angles. You can see that the angle $125^{\circ}$ is labeled. This means that the angle vertical to this labeled angle is also $125^{\circ}$. The angle $y$ forms a straight line with that angle and therefore is $55^{\circ}$.

See if you can figure out one more way to solve for this missing angle.

IV. Describe and Classify Triangles Found in Real – World Objects and Architecture

There are many triangles out there in the real – world. One of the big places that you will see triangles is in bridge construction. Look at this picture of a truss bridge.

Looking at this bridge, you can see how the basic shape of a triangle is fundamental to the design of the bridge. The triangle helps to keep the bridge stable because of the strength of its foundation. The triangle is a shape that because of its base is very stable and won’t give to pressure. It is a balanced figure.

We can also see triangles in roof design. Take a look at this picture.

The roof construction of this home is using a triangle for stability. It also allows for water run-off during rain or when snow melts. Once again, using triangles makes sense.

Look around your home and neighborhood-where are some other places where triangles are very useful? Make a short list of five other places where you can find triangles.

## Real-Life Example Completed

Tipi Decoration

Here is the original problem once again. Reread it and then answer the questions for the solution.

Jaime has been working hard on her tipi. She has decided to use the following pattern as part of the design on the material. She thinks that if she uses cloth triangles, that she can sew them onto the cloth of the tipi to make a pattern. This red, black, red pattern is going to stretch along the outside bottom edge of the tipi.

“That is very cool,” her sister Lily said admiring Jaime’s sketch.

“I think so too. It should look very pretty on the brown material,” Jaime added.

“Yes. How do you know that the triangles will be the exact same size?”

“That’s easy. I can measure the top angle of the triangle using a protractor and then create each one by tracing. This one will have an angle of $120^{\circ}$ when I am done,” Jaime explained to Lily.

“What kind of triangle is that?” Lily asked.

Answer the questions on the following page.

Solution to Real – Life Example

Here are the questions from the introduction. What type of triangle is in the pattern? Can you justify your answer?

The triangle in the pattern is an obtuse triangle. It is an obtuse triangle because it has an angle that is greater than $90^{\circ}$. The other two angles are less than $90^{\circ}$. Because the largest angle is $120^{\circ}$, it is an obtuse triangle.

## Vocabulary

Here are the vocabulary words that are found in this lesson.

Acute Triangle
A triangle where all three angles are less than $90^{\circ}$.
Right Triangle
A triangle with one $90^{\circ}$ angle and two acute angles.
Obtuse Triangle
a triangle with one angle that is greater than $90^{\circ}$.
Equilateral Triangle
all three side lengths and all three angles are congruent.
Isosceles Triangle
two side lengths are the same.
Scalene Triangle
all three side lengths are different
Congruent
means exactly the same, having the same measure.

## Time to Practice

Directions: Classify each triangle by the describe angle measures as acute, obtuse or right.

1. A triangle with three $60^{\circ}$ angles.
2. A triangle with one $110^{\circ}$ angle.
3. A triangle with one right angle and two acute angles.
4. A triangle with one $130^{\circ}$ angle.
5. A triangle with three acute angles.

Directions: Identify each triangle by the side lengths described. Identify them as equilateral, isosceles or scalene.

1. A triangle with side lengths of 6 in, 6 in and 4 inches.
2. A triangle with side lengths of 3 ft, 4 ft, and 5 ft.
3. A triangle with side lengths of 8 inches.
4. A triangle with side lengths of 7 inches, 8 inches and 8 inches.
5. A triangle with side lengths of 6 meters, 8 meters and 10 meters.
6. A triangle with side lengths of 10 mm.

Directions: Using what you have learned about the interior angles of a triangle, determine the missing angle in each triangle.

1. $45^{\circ}, 45^{\circ}, ?$
2. $60^{\circ}, 60^{\circ}, ?$
3. $90^{\circ}, 50^{\circ}, ?$
4. $100^{\circ}, 40^{\circ}, ?$
5. $110^{\circ}, 30^{\circ}, ?$
6. $50^{\circ}, 10^{\circ}, ?$

Directions: Identify three triangles in the room around you.

Jan 14, 2013

Sep 23, 2014


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# Math Help - Converting partial sum to infinite series

1. ## Converting partial sum to infinite series

If a partial sum of a series is given this way: $S_{n}&space;=&space;\frac{3^{n+5}&space;+&space;8n^{2}&space;-&space;9&space;}{3^{n}}$

Then, can I say since the $\lim_{n\rightarrow&space;\infty&space;}S_{n}&space;=&space;\frac{3^{n+5}&space;+&space;8n^{2}&space;-&space;9&space;}{3^{n}}=243$,

Is this right? Can I do this to find the series of the partial sum equation of ?

Thanks!

2. Originally Posted by xEnOn
If a partial sum of a series is given this way: $S_{n}&space;=&space;\frac{3^{n+5}&space;+&space;8n^{2}&space;-&space;9&space;}{3^{n}}$

Then, can I say since the $\lim_{n\rightarrow&space;\infty&space;}S_{n}&space;=&space;\frac{3^{n+5}&space;+&space;8n^{2}&space;-&space;9&space;}{3^{n}}=243$,

Is this right? Can I do this to find the series of the partial sum equation of ?

Thanks!
No; the series isn't geometric.

Find $T_n$ the traditional way. ($T_n=S_n-S_{n-1}$)

3. But if say I want to find Xn of this:
$\lim_{n\rightarrow&space;\infty&space;}\frac{3^{n+5}+8n^{2}-9}{3^{n}}=\sum_{n=0}^{\infty&space;}X_{n}=243$

Is there a way to solve for $X_{n}$ in the summation?

4. Originally Posted by xEnOn
But if say I want to find Xn of this:
$\lim_{n\rightarrow&space;\infty&space;}\frac{3^{n+5}+8n^{2}-9}{3^{n}}=\sum_{n=0}^{\infty&space;}X_{n}=243$

Is there a way to solve for $X_{n}$ in the summation?
Use my formula in post #2 to find the nth term.

5. hmm...I am not very sure how I can use the formula. I haven't tried this before. Is the ?

And then I have something like this: ?
But this looks weird because Tn will become 0.

6. Originally Posted by xEnOn
hmm...I am not very sure how I can use the formula. I haven't tried this before. Is the ?

And then I have something like this: ?
But this looks weird because Tn will become 0.
The nth term $T_n=S_n-S_{n-1}$

=$\frac{3^{n+5}+8n^2-9}{3^n}-\frac{3^{n+4}+8(n-1)^2-9}{3^{n-1}}$

=$\frac{3^{n+5}+8n^2-9-3^{n+5}-24(n-1)^2+27}{3^n}$

=$\frac{18+8n^2-24(n^2-2n+1)}{3^n}$

=$\frac{-16n^2+48n-6}{3^n}$

7. But the difference between the sum of (0 to n) and sum of (0 to n-1) is not always the same. So this isn't the kind of usual arithmetic series and I cannot just say this is it $\sum_{n=0}^{\infty&space;}\frac{2(-8n^{2}+24n-3)}{3^{k}}$ right?

And since it is going to be an infinite series, don't I have to take some kind of limits towards infinity?

8. Originally Posted by xEnOn
But the difference between the sum of (0 to n) and sum of (0 to n-1) is not always the same. So this isn't the kind of usual arithmetic series and I cannot just say this is it $\sum_{n=0}^{\infty&space;}\frac{2(-8n^{2}+24n-3)}{3^{k}}$ right?

And since it is going to be an infinite series, don't I have to take some kind of limits towards infinity?
This series is neither arithmetic nor geometric.

$S_n=\sum_{n=1}^{n}\frac{2(-8n^{2}+24n-3)}{3^{n}}$

9. ohh.. So it could just be like a summation of the difference between the previous and current term to get the eventual sum of the equation. But would this be very accurate compared to the actual equation?

Thank you so much!!

10. Originally Posted by xEnOn
ohh.. So it could just be like a summation of the difference between the previous and current term to get the eventual sum of the equation. But would this be very accurate compared to the actual equation?

Thank you so much!!
More than accurate...it is exact!

11. great! thanks for all the help!


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Last time we looked at the break points of where your win percentage should be to play in 8-4s vs. 4-3-2-2s. There is a third draft option, however, and that is Swiss.In Swiss drafts, you play 3 matches and win a pack for each match. Figuring out the expected value of a Swiss draft is pretty easy. There are 3 independent trials and the probability of winning each trial is $p$. There are four possible outcomes for the Swiss drafts. They and the associated probabilities are:

• Win 0 matches: $-p^3+3p^2-3p+1$. $1-p$ times $1-p$ times $1-p$ gives you $(1-p)^3$. Expanding the cube out gives you $-p^3+3p^2-3p+1$.
• Win 1 match: $p^3-2p^2+p$. $p$ times $1-p$ times $1-p$ gives you $p(1-p)^2$. Expanding the square out and multiplying by $p$ gives you $p^3-2p^2+p$.
• Win 2 matches: $-p^3+p^2$. $p$ times $p$ times $1-p$ gives you $p^2(1-p)$. Multiplying the two factors gives you $-p^3+p^2$.
• Win 3 matches: $p^3$. $p$ times $p$ times $p$ is $p^3$.

Now we determine the expected value of those outcomes:

• Win 0 matches: 0 value.
• Win 1 match: $3p^3-6p^2+3p$. Since there are three different ways you can win 1 match in a Swiss draft (win the first round or win the second round or win the third round), we multiply by 3 to determine the expected value of this event.
• Win 2 matches: $-6p^3+6p^2$. Since there are three different ways you can win 2 matches in a Swiss draft (lose the first round or lose the second round or lose the third round), we multiply by 3 to determine the expected value of this event.
• Win 3 matches: $3p^3$.

$$3p^3-6p^2+3p-6p^3+6p^2+3p^3$$

$$3p$$

The expected value of $n$ independent events each with the same probability $p$ is $np$. For this case, it's $3p$. Now we need to compare the expected value of a Swiss draft to the expected values of 4-3-2-2 and 8-4 drafts. From the last article, we determined that the point where 4-3-2-2 and 8-4 drafts had the same return for a player is if their probability of winning a match is:

$$p=\frac{\sqrt{33}-3}{6}$$

Or approximately .457. Below that probability, playing in 4-3-2-2s is better. Above that, playing in 8-4s is better. So below that line, we'll compare the expected value of a Swiss event to a 4-3-2-2 event:

$$p^3+p^2+2p=3p$$

$$p^3+p^2-p=0$$

$$p(p^2+p-1)=0$$

Discarding the $p=0$ case and using the quadratic formula, we get:

$$p=\frac{-1\pm\sqrt{1^2-4(1)(-1)}}{2(1)}$$

$$p=\frac{-1\pm\sqrt{1+4}}{2}$$

$$p=\frac{-1\pm\sqrt{5}}{2}$$

We can throw out the minus case as it causes $p$ to be less than 0, so we end up with:

$$p=\frac{\sqrt{5}-1}{2}$$

This approximates out to .618! This means that across the entire range where you would play 4-3-2-2s, only one of 4-3-2-2 or Swiss is right from an EV perspective. Which is it? Let's try a sample case...

For $p=.1$:

$$3p\,?\,p^3+p^2+2p$$

$$3(.1)\,?\,.1^3+.1^2+2(.1)$$

$$.3\,?\,.001+.01+2(.1)$$

$$.3\,?\,.001+.01+.2$$

$$.3\,>\,.211$$

Very interesting! For the entire range where you would play 4-3-2-2 events, you should play Swiss events instead if your probability of winning is the same in both! Now what about where you should play 8-4 events, above the .457 probability?

$$4p^3+4p^2=3p$$

$$4p^3+4p^2-3p=0$$

$$p(4p^2+4p-3)=0$$

Discarding the $p=0$ case and using the quadratic formula, we get:

$$p=\frac{-4\pm\sqrt{4^2-4(4)(-3)}}{2(4)}$$

$$p=\frac{-4\pm\sqrt{16+48}}{8}$$

$$p=\frac{-4\pm\sqrt{64}}{8}$$

$$p=\frac{-4\pm8}{8}$$

Discarding the minus case as usual, we get:

$$p=\frac{8-4}{8}$$

$$p=.5$$

We have two segments to test, between .457 and .5 and between .5 and 1. The first segment:

For $p=.46$:

$$4p^3+4p^2\,?\,3p$$

$$4(.46)^3+4(.46)^2\,?\,3(.46)$$

$$4(.097336)+4(.2116)\,?\,1.38$$

$$.389344+.8464\,?\,1.38$$

$$1.235744\,<\,1.38$$

So up to a probability of .5, you should be playing Swiss. Above that?

For $p=.6$:

$$4p^3+4p^2\,?\,3p$$

$$4(.6)^3+4(.6)^2\,?\,3(.6)$$

$$4(.216)+4(.36)\,?\,1.8$$

$$.864+1.44\,?\,1.8$$

$$2.304\,>\,1.8$$

Above a probability of .5, you should play 8-4s.

So what we've learned so far is that, given equal probability of winning a match across all three draft formats, if your probability of winning a match is less than 50%, you should play in the Swiss queues, and above 50%, you should play in the 8-4 queues, and you should never play 4-3-2-2s. But the reality is that skill level across the queues are different. 8-4 drafts typically attract better players than 4-3-2-2s and Swiss queues. Next, we'll attempt to take a look at how much better each queue is than the other.


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×

# really complex....

Note by Trishit Chandra
2 years, 4 months ago

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My text book named additional mathematics quoted that a and b must be real numbers for all convenience you mentioned. Basically, imaginary numbers or complex numbers posses many possibilities while staying in a function but ought to be an only value when they are not. Example: x^2 = -1, x can be + j or - j but Sqrt (-1) or j itself cannot choose to own both values. Another example: e^x = -1, x can be -3 Pi j, - Pi j, Pi j, 3 Pi j, 5 Pi j and etc. Here, we define principal value for Ln -1 which is j Pi.

Complex number undoubtedly have been studied by many people including mathematicians of course and it is found to be something important that unable to be denied by people. Without complex number, there are many logical outcomes that cannot be explained and topics remained incomplete. The most significant example is solution to x^3 + p x + q = 0. We may refuse quadratic formula when j appears, but we cannot do the same for cubic formula.

When we realize that something ought to survive, we shall have thousands of reasons to prove them right; on the other hand, we could have ten thousands of reasons instead just to reject them. Towards the surviving path, we ought to study and understand the feature of complex number rather than to go for ways that may find them some flaws.

When we do in such a way that j suddenly becomes -j, we ought to realize its nature and also the meaning of a principal value. In other words, we shall find NO TROUBLE at all when we don't do in a way that purposely try to reveal their nature of many values. Sincerely, I personally find that complex numbers are as true as they can be.

Just try our best to cope with their features. Then, we shall find that Complex Number is the way towards all truth. · 2 years, 4 months ago

Indeed. A lot of times, we memorize rules without recalling what are the conditions under which it holds. In this case, as you pointed out, the rule only applies for real numbers, and cannot be applied to complex numbers.

This is why in Rules of Exponents, we added in a warning about when these rules hold:

Other common instance of forgetting conditions is applying AM-GM to negative numbers. Staff · 2 years, 4 months ago

I meant not all convenience are applicable if not real number, not meant for not applicable to complex number. I concluded from analysis that when there comes with fractional index with simplest co-prime, example (-1)^(5/ 3) or (-1)^(7/ 2), always take the effect of its denominator before numerator. Without list of black and white one by one, we just need to emphasize that we conclude according to whatever reasonable. I think 0^0 = 1 should be included in the list. As its limit tells, 0^0 = 0^(1 - 1) = 0/ 0 doesn't deny the fact for it to be 1. 0/ 0 includes 1 but generally something instead of invalid. Indeterminate is just something but we cannot know one of them in general to satisfy particular need to be sole definable value. Thanks anyway for providing the list for me to do some revision and thinking. · 2 years, 4 months ago

thank you sir Lu Chee Ket for helping me to clear my doubt and also thanks to calvin lin. · 2 years, 4 months ago

I can write -

$$i^2 = \sqrt{-1}\times\sqrt{-1}$$

According to you ,

$$i^2 = \sqrt{-1}\times\sqrt{-1} = \sqrt{-1\times-1} = 1$$ - this is a contradiction as $$i^2 = -1$$ · 2 years, 4 months ago

yes exactly i've the confusion with the contradiction. · 2 years, 4 months ago

Good for doubting. This is the spirit that we should request to ourselves. If you truly prove it incorrect, then I personally feel happy to admit the fact. This should be our correct attitude. · 2 years, 4 months ago

thankkkk u sir Lu Chee Ket for encouraging me...i will be highly obliged if you kindly give a link of your book. · 2 years, 4 months ago

Didn't get your exact meaning. Anyway, no book from me that helps in this topic. I only wrote about general number. · 2 years, 4 months ago

In the first comment you wrote that your text book named additional mathematics quoted.........Thats why I thought that you wrote a book about this topic. But if you write about general number then you can also send me that link. · 2 years, 4 months ago

The only book published contained mistakes despite the concept. I had stopped giving to people since long ago. Upon my own need to work on it, only then I gave to one or two people. I may produce new version in future. I think this has not been the proper moment that I could introduce to you. Not easy to get help to start developing. Anyway, thanks for your concern. Calvin Lin had posted some immediate introduction to questions of your most concern. · 2 years, 4 months ago

Ok thank you sir. And nice to meet you. · 2 years, 4 months ago