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## Desert Bus donation formula analysis

Drop by and talk about anything you want. This is where all cheese-related discussions should go
MarkVI
Posts: 181
Joined: 14 Jul 2008, 19:18
First Video: iPhone Alternative
Location: New Jersey

### Desert Bus donation formula analysis

For those who are interested, I decided to put up a short summary of how to compute the total donations required to achieve k hours of game play. Conversely, I've derived the formula to compute how many hours of Desert Bus would have to be played given C dollars of total donations.

The basic idea: Hour 1 costs $1 and subsequent hours cost 1.07 times the previous hour. Let H(k) be the cost of hour k. Then H(k) = (1.07)^(k-1). Moreover, the total cost C(k) of hours 1 through k will be C(k) = H(1) + H(2) + ... + H(k) = \sum_{j=1}^{k} H(j) = \sum_{j=1}^{k} (1.07)^(j-1) = \sum_{j=0}^{k-1} (1.07)^j. This simplifies to: C(k) = (1.07^(k) - 1)*(100/7) Now to find the total hours achieved for a given donation amount C. It turns out that k = ln( 7C/100 + 1) / ln(1.07) = log_{1.07}(7C/100 + 1) Lyinginbedmon Posts: 10808 Joined: 20 Dec 2007, 18:08 First Video: BioShocked Location: Darlington, Co. Durham Contact: ### Re: Desert Bus donation formula analysis This is cool, but I must ask: How bored were you? Morgan wrote:Lyinginbedmon is short, but he makes up for it in awesomeness Jillers Posts: 3006 Joined: 14 Oct 2008, 19:26 First Video: How to Talk LIke a Pirate Location: Somewhere on top of garbage Contact: ### Re: Desert Bus donation formula analysis Nope, still hate math. Team Cybershark [subliminal] visit Geeking Out About [/subliminal] ~End Transmission~ Master Gunner Defending us from The Dutch! Posts: 19383 Joined: 29 Oct 2006, 12:19 First Video: How To Talk Like A Pirate Location: In Limbo. Contact: ### Re: Desert Bus donation formula analysis Lyinginbedmon wrote:This is cool, but I must ask: How bored were you? Probably not that bored, the formulas are standard, you just need to plug in the numbers that apply here. I did that last year to calculate for the last few hours. Crusher did it the first year to come up with his list of the first 120 hours. As well, last year somebody came up with a few charts to show how different increases would work out. Its not hard, and the formulas are easy to look up. TheRocket wrote:Apparently the crotch area could not contain the badonkadonk area. Twitter | Click here to join the Desert Bus Community Chat. steelfox Posts: 215 Joined: 22 Aug 2008, 10:00 Location: Spokane, WA ### Re: Desert Bus donation formula analysis I thought the formula was: Give money until they all go mad from driving... Also ask for ridiculous things. Matt wrote:AND IF SHE'S DEAD SHE CAN'T SAY NO. -M Master Gunner Defending us from The Dutch! Posts: 19383 Joined: 29 Oct 2006, 12:19 First Video: How To Talk Like A Pirate Location: In Limbo. Contact: ### Re: Desert Bus donation formula analysis Well that too. TheRocket wrote:Apparently the crotch area could not contain the badonkadonk area. Twitter | Click here to join the Desert Bus Community Chat. Arius Posts: 7228 Joined: 23 Jun 2009, 03:38 First Video: Max Effect Location: Inside You Contact: ### Re: Desert Bus donation formula analysis steelfox wrote:I thought the formula was: Give money until they all go mad from driving... Also ask for ridiculous things. Don't forget to compensate for when they have time away from DB, gotta make sure to break their willpower when they get back. Otherwise they resist. H̼̮̖͓̻ͮ̀ͬ̓e̟̦͉̾̔̀ͣ͆̄ ͚̤̈̉ͦ̎ͭ̚c̰̠͚̜̹ͪ̐̎̃ͅo̗͌͛ͥ͑m̍ͬͥ̚e͍̱̲̤͚̹͔͛s͚̱̤͚̲̭̗̃̎ͭ̚.̘̫̖̮̠͒̔.̝̹̟̳͚̂̆̋͌̐̚.̬͓̰̃̑ Evil Jim Posts: 7265 Joined: 14 Jul 2007, 00:39 First Video: Shake Your Hands Location: R'lyeh, Wisconsin Contact: ### Re: Desert Bus donation formula analysis Isn't it, each hour costs seven or eight percent more than the previous one, with the first hour costing a dollar? Arius wrote:People were just so awestruck by your awesomeness that they became catatonic. ThrashJazzAssassin wrote:BURN HIM! BURN THE HERETIC! DEATH TO ALL WHO SCORN THE AWESOMENESS OF EVIL JIM! MarkVI Posts: 181 Joined: 14 Jul 2008, 19:18 First Video: iPhone Alternative Location: New Jersey ### Re: Desert Bus donation formula analysis I didn't have to look anything up other than the 1.07 multiplier. Everything else is easy to derive from scratch in 5 min. Graham Super Moderator Posts: 15038 Joined: 09 Mar 2004, 19:37 Location: Victoria, BC Contact: ### Re: Desert Bus donation formula analysis Evil Jim wrote:Isn't it, each hour costs seven or eight percent more than the previous one, with the first hour costing a dollar? That's what multiplying by 1.07 will do—a 7% increase. Evil Jim Posts: 7265 Joined: 14 Jul 2007, 00:39 First Video: Shake Your Hands Location: R'lyeh, Wisconsin Contact: ### Re: Desert Bus donation formula analysis Ah, whups. I skipped over the equations in the first post. Arius wrote:People were just so awestruck by your awesomeness that they became catatonic. ThrashJazzAssassin wrote:BURN HIM! BURN THE HERETIC! DEATH TO ALL WHO SCORN THE AWESOMENESS OF EVIL JIM! Lord Chrusher Can't Drink Possible Beers Posts: 8913 Joined: 29 Apr 2005, 22:53 First Video: Door to Door Location: In England. ### Re: Desert Bus donation formula analysis Master Gunner wrote: Lyinginbedmon wrote:This is cool, but I must ask: How bored were you? Probably not that bored, the formulas are standard, you just need to plug in the numbers that apply here. I did that last year to calculate for the last few hours. Crusher did it the first year to come up with his list of the first 120 hours. As well, last year somebody came up with a few charts to show how different increases would work out. Its not hard, and the formulas are easy to look up. I will have to post a bigger table for this year. We are all made of star dust. However we are also made of nuclear waste. Remember to think before you post. Lord Chrusher Can't Drink Possible Beers Posts: 8913 Joined: 29 Apr 2005, 22:53 First Video: Door to Door Location: In England. ### Re: Desert Bus donation formula analysis Ok here is this year's version of the donation table. It goes for seven days - over a million dollars would have to be raised for Desert Bus for it to go that long, roughly the total amount of money Child's Play raised in total last year. Code: Select all Hours Total To Next Hour0 0.00 1.001 1.00 1.072 2.07 1.143 3.21 1.234 4.44 1.315 5.75 1.406 7.15 1.507 8.65 1.618 10.26 1.729 11.98 1.8410 13.82 1.9711 15.78 2.1012 17.89 2.2513 20.14 2.4114 22.55 2.5815 25.13 2.7616 27.89 2.9517 30.84 3.1618 34.00 3.3819 37.38 3.6220 41.00 3.8721 44.87 4.1422 49.01 4.4323 53.44 4.7424 58.18 5.0725 63.25 5.4326 68.68 5.8127 74.48 6.2128 80.70 6.6529 87.35 7.1130 94.46 7.6131 102.07 8.1532 110.22 8.7233 118.93 9.3334 128.26 9.9835 138.24 10.6836 148.91 11.4237 160.34 12.2238 172.56 13.0839 185.64 13.9940 199.64 14.9741 214.61 16.0242 230.63 17.1443 247.78 18.3444 266.12 19.6345 285.75 21.0046 306.75 22.4747 329.22 24.0548 353.27 25.7349 379.00 27.5350 406.53 29.4651 435.99 31.5252 467.50 33.7353 501.23 36.0954 537.32 38.6155 575.93 41.3256 617.24 44.2157 661.45 47.3058 708.75 50.6159 759.36 54.1660 813.52 57.9561 871.47 62.0062 933.47 66.3463 999.81 70.9964 1070.80 75.9665 1146.76 81.2766 1228.03 86.9667 1314.99 93.0568 1408.04 99.5669 1507.60 106.5370 1614.13 113.9971 1728.12 121.9772 1850.09 130.5173 1980.60 139.6474 2120.24 149.4275 2269.66 159.8876 2429.53 171.0777 2600.60 183.0478 2783.64 195.8579 2979.50 209.5680 3189.06 224.2381 3413.30 239.9382 3653.23 256.7383 3909.95 274.7084 4184.65 293.9385 4478.58 314.5086 4793.08 336.5287 5129.59 360.0788 5489.66 385.2889 5874.94 412.2590 6287.19 441.1091 6728.29 471.9892 7200.27 505.0293 7705.29 540.3794 8245.66 578.2095 8823.85 618.6796 9442.52 661.9897 10104.50 708.3198 10812.81 757.9099 11570.71 810.95100 12381.66 867.72101 13249.38 928.46102 14177.83 993.45103 15171.28 1062.99104 16234.27 1137.40105 17371.67 1217.02106 18588.69 1302.21107 19890.90 1393.36108 21284.26 1490.90109 22775.16 1595.26110 24370.42 1706.93111 26077.35 1826.41112 27903.76 1954.26113 29858.03 2091.06114 31949.09 2237.44115 34186.52 2394.06116 36580.58 2561.64117 39142.22 2740.96118 41883.18 2932.82119 44816.00 3138.12120 47954.12 3357.79121 51311.91 3592.83122 54904.74 3844.33123 58749.07 4113.44124 62862.51 4401.38125 67263.88 4709.47126 71973.36 5039.13127 77012.49 5391.87128 82404.37 5769.31129 88173.67 6173.16130 94346.83 6605.28131 100952.11 7067.65132 108019.75 7562.38133 115582.14 8091.75134 123673.89 8658.17135 132332.06 9264.24136 141596.30 9912.74137 151509.04 10606.63138 162115.68 11349.10139 173464.77 12143.53140 185608.31 12993.58141 198601.89 13903.13142 212505.02 14876.35143 227381.37 15917.70144 243299.07 17031.93145 260331.00 18224.17146 278555.17 19499.86147 298055.04 20864.85148 318919.89 22325.39149 341245.28 23888.17150 365133.45 25560.34151 390693.79 27349.57152 418043.36 29264.04153 447307.39 31312.52154 478619.91 33504.39155 512124.30 35849.70156 547974.01 38359.18157 586333.19 41044.32158 627377.51 43917.43159 671294.93 46991.65160 718286.58 50281.06161 768567.64 53800.73162 822368.37 57566.79163 879935.16 61596.46164 941531.62 65908.21165 1007439.84 70521.79166 1077961.62 75458.31167 1153419.94 80740.40168 1234160.33 86392.22 We are all made of star dust. However we are also made of nuclear waste. Remember to think before you post. Iqy Posts: 6 Joined: 25 Nov 2010, 03:04 First Video: Le Cafe ### Re: Desert Bus donation formula analysis What would that mean if someone donated a billion dollars? I make it about 230 years. Last edited by Iqy on 28 Feb 2013, 11:33, edited 1 time in total. Jamfalcon Posts: 3325 Joined: 08 Aug 2011, 13:59 First Video: Killer Instinct Location: Somewhat nearish Vancouver. Kind of. Contact: ### Re: Desert Bus donation formula analysis It would mean a lot of entertainment. "Jamfalcon's a super weird name" - Graham I wrote a book! MarkVI Posts: 181 Joined: 14 Jul 2008, 19:18 First Video: iPhone Alternative Location: New Jersey ### Re: Desert Bus donation formula analysis Not too bad actually. It would be just 267 hours. And to make the team bus even for 1 year (8760 hours) would cost in the range of 3.6*10^(258) dollars (many orders of magnitude more than the number of elementary particles in the universe) JackSlack Posts: 4572 Joined: 15 Oct 2010, 19:46 First Video: ENN, but I forget which. Location: Sydney, Australia ### Re: Desert Bus donation formula analysis Now a better question would be: What's the best formula for predicting a new year's DBfH donation total? I remember Fugiman and I had a go at it last year, only to way, way undershoot. Master Gunner Defending us from The Dutch! Posts: 19383 Joined: 29 Oct 2006, 12:19 First Video: How To Talk Like A Pirate Location: In Limbo. Contact: ### Re: Desert Bus donation formula analysis I have in the past tried it with software that uses genetic algorithms to try and derive the best equation to describe a data set. The problem is that there anywhere near enough data points to get a reliable result. It was a few years ago the last time I tried it, and the range between the best-fitting equations was around$100,000.
Lord Hosk
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Location: Half and inch below the knuckle of the ring finger. MI

### Re: Desert Bus donation formula analysis

He predicted based on the power law that they would raise $536 917. Based on a linear progression taking DB5 as a aberration in the math I predicted$422,550. I was off by less than one desert buck.

Desert Bus 6 total $443,630 I will predict right now that DB7 will raise$512,560 and run for 155 hours They will not have purchased the 155th hour yet but will run over because "they are so close".
Beware Bering Crystal Bears, Bearing Crystals. (Especially if the crystals they are bearing are, themselves, Bering Crystal Bears.) -Old, Stupid Proverb

[–]Graham_LRR
You hear that Khoo? We're almost better than the comic!
Lord Chrusher
Can't Drink Possible Beers
Posts: 8913
Joined: 29 Apr 2005, 22:53
First Video: Door to Door
Location: In England.

### Re: Desert Bus donation formula analysis

I ran my analysis again with the DBFH total and I got this:

The linear model predicts that they will raise $522 450 (155 hours). The power law model predicts that they will raise$599 864 (157 hours).

We are all made of star dust. However we are also made of nuclear waste.
Remember to think before you post.
the amativeness
Posts: 3735
Joined: 31 Jul 2008, 19:53
Location: America, where everyone sues everyone, always, for everything.

### Re: Desert Bus donation formula analysis

Now, random thought:

If you subtract 100,000 from both DB5 and DB6, you end up with what looks like a perfectly straight line. Based on this line, what is the DB7 prediction (adding back in that 100K that we removed)?

y=66.0008(7)+-52.8888 (+100)

Based on these numbers, I predict \$509,116.60. This means that they will bus for 154 hours, and possibly push to 155.
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Lord Chrusher
Can't Drink Possible Beers
Posts: 8913
Joined: 29 Apr 2005, 22:53
First Video: Door to Door
Location: In England.

### Re: Desert Bus donation formula analysis

<GIF of Kate shaking her head>Nope</GIF>

When you are making projections you can not just modify data to fit a trend.

We are all made of star dust. However we are also made of nuclear waste.
Remember to think before you post.
Keab42
Posts: 6662
Joined: 27 May 2009, 16:38
Location: In England

### Re: Desert Bus donation formula analysis

Unless your study is designed to show that homoeopathy actually works.
Keab42: Keeb-forty-two Deal with it. @Keab42 | | +Keab42 | | Tumblr
Dave-O_Boy
Posts: 2679
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Location: Seattle

### Re: Desert Bus donation formula analysis

MarkVI wrote:Not too bad actually. It would be just 267 hours.

And to make the team bus even for 1 year (8760 hours) would cost in the range of 3.6*10^(258) dollars (many orders of magnitude more than the number of elementary particles in the universe)

So you're saying it's probably an unrealistic goal for next year then?
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Oh and you can just call me Dave.
No need for the whole screen name.
Yaxley
Posts: 2389
Joined: 02 Nov 2008, 14:38
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Location: Pennsylvania

### Re: Desert Bus donation formula analysis

Lord Chrusher wrote:<GIF of Kate shaking her head>Nope</GIF>

I didn't have that saved, and for some reason I tried to google it. I got as far as "kate nop" before Google suggested "kate stark nope gif". I'm a little scared that it knew that.


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# What is the use of Grothendieck universes in category theory?

First of all, I have to mention that I'm truly sorry if this question would seem inappropriate for this site for some people. Still, I think it is better to ask here rather on math.stackexchange.

I recently grasped basic notions of Grothendieck universes for category theory. We most use the axioms of Zermelo and Fraenkel, Axiom of Choice and an additional axiom saying that any set is contained within some universe. What is more, I understood that the purpose of universes, intuitively, is that we can "do set theory" within the universe. That is, we can construct new sets out of old ones using the axioms of Zermelo and Fraenkel, and still stay within universe. But my understanding may be false, please, correct me, if I'm wrong. Many of the sources I consulted are either very technical or very brief, so my understanding of the matter is somewhat vague.

Still, as of now, I don't think I understand what Grothendieck universes and the axiom (the one about every set being in a universe) brings to the table of category theory. Classically, people wanted to talk, for example, about all categories of sets with (or without) structure (e.g. $\mathrm{Grp}, \ \mathrm{Ring}, \ \mathrm{R-Mod}$ or even $\mathrm{Set}$). So, if we use classes, the collections of objects in the aformentioned categories would be proper classes. Now, if we use universes, we fix a Grothendieck universe $\mathbb{U}$ and assume that collections of objects and morphisms in any category are actually sets. Then we call the members of $\mathbb{U} \ \ \mathbb{U}$-sets, the subsets of $\mathbb{U} \ \ \mathbb{U}$-classes, and we call a $\mathbb{U}$-class proper if it is not a member of of $\mathbb{U}$.

After that, we call a category $\mathrm{C}$ a $\mathbb{U}$-small if the set of objects and the set of morphisms of $\mathrm{C}$ are $\mathbb{U}$-sets. And we call $\mathrm{C}$ a locally small if the set of objects and the set of morphisms of $\mathrm{C}$ are $\mathbb{U}$-classes, and $\forall X,Y \in \mathrm{C}: \ \mathrm{Hom_C} (X,Y)$ is a $\mathbb{U}$-set.

But I have to wonder what is the motivation? I realise that in the end it is supposed to ease the foundations of category theory on set theory by erasing the need on distinguishing between sets, classes, conglomerates and whatever the "higher" notions are called (sorry, I haven't heard of them). But how is this supposed to make up for the fact that we don't study "the whole" category (such as the category of all sets, for example) anymore? And what is the motivation behind the notion of Grothendieck universe in respect of category theory? So, we can "do set theory" in a universe, and what it does for us as far as category theory goes?

To sum up: my question is what exactly the use of Grothendieck universes gives us so we don't need to study categories with proper classes of objects and morphisms, or proper conglomerates of objects and morphisms, or other collections of objects and morphisms that are not sets? If we assume that any category has only a set of objects and morphisms, but this time use the help of Grothendieck universes, why this works? Why can we forsake collections of elements that are too big to be a set in favor of sets and only sets if we fix a universe $\mathbb{U}$ and study whether sets of objects and morphisms of a category belong to $\mathbb{U}$? What is the "magic" of a Grothendieck universe?

P.S. I consulted several sources before asking this question, they were helpful to me in other aspects, but in the end they didn't resolve my confusing on this specific matter. In particular, I consulted Borceux'. Mac Lane's and Kashiwara's (coathored with P.Schapira) books on category theory, the articles by D.Murfet and M.Shulman. I know there is supposed to be an extensive exposition in SGA 4, but, unfortunately, I can't read French.

• An example where it is useful to either have universes or use some appropriate set theory is to prove a functor is fully faithful and essentially surjective if and only if it is an equivalence. One uses choice on the object set to construct the quasi-inverse and so it is convenient to have them form a set. – Benjamin Steinberg Nov 23 '16 at 16:58
• I don't quite understand what you're asking, but I think you might find some motivation and enlightment in Mike Shulman's post on universe polymorphism and typical ambiguity. – Andrej Bauer Nov 23 '16 at 17:07
• You should look at SGA 4 (especially Bourbaki's Appendix) even if you can't read French. – Fred Rohrer Nov 23 '16 at 20:48
• See my post mathoverflow.net/a/28913/1946 on how uses of Grothendieck universes can often be softened without loss to various weaker universe concepts. – Joel David Hamkins Nov 24 '16 at 0:23
• Dear Jxt921, I just remembered some old notes of mine that are more or less a translation of the basic universe stuff in SGA. – Fred Rohrer Nov 24 '16 at 15:44

Universes provide a convenient framework for stacking many levels of largeness. But a class in NBG cannot contain any other class. The most basic application which shows that universes are more convenient is that the category of functors $[\mathcal{C},\mathcal{D}]$ between two $U$-categories $\mathcal{C},\mathcal{D}$ is again a $V$-category if $U \in V$ are two nested universes. You cannot even write down this construction in NBG (unless you use inaccessible cardinals, which then is more or less equivalent to the universe approach).
• I'm beginning to develop a vague understanding and intuition. Still, why can we forsake such classical categories as $\mathrm{Set}, \ \mathrm{Grp}, \ \mathrm{Ring}, \ \mathrm{Top}$ in favor of the category of all $\mathbb{U}$-sets, the category of all $\mathbb{U}$-groups, the category of all $\mathbb{U}$-rings and the category of all $\mathbb{U}$-topological spaces? Don't we need the "entire" category which would have proper classes of objects and morphisms in NBG? – Jxt921 Nov 24 '16 at 8:00

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# Problems with in_header, \usepackage{cancel} for Rmarkdown. Cancel package not working/recognized

I have followed the instructions here among other places.

In other words, in_header: followed by -\usepackage{cancel} includes the latter in the preamble, thus presumably loads the package when creating the tex document.

When reviewing the tex doc, I see that it is in fact added. However, I ran into 2 problems:

1. Originally, for whatever reason, it doesn't find my cancel.sty file even though it's installed. Consequently, I had to add a .sty from CTAN to the same directory as the rmd file so it could knit. I also updated my tex to 2021.

2. Even after doing this, when I try to use commands from the package in my rmd file, they don't work nor display under the chunk like the other latex code (i.e. if I do $$\cancel{2}$$ it doesn't work).

Any thoughts on why this is happening?

I haven't experimented with other packages to see if this problem is package dependent nor am I familiar with the behavior of Rmarkdown within RStudio. In other words, what happens when packages are added to the preamble using the process above vs already in the template I'm using.

For example, I obviously need certain math packages (amsmath) to write out my equations and those packages are already in the default template (\usepackage{amsmath}). I'm using an APA 6th edition template.

Therefore, when I'm writing equations in my Rmarkdown file within RStudio , they pop up in the display which signifies to me packages are loaded and all things are well.

Hi,

If you can share a reproducible example, that would help us look into it.

It will help others help you if you provide some content that can be used directly; otherwise, one as to create an example themselves on what you are describing; It is time consuming and prone to error as we are not sure about what you have tried.

Thanks!

1 Like

I did some more troubleshooting and did the following:

1. I'm on MacOS, so while I updated my TeX Live to 2021 using MacTeX, I updated tlmgr just in case.

2. MacTeX adds /Library/TeX/texbin to PATH, but per the recommendation here, I added the full path to my 2021 Tex LIVE to PATH.

After doing these two things, my rmd document knits without needing a local cancel.sty package and the output works. As before, all I did was add the following to my rmd document:

header-includes   :
- \usepackage{cancel}

I'm not sure exactly what provided the solution, but nevertheless it is fixed.

Final Question

Despite the solution, I was wondering how RStudio handles preview of latex when wrapped by $$...$$. I'm curious because for the cancel package it doesn't preview its functionality, so one can only see the proper display once the document is knitted. I have attached two comparison images to make my point.

Is it that RStudio preview only relies on a certain collection of "base" packages so any other packages won't update the display?

My RStudio is Version 1.2.5033.

About the preview, this is HTML math I believe using MathJax. Not all packages and command are known by default. So this \cancel is not recognized by mathjax.

Mathjax support some packages including this ones cancel — MathJax 3.2 documentation

However for the preview it is not easy to setup to make it work.

Easy way is to use

$$\require{cancel} \cancel{\binom{n}{s}}$$

and this should preview ok but no compile in PDF.

To change the mathjax configuration of the preview is not something I know how to do. This would probably be a feature request for the IDE.

1 Like

This topic was automatically closed 7 days after the last reply. New replies are no longer allowed.

If you have a query related to it or one of the replies, start a new topic and refer back with a link.


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# The value of 0/0

## Recommended Posts

OK I admit I thought I was losing my sanity when I thought of this, but then I realized people thought Galileo was crazy when he said that the Earth went around the Sun. I am a (fairly)good mathematician and I know better than to question integer division by zero is undefined, but what about 0/0. Basically, for the past one week I was thinking about 0/0. Not integer division by zero but only about zero divided by zero. To prevent "You must be out of your mind" replies, I would like to state the.... uh.... rules first.

Here's my theory. f(x) = x/x is a constant function giving the value 1. Therefore, 0/0 is also 1. IF you're going "OMG, Gasp.... what a sacrilege!!" wait... I have googled this item and have seen the replies. None of them are... uh... convincing...(I don't know the exact word for that feeling). So here's the deal

1. This is only a discussion/debate. Not a correctly-verified-and-proven theory.

2. First, I get to twist the facts to suit my theory. You mathematicians are going to state a theory without twisting the facts to REALLY disprove my theory.

3. If you can't do that you're going to get the chance to twist the facts to disprove my theory. (I have no doubt that's going to be as easy as "pi")

OK, before you start chewing my tail I'll admit a few things first. Granted, 0/0 IS undefined when taken in elementary maths. (Come on how can you divide "nothing" sweets among "nobody" persons, to quote the classic textbook example). But as a function. Like f(x) = x/x. Simplifying we get f(x) = 1; Which is a constant function; Meaning it IS 1 for all x belongs to R. Its like the other classic example for limits; f(x) = x2 - 1/ x-1 is not defined when x = 1. But we KNOW it IS 2 when x = 1; Simplifying f(x) = x2 - 1/ x - 1 gives f(x) = x + 1. I'm stating my case on this principle.

Ok, let the waves of opposition begin!

EDIT: Oops, I ought to have read the forum sticky before I posted this. Moderators can move this to the "Speculations" section if they want to.

Edited by Raptor115

• Replies 56
• Created

#### Popular Days

In computer floating point operations, dividing by zero is defined to result in either positive or negative infinity.

If you look at the algebraic equivalent of $\frac{0}{0} = x$ it's $x\times0 = 0$. In this case, x can literally be any number since anything times 0 is 0. A graph of the result set would be meaningless, since it would, quite literally include every number in existence.

More appropriately, we could write the formula for such a graph as $f(x) = 0 \times x$ which would result in a straight line from negative to positive infinity right across the y-axis of the graph. This is the reason the operation is undefined - it doesn't result in any useful answers.

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This is something I posted in an earlier thread:=

"f you calculate (8-x^3)/(2-x) when x=2 you get 0/0........................... I'll just say I make this example of 0/0 equal to 12........................"

Are you suggesting this particular case has two answers?

Edited by Joatmon
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@Joatmon

LOL you got the point wrong. Actually by the f(x) = x2-1/ x-1 example, I wasn't trying to equate 0/0 with something. I thought I said it was meaningless to equate 0/0 with anything. OK if i didn't I'll say it at the end of this reply. Actually I meant that, like we could simplify f(x) to give x+1, we could also simplify f(x) = x/x to f(x) = 1, which makes even f(0) to be 1.

@Greg

you said X x 0 =0. Basically, we know that if ax + c = bx + c, then a = b. then putting x= 0, a =1 , b =2, c= 3 we get 1x0 + 3= 2x0 + 3 => 3 =3 implying 1= 2. @Joatmon, like this we could equate 0 to all the numbers known. (Agreed, I AM twisting the facts now) All I'm saying is that, saying 0/0 = 1 doesn't change things that radically, cause you see we can prove 1 = 2 without proving 0/0 = 1. Besides I wasn't doing a graph of "Find x such that x x 0 = 0." I said it was f(x) = x/x.

OK, now for the divide-multiply headache.

0/0 = 1; 1 x 0 = 0; Finished. No violation of the law.

If you think I missed any of your points please point it out.

Also I do know 0/0 is meaningless in elementary arithmetic. As I said I probably was out of my mind last week and I couldnt rest till I started a debate.

BTW I didn't think there would be 2 responses in 15 mins (Responsive website this. Some websites you don't get replies for weeks.....)

Also, keep the arguments coming. I'm not going to stop till I find one that'll overload my brain. (Meaning it can convince my crazy half that this debate's a waste)

Edited by Raptor115
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@Greg

you said X x 0 =0. Basically, we know that if ax + c = bx + c, then a = b. then putting x= 0, a =1 , b =2, c= 3 we get 1x0 + 3= 2x0 + 3 => 3 =3 implying 1= 2. @Joatmon, like this we could equate 0 to all the numbers known. (Agreed, I AM twisting the facts now) All I'm saying is that, saying 0/0 = 1 doesn't change things that radically, cause you see we can prove 1 = 2 without proving 0/0 = 1. Besides I wasn't doing a graph of "Find x such that x x 0 = 0." I said it was f(x) = x/x.

The fact that it implies that 1 = 2 seems to only reinforce my point that it results in no useful answers. It's like saying that 1 + 2 = blue. Sure it's an answer. It may even be correct if blue = 3, but the answer, as given, doesn't really help us in any way.

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@Greg

I just re-read your first post and I realized I missed a big chunk of what you were trying to say. Also, I thought that you hadn't (I don't want to sound rude) read my post fully at first. I just told at the starting that 0/0 was and is meaningless (I don't know about "will be"). I was saying that f(x) = x/x could be simplified to f(x) = 1 and therefore f(0) is 1.

You just took it as f(0) = 0/0 and said it would make a point of (0, y-axis(more specifically -infinity to +infinity)). OK that's like saying f(x) = x2 - 1/ x- 1 gives (1, y-axis) when x = 1, when it gives (1,2) when simplified. I'm applying the same simplifying concept to f(x) = x/x to say f(0) = 1.

Umm... about your saying "it just doesn't help us in any way"..... I just realised its true. But it still doesn't convince my crazy half that this debate's a waste

Edited by Raptor115
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Additionally, let's consider fractions as proportions (since, that's really what they are).

If I have ten widgets and I give two away, it's easy to see that

a) I have 8 left, and

b) I have given away 20% of my initial widgets.

But if I start with 0 widgets, and I give 0 away, I still have what I started with, which is none. Can I even break up 0 into fractional parts? That's what you're really trying to do when you divide - break something up into smaller chunks.

Can I really break a pie up into 0 pieces? (Well I suppose if I eat the whole thing...) but it doesn't leave me with pieces of pie, it leaves me with no pie at all. Logically it makes no sense.

And how can I break up no pie into smaller pieces of no pie? (i.e. 0/0) If I have nothing to begin with, how do it make it smaller nothings? An answer of 1 implies that there is a whole "something" there. So what is the whole "something" that results from nothing divided by nothing?

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Can I really break a pie up into 0 pieces? (Well I suppose if I eat the whole thing...) but it doesn't leave me with pieces of pie, it leaves me with no pie at all.

LOL that's a nice one. OK consider --- The pie's in your stomach. How many pies have you in your hand --- 0. The stomach divides the pie into infinite (well, to be precise its as many atoms as were in the pie originally added with the atoms in the saliva and digestive juices that went with it) pieces using HCl. How many pies does it leave in your hand --- 0. Division happens...... OK that was a joke.

Basically that's exactly what I told in my first post. I mean, please read my post carefully once and you'll find that I SAID that 0/0 DOESN'T make sense in elementary arithmetic. I only said that in relation to the function f(x) = x/x. Its like a theoretical question. Like we know i2 = -1. We all know there ain't a real number like that. Yet we still don't disown it. Maybe my question comes along these lines.

i'm sorry for writing this para and I'll edit n' remove this tomorrow. But for Greg, sorry but i gtg sleep. Its 11.45 pm and I have to go school tomorrow and work with my friend on the trinomial theorem (a+b+c)n . We don't know if it's already been done, but we certainly want to do it ourselves without external help. We got our models right today and we need to come up with a gen. formula by this week. Keep the arguments coming --- and please include a bit of humour. That way, the debate won't be dry. Good luck at some other arguments Greg.

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0/0=

= π

∴ 0/0 = π

π = The Universe

0/0 = The Universe

lol.

Edit: Pi is exactly 3.

Re-Edit: But no, seriously. If you take 0 to be 'something', it is said to be able to fit into 'something else' indefinitely or infinitely; e.g. 8/0 = .

That's according to a section of an OCR mathematics book I read anyway.

It depends if you read it as dividing 8 into 0 sections... which would lead you to think it's disappeared. Or if you look at it as how many times 0 fits inside 8 (as many times as it pleases).

Re-Re-Edit: (Cannot sleep tonight.) If 1x0= 0 and 2x0=0, then 1x0=2x0. Therefore, if (1x0)/0= (2x0)/0, the zeros cancel to conclude 1 can equal 2. Maths is broken.

Edited by Iota
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Here's my theory.

The word theory means something special in the sciences, a well-substantiated, coherent explanation of some aspect of reality. In mathematics, the word means a consistent body of knowledge, such as knot theory, Galois theory, set theory, and so on. This is a scientific forum. The one thing "theory" does not mean is "a wild guess I came up with after finishing off a quart of tequila."

f(x) = x/x is a constant function giving the value 1. Therefore, 0/0 is also 1. IF you're going "OMG, Gasp.... what a sacrilege!!"

Yep. Sacrilege. And nonsense. First off, what made you pluck that function out of the clear blue sky? What about f(x)=2x/x? or f(x)=-3x/x?

Secondly, that's not how division is defined. a/b=c means that a-b*c=0. Now let's look at a=b=0. We're trying to find a number c such that 0-0*c=0. Any and every number satisfies this expression. 0/0 is indeterminate. It has to be. Arbitrarily assign it a value, any value, and you have just opened the mathematical door of death. It allows you to prove that 1=2. The first thing that a mathematical theory (remember the definition) must be is consistent, or contradiction-free. An axiomatic system that is known to allow consistencies isn't a theory. It's trash.

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0/0=

Not quite right on that point. I don't see 0/0 being anywhere. Its basically indeterminate

= π

∴ 0/0 = π

π = The Universe

0/0 = The Universe

lol.

LOL that's a good definition actually. Just kidding..... Don't take this seriously.

Edit: Pi is exactly 3.

Is this a joke?? OK lemme see Pi is exactly what you get when you cut the magic number 'e' off a "pie".

Re-Edit: But no, seriously. If you take 0 to be 'something', it is said to be able to fit into 'something else' indefinitely or infinitely; e.g. 8/0 = .

That's according to a section of an OCR mathematics book I read anyway.

It depends if you read it as dividing 8 into 0 sections... which would lead you to think it's disappeared. Or if you look at it as how many times 0 fits inside 8 (as many times as it pleases).

Bah, doesn't anyone ever read my first post fully? I've repeated it like 5 times ---- 0/0 doesn't make sense in elementary arithmetic. So don't use real life examples. Maybe if you can demonstrate the value of i in i2 = -1 with real-life examples, then I'll accept that 0/0 is NEVER 1 in any circumstance

Re-Re-Edit: (Cannot sleep tonight.) If 1x0= 0 and 2x0=0, then 1x0=2x0. Therefore, if (1x0)/0= (2x0)/0, the zeros cancel to conclude 1 can equal 2. Maths is broken.

I demonstrated that Maths is broken without even saying 0/0 = 1 in a previous reply. So I disagree. On the other hand you've taken (1x0)/0 = (2x0)/0. Well let's put it in a different way. We are going to take 0/0 = 1. Then, 1x0 = 0; 2x0 = 0. The whole thing becomes 0/0 = 0/0; 1=1. Maths is not broken by saying 0/0 =1. Its already broken if you consider my ax + c = bx +c example.

The word theory means something special in the sciences, a well-substantiated, coherent explanation of some aspect of reality. In mathematics, the word means a consistent body of knowledge, such as knot theory, Galois theory, set theory, and so on. This is a scientific forum. The one thing "theory" does not mean is "a wild guess I came up with after finishing off a quart of tequila."

OK. I'll change that to "hypothesis". Besides I didn't drink tequila, I drank too much coffee last week.... couldn't sleep much...... Just kidding.

Yep. Sacrilege. And nonsense. First off, what made you pluck that function out of the clear blue sky? What about f(x)=2x/x? or f(x)=-3x/x?

Secondly, that's not how division is defined. a/b=c means that a-b*c=0. Now let's look at a=b=0. We're trying to find a number c such that 0-0*c=0. Any and every number satisfies this expression. 0/0 is indeterminate. It has to be. Arbitrarily assign it a value, any value, and you have just opened the mathematical door of death. It allows you to prove that 1=2. The first thing that a mathematical theory (remember the definition) must be is consistent, or contradiction-free. An axiomatic system that is known to allow consistencies isn't a theory. It's trash.

First off, I didn't pick that function out of the clear blue sky, I picked it from my sky-blue Mathematics Textbook Volume - 2 (The first volume is orange-orange). Well there ain't a difficulty about f(x) = 2x/x which is f(x) = 2 and f(x) = -3x/x which is f(x) = -3. If 0/0 = 1, it would satisfy all the conditions whereas if 0/0 = something else it would not.

How? you ask? OK lets see; f(x) = 2x/x. Then, f(0) = 2 x 0/0 = 2 x1 = 2, which gives us the right answer. Same method to prove f(0) in -3x/x would give -3. If you'd started this hypothesis with 0/0 = 2 ( or 3 or 4 or -10) then you wouldn't be getting 2 x 0/0 = 2 x 1 = 2. you would get 2 x0/0 = 2x2 = 4 ( or 6 or 8 or -20).

Next, if 0/0 = 1, then 0 - (0x1) must be 0, right? Oh well, lets see; 0x1 = 0; Therefore 0 - 0; That's 0, right? So what's wrong? Well, you're wrong about assigning 0/0 opening the mathematical door of death. I already pointed out we could prove 1=2 without needing 0/0 to be 1. If you really believe assigning 0/0 opens the mathematical door of death, please explain more clearly.

OK, till now, except for D H, everybody posted why 0/0 isn't 1. I'm not asking why it isn't. I'm only asking why it shouldn't be 1.

If you can't understand the difference between those statements, the first one's like saying "why aren't we cloning humans?" The second one's like "what's gonna be the problem if we clone humans?"

Also, if you find my English difficult to understand, I'm sorry, I'm an Indian.

Edited by Raptor115
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Is this a joke?? OK lemme see Pi is exactly what you get when you cut the magic number 'e' off a "pie".

It is quite clearly a joke, yes.

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Bah, doesn't anyone ever read my first post fully? I've repeated it like 5 times ---- 0/0 doesn't make sense in elementary arithmetic.

It doesn't make sense, period.

I demonstrated that Maths is broken without even saying 0/0 = 1 in a previous reply.

You demonstrated nothing. Instead you made a bare assertion, a fallacy. Demonstrate this please, and do so without using an illegal operation such as dividing by zero or assuming that the only solution to x2=1 is one.

OK, till now, except for D H, everybody posted why 0/0 isn't 1. I'm not asking why it isn't. I'm only asking why it shouldn't be 1.

It's quite simple. Saying that 0/0 is any particular value opens the door to contradictions. Mathematics must be contradiction free.

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OK, till now, except for D H, everybody posted why 0/0 isn't 1. I'm not asking why it isn't. I'm only asking why it shouldn't be 1.

Also, if you find my English difficult to understand, I'm sorry, I'm an Indian.

First, your English is quite good. Better than some people for whom it's the native tongue, so don't worry about that.

Second, the reason it shouldn't be 1 are the same reasons it's not 1 - it leads to contradictions and problems in the most basic of maths, as DH pointed out. Whether you want to call it indeterminate or undefined, division by zero (which is really what you're talking about) renders a lot (if not all) of mathematics essentially unstable and meaningless.

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It doesn't make sense, period.

OK, maybe I'm ignorant, or maybe its just my mind. 0/0 doesn't make sense naturally. I concede. Then tell me how i = square root of -1 makes sense naturally.

You demonstrated nothing. Instead you made a bare assertion, a fallacy. Demonstrate this please, and do so without using an illegal operation such as dividing by zero or assuming that the only solution to x2=1 is one.

Probably you didn't read all my replies. I did demonstrate that Maths is broken if you consider equating with "multiply by zero" operations. Consider two equations ax + c and bx +c. We know that if ax + c = bx + c, then a = b. Lets try putting a few values. x = 3; c = 1; so we get 3a + 1 and 3b + 1. If we get 3a + 1 = 3b + 1, then a = b right? OK lets put different values. x = 0(GASP!!); a = 1; b =2; c =3. We get (1x0) + 3 and (2x0) + 3 => 0 + 3 and 0 + 3 => 3 and 3. Since 3 =3 we say a=b and we get 1 = 2. See, I neither divided by zero nor did I assume the solution of x2 = 1 is one (What does that have to do with 0/0??)

It's quite simple. Saying that 0/0 is any particular value opens the door to contradictions. Mathematics must be contradiction free.

I think I proved Maths has contradictions without assigning a particular value to 0/0.

First, your English is quite good. Better than some people for whom it's the native tongue, so don't worry about that.

Thanks. I just happen to be good at written English. When it comes to spoken English................

Second, the reason it shouldn't be 1 are the same reasons it's not 1 - it leads to contradictions and problems in the most basic of maths, as DH pointed out. Whether you want to call it indeterminate or undefined, division by zero (which is really what you're talking about) renders a lot (if not all) of mathematics essentially unstable and meaningless.

Again, I think I just proved that Maths is already unstable with "multiplication with zero" as it is with "division by zero". Maybe the solution is to say multiplication and division operations with 0 are indeterminate??

Edited by Raptor115
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Probably you didn't read all my replies. I did demonstrate that Maths is broken if you consider equating with "multiply by zero" operations. Consider two equations ax + c and bx +c. We know that if ax + c = bx + c, then a = b. Lets try putting a few values. x = 3; c = 1; so we get 3a + 1 and 3b + 1. If we get 3a + 1 = 3b + 1, then a = b right? OK lets put different values. x = 0(GASP!!); a = 1; b =2; c =3. We get (1x0) + 3 and (2x0) + 3 => 0 + 3 and 0 + 3 => 3 and 3. Since 3 =3 we say a=b and we get 1 = 2. See, I neither divided by zero ...

You did divide by zero. You did so by assuming ax=bx implies that a=b. This is not valid with x=0 because you are dividing by zero.

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You did divide by zero. You did so by assuming ax=bx implies that a=b. This is not valid with x=0 because you are dividing by zero.

OK I'll admit you stopped me in my tracks for a minute there. Then I got a curious kind of doubt. You know when you have two equations like ax2 + bx + c and fx2 + gx + h and when they're both equal, we have this "equating co-efficients of x2" and "equating co-efficients of x" stuff. How do you say they equate coefficients without dividing anything? Am I ignorant of any intermediate steps. I'm only 15 so I'm not that expert in Maths, but I can hold my own.

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Here is a different viewpoint to consider.

Let us say that I travel 0 miles in zero seconds.

What is my speed.

You assert it is 1 mph - which is true of course since at 1 mph the distance I travel in zero seconds = speed times time =1*0 = 0 miles.

However the same would also be true if I travelled at 2 mph or 20 mph or any other speed.

Therefore I can say that speed = distance /time for the case of 0/0 can be anything I want.

Which is another way of saying it is indeterminate.

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Here is a different viewpoint to consider.

Let us say that I travel 0 miles in zero seconds.

What is my speed.

You assert it is 1 mph - which is true of course since at 1 mph the distance I travel in zero seconds = speed times time =1*0 = 0 miles.

However the same would also be true if I travelled at 2 mph or 20 mph or any other speed.

Therefore I can say that speed = distance /time for the case of 0/0 can be anything I want.

Which is another way of saying it is indeterminate.

SIGH......... I tell it and tell it and tell it......... Please read the original post and all the replies. This is purely a hypothetical question which has no real answer. Its like asking to define i = square root of -1 with a real life example. I have seen this same viewpoint in many other websites.

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OK I'll admit you stopped me in my tracks for a minute there. Then I got a curious kind of doubt. You know when you have two equations like ax2 + bx + c and fx2 + gx + h and when they're both equal, we have this "equating co-efficients of x2" and "equating co-efficients of x" stuff. How do you say they equate coefficients without dividing anything?

Simply put, that is illegal at x=0.

Am I ignorant of any intermediate steps.

Let's look at ax=bx. Here's a couple of ways to look at this. One is to rewrite this as (a-b)*x = 0. This equation has two solutions, a=b or x=0. At x=0 you cannot make any inference about the relationship between a and b. Any values will satisfy (a-b)*0 = 0. Another way to look at is to multiply both sides of ax=bx by 1/x, yielding a=b. This multiplication by 1/x (division by x) is illegal at x=0. So once again the inference that ax=bx implies a=b is invalid at x=0.

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Simply put, that is illegal at x=0.

OK, but why? I actually meant how do you equate co-efficients of x2 of two equations when x is not zero. Am I ignorant of any intermediate steps in such a procedure. I know well enough the intermediate steps in equating ax = bx LOL.

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That's OK, I can take a hint, since you don't value anyone else's thoughts I will leave you to yours.

go well

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Ok, I'm only going to give this one shot.

Suppose $\frac{0}{0} = 1$. Because of the property of $0$ it follows that:

$0 = 0 \cdot 1 = 0 \cdot \frac{0}{0} = \frac{0 \cdot 0}{0} = \frac{0}{0} = 1$.

So by claiming that $\frac{0}{0} = 1$ you are actually claiming that $0 = 1$. This is a contradiction, because we know that they have different properties in the ring of the real numbers, and as such they are unique.

So if you want to keep the algebraic structure of the real numbers as a ring, then you can not claim that $\frac{0}{0} = 1$. And if you are thinking in the lines of "Fine, I can live with losing the ring structure of the real numbers", then you are not really defining $\frac{0}{0}$ as we know the number $0$, because, well, you just lost the meaning of the number $0$, so you have nothing to define.

Makes sense?

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Ok, let the waves of opposition begin!

...Just a story...

When I was your age or just a little older I developed a truly beautiful 58 step proof that anything divided by zero equals infinity. I showed it to all my teachers and anyone I could and they mostly just shrugged. While checking it for about the 100th time I found the 45th step was a very subtle assumption of the conclusion so the proof meant nothing.

People don't think about underpinnings and definitions of our ideas nearly enough so lose sight of what the results really mean and most of us get worse with age. Never lose your doubt. Try to keep as broad a perspective as possible.

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Ok, I'm only going to give this one shot.

Suppose $\frac{0}{0} = 1$. Because of the property of $0$ it follows that:

$0 = 0 \cdot 1 = 0 \cdot \frac{0}{0} = \frac{0 \cdot 0}{0} = \frac{0}{0} = 1$.

So by claiming that $\frac{0}{0} = 1$ you are actually claiming that $0 = 1$. This is a contradiction, because we know that they have different properties in the ring of the real numbers, and as such they are unique.

So if you want to keep the algebraic structure of the real numbers as a ring, then you can not claim that $\frac{0}{0} = 1$. And if you are thinking in the lines of "Fine, I can live with losing the ring structure of the real numbers", then you are not really defining $\frac{0}{0}$ as we know the number $0$, because, well, you just lost the meaning of the number $0$, so you have nothing to define.

Makes sense?

With all due respect to all others who replied to this thread, I have to say this is by far the best one. I was going along with the OP's replies for second, but this, this completely puts the issue to rest. Well played tmpst.

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text:
11-129 Jacques Fejoz, Marcel Guardia, Vadim Kaloshin, Pablo Roldan
Diffusion along mean motion resonance in the restricted planar three-body problem (2938K, Postscript) Sep 12, 11
Abstract , Paper (src), View paper (auto. generated ps), Index of related papers

Abstract. We study the dynamics of the restricted planar three-body problem near a mean motion resonance, i.e. a resonance involving the Keplerian periods of the two lighter bodies revolving around the most massive one. This problem is often used to model Sun--Jupiter--asteroid systems. For the primaries (Sun and Jupiter), we pick a realistic mass ratio $\mu=10^{-3}$ and a small eccentricity $e_0>0$. The main result is a construction of a variety of diffusing orbits which show a drastic change of the osculating eccentricity of the asteroid, while the osculating semi major axis is kept almost constant. The proof relies on the careful analysis of the circular problem, which has a hyperbolic structure, but for which diffusion is prevented by KAM tori. In the proof we verify certain non-degeneracy conditions numerically. Based on the work of Treschev, it is natural to conjecture that the time of diffusion for this problem is at least $\sim -\ln (\mu e_0)/ (\mu^{3/2} e_0)$. We expect our instability mechanism to apply to realistic values of $e_0$ and we give heuristic arguments in its favor. If so, the applicability of Nekhoroshev theory to the three-body problem as well as the long time stability become questionable. It is well known that, in the Asteroid Belt, located between the orbits of Mars and Jupiter, the distribution of asteroids has the so-called {\it Kirkwood gaps} exactly at mean motion resonances of low order. Our mechanism gives a possible explanation of their existence. To relate the existence of Kirkwood gaps with Arnold diffusion, we also state a conjecture on its existence for a typical $\eps$-perturbation of the product of the pendulum and the rotator. Namely, we predict that a positive conditional measure of initial conditions concentrated in the main resonance exhibits Arnold diffusion on time scales $- \ln \eps / \eps^{2}$.

Files: 11-129.src( 11-129.keywords , Elliptic-Diffusion2.ps )


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# The curve vec{r}(t) = {t, t, t^2} has one point with maximal curvature. Find this point and...

## Question:

The curve {eq}\vec{r}(t) = \left \langle {t, t, t^2} \right \rangle {/eq} has one point with maximal curvature. Find this point and its curvature.

## Maximal Curvature:

The function for curvature, {eq}\kappa (x) = \frac{|f''(x)|}{(1 + (f'(x))^2)^{\frac{3}{2}}} {/eq}. Once we have found the first and second derivatives of our original function, {eq}f(x) {/eq}, we substitute them into the curvature function. Then, we find the critical points of {eq}\kappa (x) {/eq} by finding the zeros of the first derivative. To classify the critical points, we use the first or second derivative test. Then, we find the corresponding curvature value of the maximum and minimum.

To find the curvature function, we must first find the first and second derivatives of our original function, {eq}f(t) = t^2 {/eq}.

\begin{align*} f(t) &= t^2 \\ f'(t) &= 2t \\ f''(t) &= 2 \end{align*}

Using the first and second derivative equations, we generate the curvature function.

\begin{align*} \kappa (t) &= \dfrac{|f''(t)|}{(1 + (f'(t))^2)^{\frac{3}{2}}} \\ \kappa (t) &= \dfrac{2}{(1 + 4t^2)^{\frac{3}{2}}} \end{align*}

Next, we find the critical points of the function by finding the zeros of the first derivative.

\begin{align*} \kappa (t) &= \dfrac{2}{(1 + 4t^2)^{\frac{3}{2}}} = 2(1 + 4t^2)^{-\frac{3}{2}} \\ \kappa'(t) &= 2(-\frac{3}{2})(1 + 4t^2)^{-\frac{5}{2}}(8t) \\ \kappa'(t) &= -24t(1 + 4t^2)^{-\frac{5}{2}} \\ 0 &= -24t(1 + 4t^2)^{-\frac{5}{2}} \\ 0 &= -24t \\ t &= 0 \end{align*}

The critical point of the function occurs at {eq}t = 0 {/eq}.

Since the domain of the function is {eq}(-\infty, \infty) {/eq} and we have already computed the first derivative, we use the first derivative test to classify the critical point as a maximum or minimum. First, we decompose the domain such that the critical point is excluded: {eq}(-\infty,0) \cup (0, \infty) {/eq}. Then, we select one value of {eq}x {/eq} from each interval and evaluate it using the first derivative. We select {eq}x = -1 {/eq} and {eq}x = 1 {/eq}. If the value of the first derivative is positive, the function is increasing over that entire interval. If the value of the first derivative is negative, the function is decreasing over that entire interval.

\begin{align*} \kappa'(-1) &= -24(-1)(1 + 4(-1)^2)^{-\frac{5}{2}} = 0.429 \\ \kappa'(1) &= -24(1)(1 + 4(1)^2)^{-\frac{5}{2}} = -0.429 \end{align*}

The function is increasing from {eq}(-\infty, 0) {/eq} and decreasing from {eq}(0, \infty) {/eq}. Therefore, the function has maximum curvature at {eq}x = 0 {/eq}.

To find the curvature at {eq}x = 0 {/eq}, we compute {eq}\kappa(0) {/eq}.

$$\kappa (0) = 2(1 + 4(0)^2)^{-\frac{3}{2}} = 2$$

Therefore, the maximum curvature of the function is {eq}(0,2) {/eq}.


text:
# Squaring modulo n — is there a systematic way to generate the fixpoints?

The natural function

$$f_k(n) = n^2 \bmod k,\quad f_k(n) \in [0;k-1]$$

has at least the fixpoints $n\in\{0;1\}$ for each $n$, but for some $k$ there are other fixpoints. For instance, $f_{100}$ has the fixpoints $\{0;1;25;76\}$. Apart from trying out every possible value for $n$, is there a systematic way to get all fixpoints for a given $k$?

-
Using the chinese remainder theorem is a good start. – Chris Eagle Oct 18 '12 at 21:09

Along with the Chinese Remainder theorem, you need to be able to factor your $k$ into prime powers. It is a theorem requiring induction that the only solutions to $$n^2 \equiv n \pmod {p^t}$$ for a prime $p$ and $t \geq 1$ are $$n \equiv 0,1 \pmod {p^t}.$$ As a result, if your $k$ is divisible by exactly $r$ distinct primes, there are exactly $2^r$ of your fixpoints. For $100,$ note that you got precisely the four numbers from $0$ to $99$ that are $0,1 \pmod 4$ and $0,1 \pmod {25}.$

Let's see, the theorem mentioned is straightforward for odd $p,$ induction on $t.$ Slightly different for $p=2$ but easy enough and still true.

Induction, odd $p$: with $0 \leq j \leq p-1,$ then $\delta = 0,1,$ and $t \geq 1,$ we are solving $$(j p^t + \delta)^2 \equiv j p^t + \delta \pmod {p^{t+1}},$$ $$j^2 p^{2t} + 2 j \delta p^t + \delta^2 \equiv j p^t + \delta \pmod {p^{t+1}}.$$ Note $\delta^2 = \delta,$ and $2t \geq t + 1.$ So $$2 j \delta p^t \equiv j p^t \pmod {p^{t+1}},$$ $$(2 \delta - 1) j p^t \equiv 0 \pmod {p^{t+1}},$$ $$(2 \delta - 1) j \equiv 0 \pmod {p},$$ $$j \equiv 0 \pmod {p},$$ $$j = 0.$$

Induction, $p=2$: with $j = 0,1,$ then $\delta = 0,1,$ and $t \geq 1,$ we are solving $$(j 2^t + \delta)^2 \equiv j 2^t + \delta \pmod {2^{t+1}},$$ $$j^2 2^{2t} + 2 j \delta 2^t + \delta^2 \equiv j 2^t + \delta \pmod {2^{t+1}}.$$ Note $\delta^2 = \delta,$ and $2t \geq t + 1.$ So $$2 j \delta 2^t \equiv j 2^t \pmod {2^{t+1}},$$ $$j \delta 2^{t+1} \equiv j 2^t \pmod {2^{t+1}},$$ $$0 \equiv j 2^t \pmod {2^{t+1}},$$ $$0 \equiv j \pmod {2},$$ $$j = 0.$$

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You would like $$n^2\equiv n\mod{k}$$ Which is the same as \begin{align} n^2-n&\equiv0\mod{k}\\ n(n-1)&\equiv0\mod{k} \end{align} So $$k| n(n-1)$$ This means each maximal prime power factor of $k$ need to divide either $n$ or $n-1$.

If I had to solve this for some explicit $k$, I'd start by taking the largest prime power of $k$, $p^m$, and enumerate every multiple of $p^m$ less than $k$. For example, with $k=100$, we would enumerate 25, 50, and 75.

Recognizing that $p^m$ either divides $n$ or $n-1$, either $n$ is one of these enumerated values, or $n-1$ is. In the example, we've already narrowed the list of possible nontrivial solutions from $\{2,3,\ldots 99\}$ to $\{25, 26, 50, 51, 75, 76\}$.

For this example, I'd just check by hand the six possible solutions. But in general for larger $k$, I'd cross this list against a similar list built using the next largest prime power dividing $k$, sieving the list of potential solutions down further. With this example, $4$ is the next prime power, so $n$ needs to be $0$ or $1$ mod $4$. This leaves only $\{25, 76\}$.

-

$n(n-1)$ must be a multiple of $k$, that is for each prime power $p^m$ dividing $k$, either $n$ or $n-1$ must be a multiple of $p^m$. However the choice may differ per prime. Hence there are $2^r$ solutions if $k$ is divisible by $r$ primes. With $k=100= 2^25^2$, the fixpoints are $0$ and $1\pmod 4$ and $0$ and $1\pmod {25}$. The combinations of these (via chinese remainder theorem) give the four solutions you mention.

-


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In February 2021 Microsoft announced this awesome new Azure service Azure Quantum. This service has two pillars:

• Quantum Computing: this allows you to run programs on actual quantum computers! These computers are provided by Microsoft partners like Honeywell and IonQ, you can write quantum programs in Q#, which you can learn here. It is really mind blowing that we can now use actual quantum computers.
• Quantum inspired optimization (QIO): this is a new way to solve complex optimization problems on normal CPU's or FPGA's. These solvers are based on quantum computers learnings. This is the perfect way to leverage learnings from quantum computers, without having acces to large scale quantum computers. Here is an earlier post with some info on this service.

In this post we'll be looking at the QIO service to solve a very old and complex optimization problem: the Traveling salesperson problem. Let's quickly walk trough the problem itself.

Assume that you are a door to door salesperson and have a number of places you want to visit. Ofcourse, you want to do this as quickly as possible. Therefore you need the shortest route between all possible locations. This problem gets exponentially hard. The number of routes increases very quickly when the number of possible locations increases.

To calculate the quickest route, currently we need to calculate all the possible routes (which gets very hard, very fast). Using a quantum computer we could do this a lot quicker by placing multiple possible routes in a super position. Currently there are no quantum computers large enough to fulfil this assignment. One of the solution lies in Quantum Inspired Optimization (QIO) (here is a quick into into QIO).

Here we will deep dive into how to solve this problem using Azure quantum. The first step is to write the complete problem as a Quadratic Unconstrained Binary Optimization (QUBO) problem. Before we deep dive into the QUBO and Azure Quantum, we first give a high overview of the complete solution and all the steps.

This is how the full solution works:

1. Get a list of all the addresses that our sales person wants to visit;
2. Use Bing API, to get a distance Matrix for all the addresses;
3. Formulate the Azure quantum problem from the cost matrix;
4. Submit the QUBO to Azure Quantum.

### Get a list of all the addresses that our sales person wants to visit

This is the most self explanatory, collect a list of all the places the salesperson wants to visit.

### Use Bing API, to get a distance Matrix for all the addresses

When we have all the locations we want to create something called a distance matrix. This a basically a table with all the distances between all of the places. For example:

New york Los Angeles Seattle Detroit Miami
New york 70 60 50 12
Los Angeles 70 60 50 12
Seattle 70 60 50 12
Detroit 70 40 60 01
Miami 70 40 60 70

This gives all the input data we need to optimize. Using this information we can calculate the distance between multiple points. For example if we want to go from Seattle --> Miami --> Detroit, we can look up all distances in the table and calculate the total distance.
You might understand that with this information we can calculate the shortest route for our sales person. However if we have to calculate all possible routes, this will be extremely time consuming.
As the number of locations grows linear and the total possible routes grow exponentially, this means when the number of locations becomes too large it will take too much time to calculate every possible route and we need to use smart algorithms to help us find the optimal route.

## Formulate the Azure quantum problem from the cost matrix

Now that we have gathered all our input data from the distance matrix, we need to calculate the optimal route. As said before, calculating every possible route is too time consuming if the number of locations becomes too large. Therefore we will write this problem as a Quadratic Unconstrained Binary Optimization problem, see my previous post for some more info on this.

The idea is that we create a cost landscape like the image below and let the QIO solver find the lowest possible point. To create this map we give the solver coordinates and costs for all coordinates. The solver will try to find the solution with the lowest cost, which will be our optimal route.

The overall cost optimization will look something like this:

$$\text{Total cost} = \sum \text{distance all visited notes} + \text{penalties if not all constraints are met}$$

If we want to minimize the Total cost we have be use the most efficient route. This means getting a short route, but with all the constraints satisfied.

What we want to do with a QUBO, is formulating the optimization problem in a way that only uses variables that can take the value $$0$$ or $$1$$.
Let's dig in, a quick warning it will get a bit mathematical. A complete walkthrough with code is here. But let's explain each part of the total cost here bit by bit:

### Total distance all visited notes

Let's start with a more generalized form of the distance matrix, our starting point:

$$C = \begin{bmatrix} c_{1,1} & c_{1,2} & \dots & c_{1,N} \\ c_{2,1} & c_{2,2} & \dots & c_{2,N} \\ \vdots & \ddots & \ddots & \vdots \\ c_{N,1} & c_{N,2} & \dots & c_{N,N} \end{bmatrix}$$

To get the distance between two points we can actually use two vectors, like below. With this we can select the distance between any two points. This would mean that all values in the vectors are zero, except for the selected location that will be one. Using the matrix below if $$x_1$$ and $$x_{N+1}$$ will let you get the value of $$c_{1,2}$$

$$\text{Cost for single trip }= \overbrace{\begin{bmatrix} x_1 & x_2 & \dots & x_N \end{bmatrix}}^{\text{origin selector} = X} \overbrace{ \begin{bmatrix} c_{1,1} & c_{1,2} & \dots & c_{1,N} \\ c_{2,1} & c_{2,2} & \dots & c_{2,N} \\ \vdots & \ddots & \ddots & \vdots \\ c_{N,1} & c_{N,2} & \dots & c_{N,N} \end{bmatrix}}^{\text{Cost matrix}=C} \overbrace{\begin{bmatrix} x_{N+1} \\ x_{N+2} \\ \vdots \\ x_{2N} \end{bmatrix}}^{\text{Destination selector} = Y}$$

Now, we want to get the total cost of driving to multiple places. Let's start by naming our origin matrix as $$X$$ and the destination vector as $$Y$$. To get the total cost of a route with $$N$$ nodes we can formulate that as follows:

$$\text{Cost of route} = \sum_{k=1}^{N+1}\sum_{i=1}^{N}\sum_{j=1}^{N} \left( X_{N(k+1)+i}\cdot C_{i,j}\cdot Y_{Nk+j} \right)$$

To see how this can be coded with Azure Quantum Inspired Optimization take a look here.

With this we can calculate the total cost of a route, however when we do not apply constraints the solver might just stay in the same place to minimize cost. To make sure we visit all the nodes and have a valid route we need to add some constraints. Below we will explain all constraints which needed to create a valid route.

### One location at a time constrain

In the previous section we described a method that can find the total cost for a set of locations. However this won't just give the right solution, since we want to penalize the solver if the solution is not valid. With we lower the probility that the solver will choose an invalid solution.

The first constrained that we want to add is that the salesperson can only be in a single position at any time.

How can we make sure that the salesperson is only in a single place at any moment in time? The answers lies within the origin and destination selector vectors. If we have a 3 node system and we are in place 1 the origin selector would look like this:

$$\begin{bmatrix} x_1 & x_2 & x_3 \end{bmatrix} = \begin{bmatrix} 0 & 1 & 0 \end{bmatrix}$$

An invalid selection would be:

$$\begin{bmatrix} x_1 & x_2 & x_3 \end{bmatrix} = \begin{bmatrix} 1 & 1 & 0 \end{bmatrix} \\ \text{or} \\ \begin{bmatrix} x_1 & x_2 & x_3 \end{bmatrix} = \begin{bmatrix} 0 & 0 & 0 \end{bmatrix}$$

To generalize we could say that we need to satisfy the following equations:

$$x_1 \cdot x_2 = 0 \\ x_1 \cdot x_3 = 0 \\ x_2 \cdot x_3 = 0$$

These equations are only valid if there's only one single  $$1$$ value and the others are $$0$$. If this is not the case we will give an extra penalty to the solver. This means that the solver will have lower probability to pick this solution.

Now we want to write this in a more generalized form and create the code to submit to the solver. The generalized form of this constrained will be:

$$\sum_{l=1}^{N+1} \sum_{i=1}^{N} \sum_{j=1}^{N} X_{i+N(l-1)} \cdot X_{j+N(l-1)}$$

Here is the code for the solver and a more detailed description on this constraint.

### At least one location at the time

In the previous constrained we tried to make sure that the salesperson was not in multiple locations at a single time. However, if you looked closely you might have noticed that the equations above would be satisfied if the sales person is not in a single location at all. Therefore we need to make a constrained that the salesperson is in at least one location at the time.

$$0 = (N+1) -\left( \sum_{k=1}^{N(N+1)} x_k \right)$$

This constraint will make sure that the salesperson is always at a location. This constraint can also be satisfied if the sales person would be in all locations at once, however the other constraints will make sure that this will not be a valid solution.

A more detailed description and code can be found here.

### Visit all locations

Now we have three cost functions, one will calculate the complete distance of the route. The second will make sure the driver is only in a single place at each time and the last will make sure we are in at least one location. Next, we need to make sure that all locations are visited.

To add this constraint we take the similar approach as the previous constraints. Let's start with our location selector:

$$\begin{bmatrix} x_1 & x_2 & x_3 \end{bmatrix}$$

We need to visit all these nodes once (let's discard for now that we need to travel back to the starting point). This means that  if we have 3 time moments which we can call $$N$$ and we want to make sure sure that we visit node $$x_1$$ only once, the following needs to be satisfied:

$$x_1 \cdot x_{1+N} =0 \\ x_1 \cdot x_{1+2N} = 0$$

If at two different time moments $$N$$ the location $$x_1$$ has value $$1$$, the equations above will not be satisfied, in this case we will give the solver a big penalty.

### Start and stop at a specific node

Lastly, to complete the route we want to say where we want to start and stop our route. This is done by adding negative weights, a positive stimuli for the solver, for these locations. Here you can see how this is done in the code.

Now we have a complete system to calculate the perfect route. Note that when calculating the perfect route the solver might not always find the perfect solution. With tuning which is discussed later, you can try to improve the solutions.

## Submit the Problem to Azure Quantum

With the cost functions defined, we can now send this to the Azure Quantum Inspired Optimization service. The service will take all the cost and the indices of each cost and solve for the most efficient route.

You will find that tweaking parameters, solver types and hardware can make a big difference on the performance of a solution. With the broad range of solvers and hardware there will be a great solver for each problem. Especially when problems grow using FPGA's will greatly reduce your solving time.

### Tuning

When looking at the code, you see that some weights are multiplied by fixed numbers. These numbers can be used to tune the problem landscape. When you are building these optimization problems you will notice that these parameters can have a huge impact on the performance of the solution and/or solving time. So always play around with these when the solution is not as desired or slow. Also keep in mind that different solver types can have different sweet spots for these parameters.

## Wrapping up

Here I have shown you how to tackle a common optimization problem using Azure Quantum Inspired Optimization. This type of optimization can be applied to lots of other problems like portfolio optimization, scheduling problems and more. All the code for this can be found here. If you have any comments please let me know below.


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# Submodules of a finitely generated module is contained in a maximal ideal

I need to prove the following fact: Let R be a unital ring, M is a finitely generated R-module. Submodules of a finitely generated $R$-module is contained in a maximal ideal. I found the following post that deals with the same problem.

Maximal submodule in a finitely generated module over a ring

However, I don't quite understand the proof given:

1. Why we can be sure that the submodules of the the $R$-module forms a chain? My interpretation of chain is that I have $N_1 \subseteq N_2 \subseteq .....$. What happens if some of the submodules is not finitely generated? Am I misunderstanding the definition of chain?

2. Am I correct to say that the maximal submodule is just the union of all the submodules of the $R$-module?

Thanks

As for the first call $X$ the collection of proper submodules (i.e. not containing $M$) that contain a given submodule $M'$. What we want to show is that $X$ is inductive, i.e. that every chain has an upper bound. Notice that we're not asserting that $X$ is a chain, but that every chain in $X$ has an upper bound. Of course for every chain the union of the elements of the chain is an upper bound (here is where you use the f.g. hypothesis, since if the union were the whole of $M$ then we can generate by finitely many elements, so these must have been included in the union of finitely many elements in the chain, a contradiction). Zorn's lemma now provides us with a maximal element, which in our context is precisely a maximal submodule.


text:
## anonymous one year ago Given f(x) = \frac (\sqrt ( x ) - 6 )(\sqrt ( x ) + 6 ). Find f'( x ) = __________________ Find the equation of the tangent line when x = 25. No decimals allowed. y = __________________

• This Question is Open
1. anonymous

Just to avoid using the quotient rule:$f(x)=\frac{\sqrt x-6}{\sqrt x+6}=\frac{\sqrt x+6-12}{\sqrt x+6}=1-12(\sqrt x+6)^{-1}$

2. anonymous

Refer to the attachment from the Mathematica computer program.


text:
42 Answered Questions for the topic ROUNDING

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text:
# Net Attraction à la Bardeen-Pines and Kohn-Luttinger

In the lead up to the full formulation of BCS theory, the derivation of Bardeen-Pines interaction played a prominent role. The Bardeen-Pines interaction demonstrated that a net attractive interaction between electrons in an electron gas/liquid can result in the presence of phonons.

The way that Bardeen and Pines derived this result can be understood by reading this paper. The result is actually quite simple to derive using a random-phase-like approximation or second-order perturbation theory. Regardless, the important result from this paper is that the effective interaction between two electrons is given by:

$V_{eff}(\textbf{q},\omega) = \frac{e^2}{\epsilon_0}\frac{1}{q^2 + k_{TF}^2}(1 + \frac{\omega_{ph}^2}{\omega^2 - \omega_{ph}^2})$

The crucial aspect of this equation is that for frequencies less than the phonon frequency (i.e. for $\omega < \omega_{ph}$), the effective interaction becomes negative (i.e. attractive).

It was also shown by Kohn and Luttinger in 1965 that, in principle, one could also obtain superconductivity in the absence of phonons. The attraction would occur using the phenomenon of Friedel oscillations whereby the effective potential can also become negative. This was quite a remarkable result: it showed that a purely electronic form of superconductivity was indeed theoretically possible.

What makes the effective interaction become attractive in these two models? In the Bardeen-Pines case, the phonons screen the electrons leading to a net attraction, while in the Kohn-Luttinger case, Fermi surface effects can again lead to a net attraction. It is important to note that in both papers, the pre-eminent quantity calculated was the dielectric function.

This is because the effective potential, $V_{eff}(\textbf{q},\omega)$, is equal to the following:

$V_{eff}(\textbf{q},\omega) = \frac{V(\textbf{q},\omega)}{\epsilon(\textbf{q},\omega)}$

In the aforementioned cases, net attraction resulted when $\epsilon(\textbf{q},\omega) < 0$.

This raises an interesting question: is it possible to still form Cooper pairs even when $\epsilon(\textbf{q},\omega) > 0$? It is possible that this question has been asked and answered in the literature previously, unbeknownst to me. I do think it is an important point to try to address especially in the context of high temperature superconductivity.

I welcome comments regarding this question.

Update: In light of my previous post about spin fluctuations, it seems like $\epsilon < 0$ is not a necessary condition to form Cooper pairs. In the s-wave channel, it seems like, barring some pathology, that $\epsilon$ would have to be less than $0$, but in the d-wave case, this need not be so. I just hadn’t put two and two together when initially drafting this post.

### 2 responses to “Net Attraction à la Bardeen-Pines and Kohn-Luttinger”

1. glassbottomblog

One (possibly trivial) way out is for the dielectric function that matters to belong to a different set of degrees of freedom than those that are actually pairing. I imagine one can write down multiband BCS-type models that have this behavior.

Liked by 1 person

• Anshul Kogar

I don’t know that I totally understand what you’re saying… Can you expand a little?

It seems to me after thinking about this further that it is possible in the d-wave case for epsilon to remain greater than 0, but not for the s-wave case barring some pathology.

Like


text:
# measuring segments practice

Access Free Answers Of Measuring Segments Geometry Prentice Hall Measuring segments (practice) | Lines | Khan Academy ★★★ Correct answer to the question: Geometry Homework: 1.3 Measuring Segments Score: 0 of 1 pt 4 of 11 (3 complete) 1.3.14 Use the number Name_ 1-1 Additional Practice Measuring Segments and Angles In … A line c. A ray with endpoint E. Solutions ⃖ ⃗ ⃗ Khan Academy is a 501(c)(3) nonprofit organization. a. PDF Download 1.3 Practice Measuring Segments Form G Answer Key. In the following examples, find the length of and the stars. Save. 0 likes. Congruent line segments. If you are at an office or shared network, you can ask the network administrator to run a scan across the network looking for misconfigured or infected devices. If GH 5 31 and HI 5 11, then GI 5 u. Practice: Measuring segments. What else could you measure? It is the midpoint of the segment. 4. To find the measure or size of a segment, you simply measure its length. 5. Enjoy the videos and music you love, upload original content, and share it all with friends, family, and the world on YouTube. Linear Measure & Line Segments problems. 6. Definition 2: Geometry: 1.3 Measuring Segments Name _____ Period ____ Date _____ Bell Work: Use the diagram to name each of the following. Measuring Segments. 1-3 Practice Form K Measuring Segments Find the length of each segment. Definition 1: A ruler can be used to measure the distance between two points. 0% average accuracy. Geometry: Common Core (15th Edition) answers to Chapter 1 - Tools of Geometry - 1-3 Measuring Segments - Lesson Check - Page 23 4 including work step by step written by community members like you. Congruent segments are essential ingredients in proofs. If you're seeing this message, it means we're having trouble loading external resources on our website. Properties of Rhombuses, Rectangles, and Squares, Interior and Exterior Angles of a Polygon, Identifying the 45 – 45 – 90 Degree Triangle. The following postulate summarizes this concept. Congruent segments are segments with the same length. As we have learnt previously the line segment can be written as $$\overline{AB}$$ While the length or the measure is simply written AB. Note that in the two preceding equations, an equal sign is used, not a congruence symbol. Textbook Authors: Charles, Randall I., ISBN-10: 0133281159, ISBN-13: … Write the Segment Addition Postulate for the points described. This is the currently selected item. inches or foot). When copying a line segment... answer choices. QS Use the number line at the right for Exercises 4–6. • Our mission is to provide a free, world-class education to anyone, anywhere. QT 3. Our books collection spans in multiple locations, allowing you to get the most less latency time to download any of our books like this one. A line segment that contains . For instance, when you figure out that a side (a segment) of one triangle is congruent to a side of another triangle, you can use that fact to help you prove that the triangles are congruent to each other. Congruent segments are essential ingredients in proofs. View 1.3 Homework.docx from BIOLOGY 101 at Lincoln High School. Completing the CAPTCHA proves you are a human and gives you temporary access to the web property. Two lines could have the same measure but still not be identical. use a straightedge and compass. GUIDED PRACTICE: EXPLORE. Another way to prevent getting this page in the future is to use Privacy Pass. To start, ! If you are on a personal connection, like at home, you can run an anti-virus scan on your device to make sure it is not infected with malware. You’ve got your short, your medium, and your long segments. Parallel and perpendicular. 1-2A Measuring Segments Objectives: G.CO.12: Make formal geometric constructions with a variety of tools and methods (compass and straightedge, string, reflective devices, paper folding, dynamic geometry software, etc. Print; Share; Edit; Delete; Report an issue; Live modes. Performance & security by Cloudflare, Please complete the security check to access. In order to quantify the measure of a segment… nd the coordinate of each endpoint. In a figure, giving different segment the same number of tick marks indicates that they’re congruent. Measuring Segments - Worksheets, Quizzes, Guided Notes and other resources to help teachers teach Measuring Segments. You may need to download version 2.0 now from the Chrome Web Store. Yves Vandermeeren MD PhD from worksheet 1 2 measuring segments day 1 , source:researchgate.net . C is between Q and R 4. Mathematics. 6. Cloudflare Ray ID: 61477c048eaf3fd8 Definition Midpoint of a segment: The point that divides the segment into two congruent segments. GeometryCoach.com En este ejemplo en el que medimos un segmento de recta, los números abarcan la parte positiva y la negativa. Fascinating stuff, right? Practice measuring segments with Khan Academy's free online exercises. For instance, when you figure out that a side (a segment) of one triangle is congruent to a side of another triangle, you can use that fact to help you prove that the triangles are congruent to each other.

measure it with a ruler, and then go the same distance.

. AB BC The length of _ AB equals the length of _ BC. Students progress at their own pace and you see a … Measuring Segments Practice DRAFT. use a ruler and protractor. Whenever you look at a diagram in a geometry book, paying attention to the sizes of the segments and angles that make up a shape can help you understand some of the shape’s important properties. If RS = 33.5 and RQ = 80, find SQ. Some of the worksheets for this concept are Geometry measuring segments form g practice key pdf epub, Geometry measuring segments form g practice key, Teacher resource sampler tools of geometry 1, Home, Unit 1 tools of geometry reasoning and proof, Identify points lines and planes, Name class date … 0. Your IP: 172.96.179.243 Ruler postulate: The points on any line can be paired with real numbers so that, given any two points P and Q on the line, P corresponds to zero, and Q corresponds to a positive number; If Q is between P and R, then QP + QR = PR; If PQ + QR = PR, then Q is between P and R Please enable Cookies and reload the page. Measuring and Constructing Segments continued Segments are congruent if their lengths are equal.

use a … Google Analytics Hour of Day & Day of Week Reports from worksheet 1 2 measuring segments day 1 , source:hallaminternet.com Copying a Segment Method Steps sketch using estimation Estimate the length of the segment. By admin in Line Segments, Linear Measure, Number Line on May 4, 2019. How to solve problems related to measuring segments involving variables. Equation practice with midpoints practice from worksheet 1 2 measuring segments day 1 , source:khanacademy.org . Section 1.2 Measuring Segments G.4.5 Prove theorems by using coordinate geometry, including the midpoint of a line segment, the distance formula, and various forms of equations of lines and circles. Practice: Congruent segments. If RS = 44.6 and SQ = 68.4, find QR. Sketch a segment that is about the same length. Refer to the figure and the given information to find each measure. In a figure, giving different segment the same number of tick marks indicates that they’re congruent. The number is called a coordinate. Segment MP, MP, consists of points M and P and all points between M and P. The measure of MP, written MP (without a bar over the letter), is the distance between M and P. Thus, the measure of a segment is the same as the distance between its two endpoints. ). 1. Exploration Recuerda que los segmentos de recta y los puntos son la base de la geometría, así que este concepto es importante. _ AB _ BC _ AB is congruent to _ BC. 5. J is between S and H 3. E b. 9th - 10th grade . measure it with a ruler, and then go the same distance. View Lesson 1-1 Additional Practice.docx from GEO 1206320 at West Port High School. a minute ago by. Find the length of each segment… The length could either be determined in Metric units (e.g. Linear Measure example question. 1 3 Measuring Segments Practice Form G - Displaying top 8 worksheets found for this concept.. Students will use the geometric analogy such that if the area of a perfect square is represented by a square number, then the side length can be determined by … Get ready for another shock: If you’re told that one segment has a length of 10 and another has a length of 20, then the 20-unit segment is twice as long as the 10-unit segment. PR 2. Lesson 1-4: Measuring Segments and Angles Page 4 of 6 A B C 1 C A B 2 D 1 C A B Consider the following: If AB = BC, what would you call point B? geometry measuring segments form g practice key is available in our digital library an online access to it is set as public so you can get it instantly. 1. If GH 5 45 and GI 5 61, then HI 5 u. Edit. jsharr_41473. 7. Title: Microsoft Word - 1-3 Assignment SE- Measuring Segments.docx Created Date: 7/28/2017 12:27:04 PM Start a live quiz . S is between D and P 2. alternatives. Name Class Form G 1-3 Date Practice Measuring Segments In Exercises 1–6, use the figure below. millimeters, centimeters or meters) or Customary units (e.g. A point corresponds to one and only one number on a ruler. T is between M and N Find QR in the following problems. Displaying top 8 worksheets found for - Practice 1 5a Measuring Segments. Next lesson. Played 0 times. If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked. After all, length is the only feature a segment has. Edit.

use a ruler and protractor

. • This is why we advise you to always see this page when you require such book 1.3 Practice Measuring Segments Form G Answer Key, every book.By online, you could not go to get the book store in your city. Geometry: Common Core (15th Edition) answers to Chapter 1 - Tools of Geometry - 1-3 Measuring Segments - Lesson Check - Page 23 3 including work step by step written by community members like you. (No, these are not technical math terms.) You know that two segments are congruent when you know that they both have the same numerical length or when you don’t know their lengths but you figure out (or are simply told) that they’re congruent. Classic . Jul 12, 2020 # Read Geometry Measuring Segments Form G Practice Key # By C. S. Lewis, practice continued form g measuring segments on a number line the coordinates of p q r and s are 12 5 0 and 7 respectively 22 draw a sketch of this number line use this sketch to answer exercises 23 26 23 which Textbook Authors: Charles, Randall I., ISBN-10: 0133281159, ISBN-13: …


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A Complete ALMA Map of the Fomalhaut Debris Disk [EPA]

We present ALMA mosaic observations at 1.3 mm (223 GHz) of the Fomalhaut system with a sensitivity of 14 $\mu$Jy/beam. These observations provide the first millimeter map of the continuum dust emission from the complete outer debris disk with uniform sensitivity, enabling the first conclusive detection of apocenter glow. We adopt a MCMC modeling approach that accounts for the eccentric orbital parameters of a collection of particles within the disk. The outer belt is radially confined with an inner edge of $136.3\pm0.9$ AU and width of $13.5\pm1.8$ AU. We determine a best-fit eccentricity of $0.12\pm0.01$. Assuming a size distribution power law index of $q=3.46\pm 0.09$, we constrain the dust absorptivity power law index $\beta$ to be $0.9<\beta<1.5$. The geometry of the disk is robustly constrained with inclination $65.!!^\circ6\pm0.!!^\circ3$, position angle $337.!!^\circ9\pm0.!!^\circ3$, and argument of periastron $22.!!^\circ5\pm4.!!^\circ3$. Our observations do not confirm any of the azimuthal features found in previous imaging studies of the disk with HST, SCUBA, and ALMA. However, we cannot rule out structures $\leq10$ AU in size or which only affect smaller grains. The central star is clearly detected with a flux density of $0.75\pm0.02$ mJy, significantly lower than predicted by current photospheric models. We discuss the implications of these observations for the directly imaged Fomalhaut b and the inner dust belt detected at infrared wavelengths.

M. MacGregor, L. Matra, P. Kalas, et. al.
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