text:
[–] 72 points73 points

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Use consecutive fibonacci numbers to approximately convert between miles and kilometers. The ratio of km to mi is 1.6 and the Golden ratio is approximately 1.618, so 8 miles is roughly 13 km, 89 km is roughly 55 miles, and so on.

Obviously the accuracy increases with larger numbers, and the first few numbers give pretty rough results - 1 mile is not really close to 1 km.

[–] 78 points79 points

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TIL memorizing an infinite sequence of numbers is easier than multiplying by 1.6

[–] 4 points5 points

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You're really just memorizing two numbers and the world's easiest 2nd degree recursion.

[–] 6 points7 points

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Eh they're in my head already for other reasons so it's easier for me. YMMV

[–]Math Education 13 points14 points

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YMMV by a factor of 1.6 when you convert to K :-)

[–] 17 points18 points

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Obviously the accuracy increases with larger numbers

Not true!

A mile is 1.609344 km. 5 miles to 8 kilometers is a better approximation than 8 to 13. The closest approximation from consecutive fibonacci numbers is 13 to 21.

[–] 6 points7 points

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Indeed, but for any consecutive Fibonacci numbers greater than 13, you would still miss by less than 10 meters per mile (or 6 yards per kilometer the other way around), which most of the time should give you a good enough estimation.

[–] 0 points1 point

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I agree, the approximation is fine, and it is probably easier for a lot of people to remember the concept of fibonacci numbers than it is for them to remember 5/8.

[–] 0 points1 point

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I think ilmmad means on average, not for each individual result.

[–] 0 points1 point

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Yes, I was interpretting "increases" quite specifically, but I suppose he did just mean that the accuracy is better for large numbers.

[–] 25 points26 points

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Squaring numbers that are between 0 and 100 ( obviously this also works for larger numbers, but that makes it harder):

(x-a)(x+a) = x2 - a2 thus x2 = (x-a)(x+a) + a2

Now it's easy to calculate 642: 60 * 68 = 3600 + 480 = 4080, 4080 + 16 = 4096 And even easier: 942 = 88 * 100 + 36 = 8836

Impressed quite a lot of people with this.

[–] 15 points16 points

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I'm not sure if it's related, but squaring a number ending in 5. Let n be a number composed of the digits preceding the 5. Multiply n by (n+1) and append a 25.

For example for 652 , 6 * 7 = 42, so 652 = 4225.

[–] 5 points6 points

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(10n+5)(10n+5) = 100n2+2*50n+25 = 100n2+100n+25 = 100n(n+1)+25

[–] 2 points3 points

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I remember Art Benjamin the mathemagician doing this.

[–] 0 points1 point

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I don't see how this is any easier than straight multiplication; not that I don't do this, but in some ways it's worse to expand the square into a longer equation because one needs to remember more intermediate numbers.

If you calculate 642 straight, you have to remember two numbers and add them. Your way requires remembering more numbers. I personally usually just multiply, but I'd suggest this alternative to your method:

642 = (60+4)2 = 602 + 2*60*4 + 42

This is similar to your way, but has the advantage of requiring less memorisation, because all constituents are contained within the original problem.

[–] 0 points1 point

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Yeah, I regularly just use a binomial expansion in my head for squaring e.g 57. People who I tutor look at me like I'm a savant ;-p

[–] 0 points1 point

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Strange, I use this in the other direction all the time, i.e. to find 56*62 I get 592 - 9 and I have a pretty good grasp on how to find squares from easier squares so 592 = 602 - 60 - 59. Never thought about using in your way. Good trick.

[–]Math Education 20 points21 points

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I remember that 1/7 = 0.142857... repeating, and that multiples of 142857 are cyclic. This lets me find, for instance, 4/7 = 0.571428....

[–] 3 points4 points

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Woah, I always knew 1/7 was about .14 but never noticed the cyclical aspect. That's awesome

[–]Math Education 5 points6 points

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[–] 2 points3 points

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I like that it doubles the 7

14 = 2 * 7 28 = 2 * 2 * 7 57 = wait for it... 2 * 2 * 2 * 7 = 56, why 57 then?

Double 57, you get 114, and the hundreds place marker is added to 56 to make the 57, and it continues back to 142857...

[–] 0 points1 point

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You can also put a backslash ('\') before any character to print it as is.

[–] 0 points1 point

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I use this a lot too. also a decent approximation for pi is 22/7 because you get 3.142857.... edit: thanks bwsullivan

[–]Math Education 1 point2 points

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22/7

[–] 1 point2 points

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i like the approximation 355/113, because iirc it gives you 7 sig figs for only 6 integer digits, and those 6 digits are easy to remember because you can just split the doubled series of the first 3 odd integers in half: 113355 -> 113\355 = 355/113 (in this non-standard but obvious notation).

[–] 7 points8 points

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56=7x8. Five, six, seven, eight.

Also:

1/7 = .14 28 57 14 28 57 repeating
2/7 = .   28 57 14       repeating
3/7 = . 4 28 57 1        repeating
4/7 = .      57 14 28    repeating
5/7 = .       7 14 28 5  repeating
6/7 = .    8 57 14 2     repeating

Same digits, same order. Whenever you divide by 7, you have to end up with one of these repeating decimals (or it's evenly divisible).

[–] 1 point2 points

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Any trick to remembering where each one lies?

[–] 5 points6 points

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ascending order of first digit:
1/7 - .1...
2/7 - .2...
3/7 - .4...
4/7 - .5...
5/7 - .7...
6/7 - .8...

[–] 1 point2 points

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Just remember 14 28 57 and then estimate the beginning digit from the size of the fraction. I don't see an obvious pattern anyway.

[–] 0 points1 point

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Aha, that's exactly what I needed (and failed) to realize. Thank you!

[–] 0 points1 point

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You've seen the other comments, so all I'll add it that the order of the digits is an almost doubling sequence: 14 doubled is 28 doubled is 56. For some reason I find it noteworthy that no multiples of 3 are in the digit string: no 3, no 6, and no 9. And every other digit appears once and only once. So 56 is wrong, and needs to be changed to 57.

Or, a different thing to notice, if you add 1/7 and 6/7 you get 1, but in the form of:

  1/7 = .14 28 57 repeating
+ 6/7 = .85 71 42 repeating
-----------------
7/7 = .99 99 99 repeating

What makes this kinda weird for me is that I completely grok why casting out nines works as a check

Side note: you use modular arithmetic, and 10 = (9 + 1) =9 1.

Then look at the place value expansion of any number, with the sum of (digits multiplied by powers of 10),

which turns into the sum of (digits times powers of 1 modulo 9)

side side note: since 10 =11 -1, you could alternately add and subtract digits to get the remainder of a number divided by 11. Start at the ones place (+) and subtract the 10's place.

But I completely fail to see why 3, 6, and 9 are not in the repeating expansion of 1/7 and why the decimal expansion is rotating through only the same digits. What does 10 =7 +3 =7 -4 explain?

[–] 0 points1 point

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I was noticing that, and I think I am going to remember the base pair as "14.2," because 14.2 rounds to 14, 28.4 to 28, and 56.8 to 57.

[–] 6 points7 points

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Calculating 10x , where x is a non-integer. The first thing is that x can always be broken down into x = k+y, where k is an integer and y is between 0 and 1. Then, obviously, 10k+y = 10k *10y. So the question becomes how do you do 10y where y is between 0 and 1? Well, you can figure it out using some tricks when y = 0.1, 0.2, ..., 0.9, and then interpolate between those. For example, say you need to get 100.3. Well 210 = 1024 ~= 103. So 210/10 ~= 103/10 = 100.3. So 100.3 ~= 2 (correct answer is 1.99). Then knowing 100.3 makes figuring out 100.6 and 100.9 easy. Also, 100.5 is easy, because 210 ~= 103, so 25 ~= 103/2 so 25 /10 = 3.2 ~= 100.5.

[–] 2 points3 points

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that's some feynman thinking right there

[–]Math Education 13 points14 points

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Honestly, just being able to do any kind of mental math, trickery or otherwise, can be impressive to people. I'll do Simplex Method examples on the board and it seems to wow students, and when I give them a quiz with "nice" integer solutions, all I hear for 10 minutes is the ratatatat of calculator keys.

[–] 3 points4 points

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could you give an example of a simplex method?

[–]Math Education 2 points3 points

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This has some good info about the method and why it works.

As a student pointed out, in jest, the simplex method is far from simple :-)

[–] 0 points1 point

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[–] 0 points1 point

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Fucking wyndor glass company!

[–] 3 points4 points

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My friends have dubbed me rain man. Never that quick, but I can see the numbers in my head while I do the long division or multiplication, so they're impressed as fuck. Got beat up a lot as a kid.

[–] 4 points5 points

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This is incredibly low level, but when i was a kid I had trouble subtracting numbers below ten from teens if the teen's one-place number was lower than the sub-10 number being subtracted. Something like 15-7=X.

I realized that you could instead subtract the ones-place number from the sub-10 number and then subtract the result from 10 and it still worked: 7-5=2; 10-2=8, which is 15-7.

I always thought it was pretty cool back then and it helped me do those damned timed tests that all elementary school teachers loved, circa 4th grade.

[–] 1 point2 points

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Wow, me too. Exact same thing.

I also do this for adding too. 15 + 7 in my brain goes 15 + 5 + (7 - 5) instead of simply 15 + 7. I usually am faster than most of my peers at mental math too, my brain seems to always just liked working with numbers less than 5, and wholes of 10.

64 + 17 (in my brain) : 60 + 10 + 7 + 4 = 70 + 10 + 1 = 81.

Maybe it's natural for this type of calculation to occur since it is most simple?

[–] 2 points3 points

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You made me login for that, but 64 + 17 is CLEARLY 63 + 17 (+1) = 60 + 20 (+1) = 80 (+1) = 81.

Had to say it, sorry xD

In my head I always add or substract from both numbers until the sum of the two resulting numbers gives a nice multiple of a power of 10 and then add/substract what I substracted/added before.

[–] 0 points1 point

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I find it easier to go from higher order digits down, thus adding extra 1000's, 100's, 10's etc. as they come.

[–] 0 points1 point

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Am I the only one who does these simple calculations visibly? If I want to add two numbers of graspable length, I imagine them beneath each other and add per place. I'd internally go:

64
17
--
11
6
1
--
81

[–] 0 points1 point

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Yes, I do it in my head visually (I find remembering the numbers as I calculate is easier if I visually think of the numbers). My visuals go horizontal though.

64 + 17, slide the 1 from the 17 to the 6 of the 64. 74 + 7 = 70 + 4 + 7 = 70 + 10 + 1. Slide again, finish.

These methods seem so convoluted if you actually think about it, but to each their own hah!

[–] 5 points6 points

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If I have to add two messy numbers, but one is close to a round number, I will subtract/add as necessary and then do the reverse at the end. For example:

998 + 3735 = (1000 - 2) + 3735 = 4375 - 2 = 4373

[–] 4 points5 points

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Here is an easy method to find square root approximations of integers:

Call the number you want to square root x. Find the highest square number below x, call it y2. Then your approximation of the square root is y + (x-y2 )/(2y+1).

For example, if x is 807, the nearest square below it is 784, or 282. So the square root approximation is 28+ 23/57.

sqrt(807) = 28.408

28+ 23/57 = 28.404

[–] 1 point2 points

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What's the point of the (2y+1) instead of the just 2y you would get from a Taylor series? For your example, you get 23.411, which is marginally better than 23.404. Also I think you mean "highest square number below..."

[–] 1 point2 points

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Because 2y+1 is the difference between y2 and (y+1)2. This matters below a 'new' square number, for example 399. With only using 2y, you would get 19+38/38, or 20, with 202 being 400 and (19+38/39)2 being approximately 398.975.

And you are right, I do mean highest square number below. Edited.

Edit: This method will always slightly underestimate the value the square, with this error peaking at the mid point between the two squares whereby the difference between your square root squared and the actual square root squared will tend to 1/4. To account for this, you can slightly increase the value of your fraction: You can make it 28 + 93/228 (aka original fraction + 1/4(2y+1) ) for values near the middle of two squares to increase accuracy.

[–] 8 points9 points

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calculating the summation of a consecutive sequence of numbers begining at 1 and up to n. f(n) = n(n+1)/2

Explains the proof. http://www.maths.surrey.ac.uk/hosted-sites/R.Knott/runsums/triNbProof.html

It's impressive if you can do multiplication and division in your head.

[–] 9 points10 points

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My favorite party trick. (I don't get invited to many parties.)

[–] 6 points7 points

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There's a good "party" explanation for this formula, actually.

Say you're at a party with n+1 people, and you want them all to shake hands. How many handshakes happen?

Well the first person shakes n hands. The second person shakes n-1 more hands (already met the first person, but needs to meet everyone else). This goes all the way to the nth person, who shakes one hand, that of the n+1st person, who has no hands left to shake.

So the number of handshakes is n + (n-1) + ... + 3 + 2 + 1.

But it's also just the number of pairs of people at the party, which is n+1 choose 2, which is

(n+1)!
--------
2! (n-2)!

which is, of course

(n+1)(n)
---------- .
2

[–] 7 points8 points

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I used this at a bar while people were playing a trivia game and they bought me a drink. Math BA paying for itself one drink at a time.

[–] 0 points1 point

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there's a few good simple bar math tricks,

Circumference of pint glass > height of pint glass + 3 ash trays is usually good for a free pint.

[–] 3 points4 points

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Recognizing that the multiplication algorithm is just the distributive property will save you a lot of pain. It's a lot easier to just use the property than to try to mentally apply the algorithm.

56 x 30 = (50 + 6) x 30 = (50 x 30) + (6 x 30) = 1500 +180 = 1680

[–] 1 point2 points

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Low dee high minus high dee low all over the square of what's below. Also one dee two plus two dee one. That shit got me through ap calc.

[–] 0 points1 point

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I never memorized the "quotient formula." I always write the denominator as an inverse product then use the product rule and chain rule. My philosophy about learning math is that it's always better to derive than to memorize. (Of course I have lots of things memorized that aren't easily derivable or because I've used them enough times that I just can't forget them.)

[–] 7 points8 points

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If you are in physics or pre calc, memorize sin/cos of 30, 45, and 60. 0.5, 0.707, 0.866. Saves you a lot of calculation.

[–] 23 points24 points

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Honestly I rather have exact values and use radians : this make it esay to find the rapidly.

[–]Number Theory 6 points7 points

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fonction

[–] 2 points3 points

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The table is in French.

[–] 1 point2 points

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hahahaha

[–] 0 points1 point

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im in both.. ty!

[–] 3 points4 points

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Memorize facts in ln(x)/ln(2). Addition and subtraction can be used instead of multiplication and division. +1 means twice as big. Makes estimates easier.

For example, I know my CPU can do 230 integer multiplications and 228 divisions a second. 30-28 = 2, which means multiplication is about 22 = 4 times faster than division.

Edit: Corrected example.

[–] 1 point2 points

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Or log_2(), or just log() to computer scientists. It doesn't take long to start thinking in base 2, to the point that 230 / 228 = 4 is just common sense.

[–] 0 points1 point

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Don't you mean 30-28 = 2, so it's 4 times faster?

[–] 0 points1 point

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Yes, thank you.

[–] 2 points3 points

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It can really come in handy to know the following powers of two:

2^8  =  256  (number of possible values a byte can take)
2^10  =  1024  (kilobyte)
2^20  ~= 1 million  (megabyte)
2^30  ~= 1 billion  (gigabyte)  (or # of operations a processor does per second)
2^40  ~= 1 trillion  (terabyte)

Example usages:

• If a rabbit population can double every month (given unlimited carrots), then how long did it take to get to a billion rabbits after Noah let them off the ark? A: 30 months.

• I want to build a 20 questions machine that can beat even the best guessers. How many different things must the machine know? A: At least a million, because the guesser can cut the space of possible correct answers in half with each question.

• I want to run a simulation that involves a brute-force search over all subsets of X. How big of a set can I make X, realistically? A: about 30 elements. There are 230 = a billion subsets, which should take no more than a few seconds to iterate over. Note: 40 elements will take 1024 times as long as 30! 50 elements will take a million times as long!

[–] 2 points3 points

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You seem to have put trillion instead of billion and billion instead of milliard :P

[–] 0 points1 point

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I don't think so -- where specifically?

[–] 0 points1 point

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[–] 0 points1 point

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Oh, haha, I didn't know that!

[–] 2 points3 points

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(X+1)2 = X2 + 2X +1

Yes, very simple if broken down, but I found it easy to get the next number's square just by adding the double the original number + 1.

432 = 422 + 2*42 + 1 = 422 + 42 + 43

Simple, I know.

[–] 1 point2 points

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02 + 1 =12
12 + 3 =22
22 + 5 =32
32 + 7 =42
.
.
.

[–] 3 points4 points

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Add up all the digits of any integer. If the sum of the digits is divisible by 3, then the original number is as well. The exact same thing is true for numbers divisible by 9.

Perform multiplication in your head by doubling one number and halving the other. Good for quick approximation as well.

[–] 4 points5 points

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There's an ungodly number of divisibility rules. Some more or less useful than others.

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

[–] 0 points1 point

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[–] 0 points1 point

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I do all sorts of refactoring for quick multiplication, not just double/half...

ex: 75 * 64... ugh, 4 times 5, carry the two... nope, no need for all that. 75 is 3/4 of 100, so 3/4 of 64 (which is 48) times 100: the answer is 4800. Too easy? well...

ex2: 42 * 28... bah that's no good, but it's the same as (pull a 7 from each) 49 * 24, and since 24 * 50 is 1200, just subtract 24: the answer is 1176

[–] 0 points1 point

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For 75*64, I like to think of that as 75 * 64 = 150 * 32 = 300 * 16 = 4800

[–] 1 point2 points

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One I learned in middle school that always stuck with me was squares of two digit numbers ending in 5. (a5)2 = (10a + 5)2 = 100a2+100*a + 25 = 100a(a+1) + 25 means that you can take the leading digit, multiply it by one more than itself and concatenate a 25. e.g. 552 = 56100 + 25 = 3025.

Now by know squares of two digits numbers ending in 5 (and 0) you can use a couple of tricks to do a sort of laborious two digit mental multiplication. I don't recommend this method for general two digits, it's just the sort of system I developed over the years.

[–] 2 points3 points

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You typed 5x6x100 but used asterisks with no spaces, so reddit formatted it as an italicized 6 and outputted 56100.

[–] 0 points1 point

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And right after I read this comment, I went back to edit my comment with many terribly positioned asterisks. Thank you

[–] 1 point2 points

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2x is roughly 10x/3. This can usually get you within an order of magnitude of the true value (because 23 is approximately 10). Not necessarily my favorite but I use it all the time.

[–]Calculus 1 point2 points

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Awesome

[–] 1 point2 points

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This might be insignificant to you more intelligent folk. When I was 9 I found an easy way to multiply by 5. You half the number then remove the decimal. Take this for example, what's 33 * 5 = ? 33 / 2 = 16.5 So, 33 * 5 = 165. Of course 33 by 5 is very simple but when you get to much larger numbers it can be quite useful.

[–]Applied Math 0 points1 point

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remove the decimal

What you're doing is taking advantage of the fact that 5 = 10/2. So you divide by 2 and then multiply by 10, which moves the decimal one place to the right. Since dividing by 5 only ever leaves one digit to the right of the decimal point, you're set.

[–] 1 point2 points

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The doomsday algorithm for finding days of the week requires a little memorization and some pretty easy mental math. Extremely fun and you can do it to anyone...usually the reply you'll get is "I guess so"

[–] 1 point2 points

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Pretty simple, but for 2 digit numbers multiplied by 11:

Add the two numbers together, and place the result between the original digits. e.g. 11x34=374

[–] -1 points0 points

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So 11 x 37 = 3107 ?

[–] 4 points5 points

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when the result of the addition exceeds 10 you carry the 1 and add it to the hundreds place, leaving the remainder in the middle.

[3 + 7 = 10]

11 X 37 = (3+1)(remaining 0)(7) = 407

[9 + 3 = 12]

93 X 11 = (9+1)(remaining 2)(3)= 1023

As long as you carry the one it will always work.

[–] 0 points1 point

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How to square a number ending in 5. Let's square 135.

You take the digits besides the one's digit: 13

Add one to that number: 14

Multiply these together: 13 x 14 = 182

Then tack on a 25 at the end: 18225

1352 = 18225

[–] 0 points1 point

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If the sum of all the digits in a number equals 9, then the number itself is divisible by 9. Example 3231. 3+2+3+1=9 and 3231 ÷ 9 = 359.

[–] 0 points1 point

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I contributed the chapter on estimating square roots to the O'Reilly book Mind Performance Hacks so I'd have to say that :)

[–] 0 points1 point

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When driving I like to calculate how long it will take me to get the the next town based on my speed and how many kilometers away it is. Pretty dumb but it makes long drives more interesting. And if I see number plates with 4 or more numbers in them I make up another number and try to make an equation containing the numbers on the plate to equal the number I made up.

[–] 0 points1 point

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Don't know if anyone's put up the 'multiply by eleven' trick yet, so I'll slap it up for completeness:

To multiply by eleven, add the two neighbouring digits together and put the result down, keeping the end digits. Remember to work in any carry.

For example:

45 x 11 = 4 (4 + 5) 5 = 495

In the case of a carry:

56 x 11 = 5 (5 + 6) 6 = 5 (11) 6 (carry the 10 column from the 11) = 616

And it works for any length of number :

345 x 11 = 3 (3 + 4) (4 + 5) 5 = 3795.

12345 * 11 = 1 (1+2) (2+3) (3+4) (4+5) 5 = 135795

[–] 0 points1 point

sorry, this has been archived and can no longer be voted on

12.52 can be written as (x+.5)2 which expanded is x2 + x + .25 .

So when squaring a (positive integer + .5) square the positive integer, add the positive integer to that square and then add a quarter.

One time I wanted to know what 960 square was in the shower (i.e. no readily available calculator) so I did something similar:

9602 = (1000-40)2 = 1000000 - 80000 + 1600 = 921600

However this took me a minute to do because I couldn't use the shower door as it hadn't properly fogged up yet.

[–] 0 points1 point

sorry, this has been archived and can no longer be voted on

Calculating compound interest in your head is impressive. and it's basic math.

[–] 0 points1 point

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A helpful trick by log differentiation gives

F'(x)= ((ea * fb * gc ) / (hd * jn * km ) )' =

F(x)[ae'/e + bf'/f + cg'/g - dh'/h -nj'/j - mk'/k]

where e,f,g,h,j,k are functions of one variable

[–] -1 points0 points

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c)/(h dj nk m))'

[–] 0 points1 point

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My math teacher last year taught us a cool trick for remembering the values of the special angles to trig functions. You take your hand and the thumbs each correspond to an angle. The thumb is 0 degrees, index finger is 30 degrees, middle finger is 45 degrees, ring finger is 60 degrees, and pinky is 90 degrees. For whatever angle you're trying to remember the value for, bend in the corresponding finger.

For the sine of any angle, it's the square root of the number of fingers above the bent one divided by two. For example, sin(60) would have three fingers above the best middle finger and then the value would be √3/2.

For cosine, it's the square root of the number of fingers below the bent one divided by 2 and for tangent, its the square root of the number of fingers above the bent one divided by the square root of the number below.

This came in real handy in trig last year.

[–] -5 points-4 points

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Equation with a given slope through a given point: point-slope form is for n00bz. If you have Ax + By = C, the slope is -A/B. So if your slope is, say, 3/4, then the equation will be 3x - 4y = C, and you just plug in the point to get C. It's much faster than point-slope!

[–] 1 point2 points

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Is it?

y = 3/4x - C, you can instantly see the slope is 3/4 with no further work.

Doing -A/B is the exact same (same amount of work) as rearranging the equation.

[–] -3 points-2 points

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It depends on what you're trying to do. If standard form is more useful to you, then this method causes unnecessary steps that make it exponentially harder to do mentally.

[–] 4 points5 points

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You only have to substract y from both sides and multiply by B, I wouldn't call this "exponentially harder"...

[–] 3 points4 points

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Not even "linearly" hard. The hardness is constant, always, for such equations.

[–] 0 points1 point

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All this math talk has gotten me hard now. I'm going over to /r/gonewild

[–] -3 points-2 points

sorry, this has been archived and can no longer be voted on

while you nerds are grinding the gears in your heads, I'll be solving problems with a computer and using the time saved to sleep with many beautiful women


text:
ISSN:2164-6376 (print)
ISSN:2164-6414 (online)
Discontinuity, Nonlinearity, and Complexity

Dimitry Volchenkov (editor), Dumitru Baleanu (editor)

Dimitry Volchenkov(editor)

Mathematics & Statistics, Texas Tech University, 1108 Memorial Circle, Lubbock, TX 79409, USA

Email: dr.volchenkov@gmail.com

Dumitru Baleanu (editor)

Cankaya University, Ankara, Turkey; Institute of Space Sciences, Magurele-Bucharest, Romania

A Survey on Sufficient Conditions for Hamiltonian Cycles in Bipartite Digraphs

Discontinuity, Nonlinearity, and Complexity 11(1) (2022) 1--8 | DOI:10.5890/DNC.2022.03.001

Huifen Ge$^1$, Shumin Zhang$^{2}$, Chengfu Ye$^1$

$^1$ School of Computer, Qinghai Normal University, Xining 810001, China

$^{2}$ School of Mathematics and Statistics, Qinghai Normal University, Xining 810001, China

Abstract

A digraph $D=(V,A)$ is called a bipartite digraph if there exists a partition $(X,Y)$ of $V(D)$ into two partite sets such that every arc of $D$ has its end-vertices in different partite sets. It is called balanced if $|X| = |Y|$. If a digraph $D$ has a directed cycle $C$ which covers all of its vertices, then $D$ is Hamiltonian. This paper mainly introduces some sufficient conditions for Hamiltonian cycles in balanced bipartite digraph.

References

1.  [1] Bang-Jensen, J. and Gutin, G. (2000), Digraphs: Theory, Algorithms and Applications, Springer.
2.  [2] Dirac, G.A. (1952), Some theorems on abstract graphs, London Math. Soc., 2(3), 69-81.
3.  [3] Diestel, R. (2016), Graph Theory, (fifth Edition), Springer.
4.  [4] Garey, M.R. and Johnson, D.S. (1979), Computers and intractability. W.H. Freeman. San Francisco.
5.  [5] Adamus, J. and Adamus, L. (2012), A degree condition for cycles of maximum length in bipartite digraphs, Discrete Mathematics, 312, 1117-1122.
6.  [6] Adamus, J., Adamus, L., and Yeo, A. (2014), On the Meyniel condition for hamiltonicity in bipartite digraphs, Discrete Math and Theoretical Computer Science, 16, 293-302.
7.  [7] Bermond, J.C. and Thomassen, C. (1981), Cycles in digraphs-a survey, J. Graph Theory, 5(1), 1-43.
8.  [8] Amar, D. and Manoussakis, Y. (1990), Cycles and paths of many lengths in bipartite digraphs, Journal of Combinatorial Theory Ser. B., 50, 254-264.
9.  [9] Manoussakis, Y. and Milis, I. (1999), A sufficient condition for maximum cycles in bipartite digraphs, Discrete Mathematics, 207, 161-171.
10.  [10] Ghouila-Houri, A. (1960), Une condition suffisante d'existence d'un circuit hamiltonien, Rendus de I' Academie des Sciences Paris Ser. A-B, 25, 495-497.
11.  [11] Nash-Williams, C.St.J.A. (1969), Hamilton circuits in graphs and digraphs, in The many facets of graph theory'', Springer, Lecture Notes, 110, 237-243.
12.  [12] Woodall, D.R. (1972), Sufficient conditions for circuits in graphs, Proceeding London Math. Society, 24, 739-755.
13.  [13] Meyniel, M. (1973), condition suffisante d'existence d'un circuit hamiltonien dans un graphe oriente, Journal Combinatorial Theory Ser. B, 14, 137-147.
14.  [14] Bang-Jensen, J., Gutin, G., and Li, H. (1996), Sufficient conditions for a digraph to be hamiltonian, J. Graph Theory, 22(2), 181-187.
15.  [15] Bang-Jensen, J., Guo, Y., and Yeo, A. (1999), A new sufficient condition for a digraph to be hamiltonian, Discrete Applied Math., 95, 61-72.
16.  [16] Wang, R. and Wu, L. (2020), A dominating pair condition for a balanced bipartite digraph to be hamiltonian, Australasian Journal of Combinatorics, 77, 136-143.
17.  [17] Wang, R., Chang, J., and Wu, L. (2020), A dominated pair condition for a digraph to be hamiltonian, Discrete Mathematics, 343, 111794.
18.  [18] H\"{a}ggkvist, R. and Manoussakis, Y. (1989), Cycles and paths in bipartite tournaments with spanning configurations, Combinatorica, 9(1), 33-38.
19.  [19] Zhong, W. (1992), A sufficient condition for hamiltonian cycles in bipartite tournaments, Australasian Journal of Combinatorics, 5, 299-304.
20.  [20] Gutin, G. (1995), Cycles and paths in semicomplete multipartite digraphs, theorems and algorithms: a survey, J. Graph Theory, 19(4), 481-505.
21.  [21] Wang, R. (2017), A sufficient condition for a balanced bipartite digraph to be hamiltonian, Discrete Mathematics and Theoretical Computer science, 19(3).
22.  [22] Darbinyan, S.Kh. and Karapetyan, I.A. (2018), On a problem of Wang concerning the hamiltonicity of bipartite digraphs, Mathematical Problems of Computer Science, 49, 26-34.
23.  [23] Darbinyan, S.Kh. (2019), Sufficient conditions for hamiltonian cycles in bipartite digraphs, Discrete Applied Math, 258, 87-96.
24.  [24] Adamus, J. (2017), A degree sum condition for hamiltonicity in balanced bipartite digraphs, Graphs and Combinatorics, 33, 43-51.
25.  [25] Chv{a}tal, V. (1973), New directions in hamiltonian graph theory, in New directions in the theory of graphs (F. Harary, ed.). Academic Press, New York. 5(1), 65-95.
26.  [26] Darbinyan, S.Kh. (1985), Pancyclicity of digraphs with the Meyniel condition, Studia Sci. Math. Hungar, 20(1-4), 95-117.
27.  [27] Darbinyan, S.Kh. (1986), On the pancyclicity of digraphs with large semidegrees, Akad. Nauk Arm. SSR Dokl., 83(3), 99-101.
28.  [28] H\"{a}ggkvist, R. and Thomassen, C. (1976), On pancyclic digraphs, Journal of Combinatorial Theory, Series B, 20(1), 20-40.
29.  [29] Overbeck-Larisch, M. (1977), A theorem on pancyclic-oriented graphs. Journal of Combinatorial Theory, Series B, 23, 168-173.
30.  [30] Thomassen, C. (1977), An Ore-type condition implying a digraph to be pancyclic, Discrete Mathematics, 19, 85-92.
31.  [31] Beineke, L.W. and Little, C. (1982), Cycles in bipartite tournaments. Journal of Combinatorial Theory Ser. B, 32(2), 140-145.
32.  [32] Zhang, K. (1981), Vertex even-pancylicity in bipartite tournaments, J. Nanjing Univ. Math. Biquarterly, 1, 85-88.
33.  [33] Gutin, G. (1989), A characterization of vertex pancyclic partly oriented k-partite tournaments, Vestsi Acad. Navuk BSSR Ser. Fiz.-Mat. Navuk, 2, 41-46.
34.  [34] Gutin, G. (1995), Characterizations of vertex pancyclic and pancyclic ordinary complete multipartite digraphs, Discrete Mathematics, 141, 153-162.
35.  [35] Darbinyan, S.Kh. and Karapetyan, I.A. (2017), On longest non-hamiltonian cycles in digraphs with the conditions of Bang-Jensen, Gutin and Li, Discrete Applied Mathematics, 216, 537-549.
36.  [36] Meszka, M. (2018), New sufficient conditions for bipancyclicity of balanced bipartite digraphs, Discrete Mathematics, 341, 3237-3240.
37.  [37] Darbinyan, S.Kh. and Karapetyan, I.A. (2017), A sufficient condition for pre-hamiltonian cycles in bipartite digraphs, Eleventh International Conference on Computer Science and Information Technologies(CSIT), 101-109.
38.  [38] Adamus, J. (2018), A Meyniel-type condition for bipancyclicity in balanced bipartite digraphs, Graphs and Combinatorics, 34, 703-709.
39.  [39] Darbinyan, S.Kh. (2018), Sufficient conditions for balanced bipartite digraphs to be even pancyclic, Discrete Applied Math, 238, 70-76.
40.  [40] Wang, R. and Guo, J. (2017), A note on cycles of maximum length in bipartite digraphs, Australasian Journal of Combinatorics, 67(1), 1-10.

format: plain


text:
## 09 Feb The Colours of Infinity: The Beauty and Power of Fractals

Edited by Nigel Lesmoir-Gordon

Springer, 2010, x+207 pp.

In 1980, at the IBM Thomas Watson Research Centre in upstate New York, the mathematician Benoit Mandelbrot was probably the first person to view a geometric picture of an unusual set of pixels on the computer screen and discover its enormous complexity and geometric richness. This set had its origins much earlier on at the beginning of the 20th century in the research of the French mathematicians in the field of complex dynamics, Pierre Fatou and Gaston Julia. In the eighties, the properties of this set were well investigated and it was named the Mandelbrot set (or commonly called the M-Set). In the following decades, applications and ramifications of the M-set began to emerge in other disciplines.

This book is an offshoot of the full-length 1995 television documentary with the same title, presented by the late science fiction writer and thinker Arthur C. Clarke and produced by the film and documentary maker Nigel Lesmoir-Gordon. While the film is a visually stunning display of computer graphics and animation that captures the imagination of its viewers, the book version is literally an intellectual eye-opener for those who have only vaguely heard of the M-Set or of fractals (a term coined by Mandelbrot himself).

There are not many mathematical equations in the book. There is basically only one important equation — presented in an iterative form $Z^2Dz^2+c$ (which Arthur

Clarke himself likened to the Einstein equation $E=mc^2$ in depth and simplicity). The emphasis is on pictures for visualising the endless process of “self-similarity” in nature. The psychedelic colours and dream-like patterns generated have a surrealistic, hypnotic effect. It is almost unbelievable that they are produced mechanically by a deterministic equation. Little wonder that the M-set is sometimes called the “thumbprint of God”.

It is not just the pictures that mesmerise the mind. A hint of the mathematics behind the apparent order emerging from chaotic processes is beautifully illustrated by an imaginary soccer game played in a definite and yet random way. The mathematically minded reader will enjoy the exposition of the chapter on “Fractal transformations” written by Michael Barnsley and Louisa Barnsley. In it you will find references to the Iterated Function System (IFS), the mathematics behind fractal image compression technology.

If you have seen the documentary, you would have heard Michael Barnsley recalling the recurrent nightmare he had for over 20 years which finally gave him the key to fractal image compression — the Collage Theorem. For the benefit of those who have not watched the film, it is an account that is reproduced in “The Film Script” which is appended at the end of the book. If you have seen the film, reading this complete film script is like watching it again in your mind. If you read it first and then see the film, you would appreciate the film even more.

Lesmoir-Gordon describes in the book how his first acquaintance with the M-Set in 1991 led him in an obsessive-compulsive and yet serendipitous way to a collaboration with Arthur Clarke on a project beyond his wildest imagination. It is an inspiring behind-thescenes account that is not usually revealed or widely known.

The serious reader will find it intellectually challenging and satisfying when digesting the 134 pages of exposition on the geometric visualisation of the M-Set, the tilings of Escher and the fractal view of financial markets — given by the masters and practitioners of the art and science of fractals like Ian Stewart, Arthur Clarke, Benoit Mandelbrot, Michael Barnsley, Will Rood, Gary Flake and David Pennock.

Y. K. Leong.

National University of Singapore, Singapore.

Source:- Asia Pacific Mathematics Newsletter, Volume 1 No. 2 (April 2011).

It has been republished here with a special permission from World Scientific.


text:
## Arrhenius Equation

Arrhenius Equation: $\ln k = - \frac{E_{a}}{RT} + \ln A$

megangeorge-1K
Posts: 59
Joined: Thu Feb 01, 2018 3:01 am

### Arrhenius Equation

What do we use the Arrhenius equation for?

Michelle Wang 4I
Posts: 63
Joined: Fri Sep 28, 2018 12:27 am

### Re: Arrhenius Equation

You would use the Arrhenius equation to relate things like the rate constant with temperature, activation energy and the pre-exponential factor A. There can be questions asking you to solve for one of them given the others.

eden tefera 2B
Posts: 39
Joined: Fri Sep 28, 2018 12:21 am

### Re: Arrhenius Equation

To add, using the arrhenius equation to solve for pre-exponential factor, A, gives you insight into the frequency of collisions that are in the correct orientation.

mbaker4E
Posts: 69
Joined: Fri Sep 28, 2018 12:26 am

### Re: Arrhenius Equation

If you need to find frequency factor of collisions (A) and you are given activation energy (Ea) you would use the Arrhenius equation.


text:
# Help me to understand the Gaussian blurring (2)

Here is an unknown luminosity function $f(x,y)$ and its integration results: \begin{align*} p_{i,j} &= \frac{1}{\Delta_{i,j}}\iint\limits_{D_{i,j}} \! f(x,y) \, dx \, dy,\\ \Delta_{i,j} &= \iint\limits_{D_{i,j}} \!dx\,dy\;. \end{align*} Let's consider the following transformations: \begin{align*} F(x,y,r,\sigma)&=\frac{1}{2 \pi \sigma^2}\int\limits_{x-r}^{x+r} \int\limits_{y-r}^{x+r} \! f(u,v) \,e^{-\frac{(u-x)^2+(v-y)^2}{2\sigma^2}} \, du \, dv\;,\\ q_{i,j}(r,\sigma)&=\frac{1}{\Delta_{i,j}}\iint\limits_{D_{i,j}} \! F(x,y,r,\sigma) \, dx \, dy\;. \end{align*} Is there a functional relationship between:

1. $p_{i,j}$ and $q_{i,j}(r,\sigma)$, where $r \in [0,+\infty)$;
2. $p_{i,j}$ and $q_{i,j}(\infty,\sigma)$?
-
When you ask a question that's closely related to an earlier question, please link to the earlier question to avoid needless duplication of efforts. – joriki Aug 10 '12 at 12:49

## 1 Answer

No, there can't be such a relationship, since $q_{i,j}$ depends on values of $f$ outside of $D_{i,j}$ and $p_{i,j}$ doesn't.

-
What if $r$ is chosen suchwise $[x-r,x+r]\times[y-r,y+r] \subseteq D_{i,j}$?! – Max Aug 10 '12 at 18:43
@Max: That doesn't make any sense. $x$ and $y$ are bound variables in the integral defining $q_{i,j}$; they range over all of $D_{i,j}$, so there's no non-zero $r$ that would lead to that inclusion. The whole idea of blurring is to mix in values from the neighbourhood, and if you do that at the boundary of $D_{i,j}$, it necessarily mixes in values from outside $D_{i,j}$. – joriki Aug 10 '12 at 20:14
What if $\;\bigcup\limits_{i,j} D_{i,j} = \mathbb R \times \mathbb R,\; \bigcap\limits_{i,j} D_{i,j} = \varnothing$ and values of $\Delta_{i,j}$ is small enough that we can assume that $p_{i,j} \approx f(x_i, y_j)$ and $q_{i,j}(r, \sigma) \approx F(x_i, y_j, r, \sigma)$? – Max Aug 11 '12 at 12:21
More precisely: $f(x,y) \approx p_{i,j}$ for any $(x,y) \in D_{i,j}$ and $q_{i,j}(r, \sigma) \approx F(x_i, y_j, r, \sigma)$ where $(x_i, y_j)$ is a centroid of $D_{i,j}$ – Max Aug 11 '12 at 12:41
@Max: Well, then either $\sigma$ is small compared to the extent of the $D_{i,j}$, in which case in this approximation $q_{i,j}=p_{i,j}$, or it isn't, and then you have the same problem that $q_{i,j}$ is affected by function values outside $D_{i,j}$. I don't see how you could possibly get around that without defeating the whole purpose of blurring and rendering the problem trivial. – joriki Aug 12 '12 at 10:13


text:
6.EE.B

Reason about and solve one-variable equations and inequalities.

Standards

6.EE.B.5 Understand solving an equation or inequality as a process of answering a question: which values...
6.EE.B.6 Use variables to represent numbers and write expressions when solving a real-world or...
6.EE.B.7 Solve real-world and mathematical problems by writing and solving equations of the form $x + p =... 6.EE.B.8 Write an inequality of the form$x > c$or$x < c\$ to represent a constraint or condition...

Tasks aligned to this cluster

Triangular Tables
Busy Day
Which Goes with Which?
All, Some, or None?


text:
statsmodels.duration.survfunc.SurvfuncRight.summary¶

method

SurvfuncRight.summary()[source]

Return a summary of the estimated survival function.

The summary is a datafram containing the unique event times, estimated survival function values, and related quantities.


text:
MCQ | Waves

QUIZ TEST – I

Question:1. The amplitude of resulting wave dues to superposition of y1 = A sin (ωt − kx) &  y2 = A sin (ωt − kx + δ) is

(A) 2A cos δ

(B) 2A tan (δ/2)

(C) A cos δ sin δ

(D) none

Ans: (D)

Question:2.  A sine wave has an amplitude A and wavelength λ . The ratio of particle velocity and the wave velocity is equal to ( 2πA = λ )

(A) ≤ 1

(B) = 1

(C) ≥ 1

(D) data insufficient.

Ans: (A)

Question:3. The equation of a wave pulse moving with a speed 1 m/sec at time t = 0 is given as y = f(x) = 1/(1 + x2) . Its equation at time t = 1 second can be given as

(A) $\displaystyle y = \frac{1}{1+(1+x)^2}$

(B) $\displaystyle y = \frac{1}{1+(1-x)^2}$

(C) $\displaystyle y = \frac{1}{1+(1+x^2)}$

(D) $\displaystyle y = \frac{1}{1+ (\frac{1}{1+x^2}))}$

Ans: (B)

Question:4. The velocity of a transverse wave in a string does not depend on

(A) tension

(B) density of material of string

(D) length of string

Ans: (D)

Question:5. The frequency of a tuning fork with an amplitude A = 1 cm is 250 Hz. The maximum velocity of any particle in air is equal to

(A) 2.5 m/s

(B) 5 π m/s

(C) 3.30 /π m/sec

(D) none of these

Ans: (B)

Question:6. In a resonance column experiment, the first resonance is obtained when the level of the water in tube is 20 cm from the open end. Resonance will also be obtained when the water level is at a distance of

(A) 40 cm from the open end.

(B) 60 cm from the open end.

(C) 80 cm from the open end.

(D) data insufficient.

Ans: (B)

Question:7. A wire of length l having tension T and radius r vibrates with natural frequency f. Another wire of same metal with length 2l having tension 2T and radius 2r will vibrate with natural frequency

(A) f

(B) 2f

(C) 2√2 f

(D) f /2√2

Ans: (d)

Question:8. Under the same conditions of pressure and temperature, the velocity of sound in oxygen and hydrogen gases are vo and vH , then

(A) vH = vo

(B) vH = 4vo

(C) vo = 4 VH

(D) vH = 16 vo

Ans: (B)

Question:9. A tuning for of frequency 600 Hz produces a progressive travelling wave having wave velocity 300 m/s. Two particles of a medium, separated by 1.5 m, vibrate being affected by the wave (A) in phase

(B) in opposite phase.

(C) 45° out of phase.

(D) none of these

Ans: (A)

Question:10. At t = 0 source starts falling under gravity and a detector is projected upwards with a velocity 10 m/s. For the vertical upward motion of detector

(A) apparent frequency received by detector = source frequency.

(B) initially apparent frequency > source frequency and finally less than source frequency.

(C) apparent frequency depends only on the detector velocity.

(D) date insufficient.

Ans: (B)

Question:11. A string is clamped on both ends. Which of the following wave equations is valid for a stationary wave set up on this string ? (Origin is at one end of string.)

(A) y = A sin kx. sin ωt

(B) y = A cos kx sin ωt

(C) y = A cos kx. cos ωt

(D) None of the above.

Ans: (A)

Question:12. A string is hanging from a rigid support. A transverse wave pulse is set up at the bottom. The velocity v of the pulse related to the distance covered by it is given as

(A) v ∝ √x

(B) v ∝ x

(C) v ∝ 1/x

(D) none of these

Ans: (A)

Question:13. The third overtone of a closed organ pipe is equal to the second harmonic of an open organ pipe. Then the ratio of their lengths is equal to

(A) 7/4

(B) 3/5

(C) 3/2

(D) none of these

Ans: (A)

Question:14. Standing waves can be produced in

(A) solid only

(B) liquid only

(C) gases only

(D) all of the above

Ans: (D)

Question:15.  If the temperature of the medium drops by 1 %, the velocity of sound in that medium

(A) increases by 5 %

(B) remains unchanged

(C) decreases by 0.5 %

(D) decreases by 2 %

Ans: (C)

Question:16. The velocity of sound through a diatomic gaseous medium of molecular weight M at 0°C is approximately.

(A) √(R / M)

(B) √(3R / M)

(C) √(382R / M)

(D) √(273R / M)

Ans: (C)

Question:17. The amplitude of a wave disturbance propagating in the positive x direction is given by $\displaystyle y = \frac{1}{(1+x^2)}$ at time t = 0 and by $\displaystyle y = \frac{1}{(1+(x-2)^2)}$ at time t = 2 seconds where x and y are in meters. The shape of the wave disturbance does not change during the propagation. The velocity of the wave is

(A) 0.5 m/s

(B) 1 m/sec

(C) 2 m/s

(D) 1.5 m/sec

Ans: (B)

Question:18. A wave is represented by the equation y = [A sin {10πx + 15π t + (π/3)}] where x is in meters and t is in seconds. The expression represents

(A) A wave travelling in positive x-direction with a velocity 1.5 m/s.

(B) A wave travelling in negative x-direction with a velocity 1.5 m/s.

(C) A wave travelling in the negative x-direction having a wavelength 2 m.

(D) A wave travelling in positive x-direction having a wavelength 2 m.

Ans: (B)

Question:19. A transverse wave is given by A sin(ωt − αx) where ω and α are constants. The ratio of wave velocity to maximum particle velocity is

(A) αA

(B) 1/αA

(C) 1

(D) none of the above.

Ans: (B)

Question:20. Two blocks, each of mass m, are connected by a massless thread Y and A represent Young’s modulus and cross sectional area of wire respectively. The strain developed in the thread is

(A) $\displaystyle y = \frac{mg(1 + sin\theta)}{2 Y A}$

(B) $\displaystyle y = \frac{m g}{ Y A}$

(C) $\displaystyle y = \frac{m g sin\theta}{ Y A}$

(D) $\displaystyle y = \frac{2 m g}{ Y A}$


text:
Geodesics on the sphere

In a few days I will be giving a talk to (smart) high-school students on a topic which includes a brief overview on the notions of curvature and of gedesic lines. As an example, I will discuss flight paths and, more in general, shortest length arcs on the 2-dimensional sphere.

Here is my question: let $$p,q$$ be distinct (and, for simplicity, non-antipodal) points on the sphere $$S^2$$. Is there an elementary proof of the fact that the unique shortest path between $$p$$ and $$q$$ is the shortest arc on the unique great circle containing $$p$$ and $$q$$?

By ''elementary'' I mean a (possibly not completely formal, but somehow convinving) argument that could be understood by a high-school student. For example, did ancient Greeks know that the geodesics of the sphere are great circles, and, if so, how did they prove'' this fact?

Here is the best explanation that came into my mind so far. Let $$\gamma\colon [0,1]\to S^2$$ be any path, and let $$\gamma(t_0)=a$$. The best planar approximation of $$\gamma$$ around $$a$$ is given by the orthogonal projection of $$\gamma$$ on the tangent space $$T_a S^2$$. This projection is a straight line (that is, a length-miminizing path on $$T_a S^2$$) only if $$\gamma$$ is supported in a great circle containing $$a$$. This sounds quite convincing, and with a bit of differential geometry it is not difficult to turn this argument into a proof. However, I was wondering if I could do better...

• Of course there are isometries fixing the planes. Fixed point sets of isometries are totally geodesic. But I like the following physics argument: If you take an elastic band and connect it from $p$ to $q$, it will go to a great circle. It takes effort to move it away from a great circle, i.e. it is not the shortest. – Thomas Rot Jan 14 at 13:00
• If one knows that geodesics are unique (at least locally), then by using the reflection with respect to the plane containing the starting point of the geodesic and its initial speed (this plane obviously contains the centre of the sphere) it would be easy to prove the statement. However, I would like not to rely on uniqueness. Also observe that my definition of geodesic is based on minimization of length. – Roberto Frigerio Jan 14 at 13:14
• @FrancescoPolizzi: conformal maps don't preserve geodesics. They mix them up with curves of constant curvature. – Ben McKay Jan 14 at 14:27
• Here’s a perverse plausibility argument. Given a curve on the sphere, let its fan be the union of all radial segments that end on the curve. The length of the curve can be measured as the twice the area of the fan. The area of the fan is minimized by “unfolding” it or flattening it as much as possible. – Deane Yang Jan 14 at 16:23
• If this is not the main aim of your talk and you just want them to understand the idea, why not try "proof by experiment" and hand them a ball and a piece of string or a rubber band? It's not mathematical, but it will be more memorable than a proof which might be to advanced for their current level. – mlk Jan 14 at 20:09

Imagine that the sphere is the surface of the earth, and that an earthquake happens at the south pole, point $$p$$. A seismic wave goes out from the point $$p$$, so that the wave front is a circle of latitude moving at a constant speed $$c$$. At some point this circle/wave will meet the point $$q$$ (let's say that $$q$$ lies within the continent of Antarctica, so the seismic wave will reach it). For any path from $$p$$ to $$q$$, flying along this path at the speed $$c$$, starting at the moment of the earthquake, will never overtake the wave, and will keep apace of it only if it coincides with the line of longitude going through $$q$$.

Underlying this physical explanation is Gauss's lemma, from which a rigorous calculus argument can be extracted. But maybe this intuition is enough to convince a high schooler, or at least make them think about why it might be true?

• I would believe that simply assuming $p'$ is at the pole as a first case is enough to see that you follow a meridian to get to $q'$. You can then rotate $p' \to p$ and $q' \to q$ anywhere on the sphere. – student Jan 17 at 12:19

Ancients did not know line integrals, but they understood that the length of the curve is the limit of the lengths of inscribed broken lines. To prove the inequality for broken lines (piecewise-great-circles), it is enough to prove that a side of a small spherical triangle is less than the sum of two other sides. This can be done using spherical trigonometry, which the ancients did know.

The spherical cosine theorem says $$\cos a=\cos b\cos c+\sin b\sin c\cos A\,$$ where $$a,b,c$$ are sides and $$A$$ is the angle between $$b$$ and $$c$$. Estimating this from below (cosine is decreasing) we replace $$\cos A$$ by $$-1$$, and obtain $$\cos a\geq \cos b\cos c-\sin b\sin c=\cos(b+c),$$ therefore $$a\leq b+c.$$

Spherical cosine theorem is ancient but relatively late (after the epoch of Archimedes), but one can also give an elementary-geometric proof of the triangle inequality for small spherical triangles. All one needs is that three angles at the vertex of a 3-faced cone (or pyramid) satisfy the triangle inequality.

If the high school students are taking, or have already taken, Calculus I, and know how to differentiate, then I would use the notion of acceleration. I would start by asking: what is a straight line in the plane? One way to characterize it is as the solution to the differential equation $$\gamma''(t)=0,$$ where $$\gamma(t)=(x(t), y(t))$$. In particular, they can check that $$(a+bt, c+dt)$$ yields a solution. So lines are curves with no acceleration. Thus on the sphere the analogue of a line segment, or the straightest curve possible, may be regarded as a curve with no acceleration in the sphere. More precisely, we want the acceleration vector to be orthogonal to the sphere, or parallel to the position vector. So we need to solve the equation $$\gamma''(t)||\gamma(t),$$ where $$\gamma(t)=(x(t), y(t), z(t))$$, and one quickly checks that great circles, e.g., $$(\cos(t),\sin(t), 0)$$ yield a solution. To drive the point home further, I would draw a straight line on a piece of paper, and then roll the paper to a cylinder to show them that the acceleration vector remains orthogonal to the surface under isometric deformations. More precisely, straight lines turn into helices $$(\cos(t), \sin(t), h\,t)$$, where again one can quickly check that $$\gamma''(t)$$ is parallel to $$(\cos(t), \sin(t),0)$$ and thus is orthogonal to the cylinder.

I always found the following argument the most appealing intuitively.

Imagine the sphere is hollow. Obviously the shortest path from $$p$$ to $$q$$ is a straight line through the interior of the sphere. (I always imagine just ramming a knitting needle through it.)

Now of course the question was not to find the shortest path in general, but the shortest path over the surface of the sphere. It is intuitively clear that this will be the path that follows the 'true' shortest path as closely as possible. How do we find that?

Well just imagine a light source at the center at the sphere. The path on the surface closest to the straight line through the interior is then simply the shadow of that straight line.

What remains to be seen is that this is part of a great circle but that is easy: the 'shadow path' is the intersection of the sphere with the plane defined by the three points $$p$$, $$q$$ and center.

• The keyword here is "gnomonic projection" (projection from the center of the sphere, which maps great circles to straight lines). But I think you're begging the question: why is it true that the geodesic on the sphere is the (gnomonic) projection of the 3D line segment? – Gro-Tsen Jan 15 at 13:17
• I admit that in my formulation it is a bit shady (haha) but the idea is that for every point $t$ on the 'true' shortest path we want to find the point $t'$ on the surface that is that is closest to it. Of course the claim that these points $t'$ together do indeed form the shortest path over the surface is itself highly questionable and perhaps needs its own justification, but it seems that your question was why $t'$ is the gnomonic projection of $t$. This is rather straightforward. (ctd in next comment) – Vincent Jan 15 at 15:59
• Imagine the shortest path from $t$ to $t'$, which is of course a straight line. When it hits the surface it must do so perpendicularly, because if it hits the surface under an angle this means we could have found a point $t''$ on the surface that is closer to $t$ than $t'$ was. This is an obvious property of spheres and we are only talking about $t$ and its projection $t'$ ignoring the rest of the path $pq$ so no circular reasoning here (haha). But if line $tt'$ hits the surface perpendicularly, this means that extending this line in the other direction it will pass through the center. – Vincent Jan 15 at 16:02

Even knowing differential geometry I've always found it unsatisfying that all of the simplest proofs of this fact that I know still involve a reasonable amount of calculus. Spheres and great circles feel like fairly elementary geometric objects, so surely there is an elementary geometric argument. The following argument still involves a bunch of calculus under the hood, but it feels a bit more geometric to me.

Take a smooth curve on the sphere, and choose a small segment $$C$$ of the curve. There is a unique plane $$P$$ which contains the endpoints $$p$$ and $$q$$ of the segment and the center of the sphere. Consider the curve $$C'$$ which is obtained by reflecting $$C$$ across this plane and reversing the direction of travel, so that it goes from $$q$$ to $$p$$; note that $$C'$$ has the same length as $$C$$ because reflecting the sphere across a plane through the origin preserves distances.

Now consider the concatenation of $$C$$ and $$C'$$ - this is a small loop based at $$p$$. Among all loops based at $$p$$ which pass through $$q$$, the shortest one is the loop which encloses the smallest area, and an area of zero is attained if $$C'$$ is just $$C$$ with reversed orientation, i.e. $$C$$ is fixed under reflection about the plane $$P$$. But the curves fixed under reflection about a plane through the origin are precisely the great circles, so we're done.

The main idea here is that geodesics are fixed point sets of isometries, if there are lots of isometries. This idea can be used to find geodesics in homogeneous spaces, for instance.

I think what is needed is a heuristic that would then motivate any further derivation you care to present. For this situation, I would offer curvature. If I have two points and a circular arc going through them, the arc length is greatest when the chord is the diameter, and approaches least when the involved circle is as large as can be managed. (Use some basic trig to back this up if needed.) On the plane, a circle passing through two points can have arbitrarily large radius. On a sphere, the largest circle is a great circle or geodesic.

You can expand on this heuristic to point out various subtleties, such as definition of arc length or area of a subregion of or on a manifold.

Gerhard "Or Just Keep It Simple" Paseman, 2019.01.14.

It might be instructive to show that great circles are the "straight lines" on a sphere: Start from a point on the northern hemisphere and (imagine to) walk due east until you have returned to your position. Then your your right foot will have walked further than your left one. On the southern hemisphere, it's the other way round. Hence, circles of constant latitude are not straight -- except, of course, the equator.

I think this gives an intuitive geometric understanding of straight lines, and it more or less directly generalises to show that other great circles are straight, and other curves are not. Also, it shows that lines that look straight on a map might not actually be. (You could follow that with a calculus-based approach if the pupils have the required tools.)

You still need to argue that this notion of straightness coincides with "shortest path", but that seems intuitively clear to me.

I don't know that these are correct or not.

1. The simplest way (in practice ) that come to my mind is the following trick. Consider two points $$p$$ and $$q$$ on sphere. connect them by an straight line (a line of powerful magnet) in $$\Bbb R^3$$ then using some iron filings you can see the trace of iron filings on sphere as the shortest arc between these two points.
2. The second way for this purpose, I think stereographic projection from north pole $$N$$ would be helpful. Rotate the sphere such that projections of $$p$$ and $$q$$ under stereographic projection ($$\sigma(p)=p'$$ and $$\sigma(q)=q'$$) together south pole $$S$$ lies in a line (i.e. $$Sp'q'$$). then the inverse image of this line segment represents the shortest arc between $$p$$ and $$q$$ (of course non-antipodality is assumed).

For a few minutes of contrast, I suggest simply drawing the geodesics on a torus of revolution, earliest I know is Bliss 1902. The facts about them can be recovered from Clairaut's relation. The nice part about this example is that you can have one with you, such as an inflated tube used with young children in a swimming pool.

do Carmo does an example, pages 262-263, a geodesic that wraps infinitely often around the neck of a hyperboloid of one sheet, never reaching it. This happens with the torus also.

Your students likely know about Pythagoras theorem. Because of the triangle inequality, we know it is the shortest distance between 2 points in a flat space. It also turn out that it is the metric of a plane surface:

$$ds^2 = dx^2 + dy^2$$

Now for a sphere you can derive the metric equation much the same way. You simply create a small triangle at the surface of the sphere. The length element $$ds^2$$ will be a small arc along the latitude (squared) + a small arc along the longitude (squared). Which gives you this simple equation for the metric on the surface of a sphere:

$$ds^2 = (R \cdot d\theta)^2 + (R \cdot \sin(d\theta) \cdot d\phi)^2$$

Take some old globes to class, and some pushpins and some string.

First step put in tall pushpins at every latitude, wind the string tight around the tops, up off the surface, so it is in 6 or 8 sections. Measure the length of sections, add. Now try to find a shorter path, again up off the surface, with other pins. Don't tell them there's no shorter path, let them figure it out, and explain why.

Now ask whatis the difference between having it up off the surface in segments, and down on the surface in a smooth curve. Will it make a difference to finding shortest path?

Some smartie pants will say you can tunnel through the earth and get the shortest path. That's an opportunity to explain what lateral thinking is and how important it is to the history of science.

Sphere is symmetric under rotation so whatever is true of the longitude great circle will be true of every other great circle.

If you can reduce the problem to that of finding the shortest path between two points at the same longitude, then you can proceed as follows.

Follow some path from your point of origin to your destination at uniform speed $$c.$$ Let $$a$$ be the component of your velocity in the direction in which only your longitude is changing. Then the component of your velocity in the direction in which only your latitude is changing is $$\sqrt{c^2 - a^2} < c.$$ You're not gaining latitude as fast as if you moved along the meridian, but your odometer is still putting distance behind you just as fast.

Draw the straigt line $$L$$ between $$p$$ and $$q$$ and parametrise your path $$\gamma(s)$$ with $$s\in L$$ such that $$s$$ is the orthonormal projection on $$L$$ of $$\gamma(s)$$. (we note $$p$$ the projection on $$L$$ so $$p(\gamma(s))=s$$. Now take your path and rotate it around $$L$$. This draw a cylinder $$C$$ around $$L$$ defined as $$C=\{x\in \mathbb{R}^3:\|x-p(x)\|=\|\gamma(p(x))-p(x)\|\}$$ Then we use that

1. On a cylinder the shorter path is to go straigh. Indeed by symetrie you don't gain anything turning left or right.
2. The cylinder drawn from the geodesic line is the smallest one, Meaning that it is inside any other cylinder that we can construct this way.
3. If a cylinder $$C_1$$ is convex and smaller than another cylinder $$C_2$$. Then the path on $$C_1$$ is smaller than the path on $$C_2$$

EDIT : second proof. Start with $$p$$ and $$q$$ the north and south pole. You can paramerise with the height $$t\in [-1,1]$$, and $$\theta(t)$$ ($$z=t,x=r(t)\cos(\theta(t)),y=r(t)sin(\theta(t))$$), with $$t^2+r(t)^2=1$$. Then by symetrie of revolution the shorter path is to go straight. And therefore $$\theta = c$$ is constant. Then we can conclude using that a smaller path of a geodesic is a geodesic.

There is an elegant way to see geodesics on any surface.

Take a piece of paper in a rectangular shape (with width short enough). The geodesic is drawn by the line in the middle of the rectangle when you can flat at most the rectangle on the surface.

It comes from the fact that by using a rectangle and flatten at most both long edges, you induce a Killing field. The codimension 1 coincides with the fact that the geodesic is of dimension 1.

A couple of answers already propose using the triangle inequality, but I think they both apply it using a calculus argument. There's a simple geometric argument. Let $$C$$ be the shortest path from $$A$$ to $$B$$. Take the point $$P$$ on $$C$$ which is furthest from the great circle $$AB$$. If $$P$$ is not on the great circle then $$AP + PB > AB$$, and since the length of $$C$$ is at least $$AP + PB$$ you have a contradiction.

• I don't see how you get that triangle inequality for points off the great circle. – Ben McKay Jan 15 at 20:04
• @BenMcKay, I don't understand what you're querying. Can you rephrase? – Peter Taylor Jan 15 at 23:26
• I don't understand why, if $P$ is not on the great circle then $AP+PB>AB$. – Ben McKay Jan 16 at 6:53
• @BenMcKay, see Alexandre's answer. – Peter Taylor Jan 16 at 8:35


text:
Warning: This document is for the development version of SANE Standard. The latest version is 1.06.

# 6 Contact Information¶

The SANE standard is discussed and evolved via a mailing list. Anybody with email access to the Internet can automatically join and leave the discussion group by sending mail to the following address.

To subscribe, send a mail with the body “subscribe sane-devel” to the above address.

A complete list of commands supported can be obtained by sending a mail with a subject of “help” to the above address. The mailing list is archived and available through the SANE home page at URL:


text:
# statistics

## An Illustrated Guide to TMLE, Part III: Properties, Theory, and Learning More

The is the third and final post in a three-part series to help beginners and/or visual learners understand Targeted Maximum Likelihood Estimation (TMLE). In this section, I discuss more statistical properties of TMLE, offer a brief explanation for the theory behind TMLE, and provide resources for learning more. Properties of TMLE 📈 To reiterate a point from Parts I and II, a main motivation for TMLE is that it allows the use of machine learning algorithms while still yielding asymptotic properties for inference.

## An Illustrated Guide to TMLE, Part II: The Algorithm

The second post of a three-part series to help beginners and/or visual learners understand Targeted Maximum Likelihood Estimation (TMLE). This section walks through the TMLE algorithm for the mean difference in outcomes for a binary treatment and binary outcome. This post is an expansion of a printable “visual guide” available on my Github. I hope it helps analysts who feel out-of-practice reading mathematical notation follow along with the TMLE algorithm.

## An Illustrated Guide to TMLE, Part I: Introduction and Motivation

The introductory post of a three-part series to help beginners and/or visual learners understand Targeted Maximum Likelihood Estimation (TMLE). This section contains a brief overview of the targeted learning framework and motivation for semiparametric estimation methods for inference, including causal inference. Table of Contents This blog post series has three parts: Part I: Motivation TMLE in three sentences 🎯 An Analyst’s Motivation for Learning TMLE 👩🏼💻 Is TMLE Causal Inference?

## Become a Superlearner! An Illustrated Guide to Superlearning

Why use one machine learning algorithm when you could use all of them?! This post contains a step-by-step walkthrough of how to build a superlearner prediction algorithm in R. HTML Image as link A Visual Guide… Over the winter, I read Targeted Learning by Mark van der Laan and Sherri Rose. This “visual guide” I made for Chapter 3: Superlearning by Rose, van der Laan, and Eric Polley is a condensed version of the following tutorial.

## Rethinking Conditional and Iterated Expectations with Linear Regression Models

An “aha!” moment: the day I realized I should rethink all the probability theorems using linear regressions. TL;DR You can a regress an outcome on a grouping variable plus any other variable(s) and the unadjusted and adjusted group means will be identical. We can see this in a simple example using the palmerpenguins data: #remotes::install_github("allisonhorst/palmerpenguins") library(palmerpenguins) library(tidyverse) library(gt) # use complete cases for simplicity penguins <- drop_na(penguins) penguins %>% # fit a linear regression for bill length given bill depth and species # make a new column containing the fitted values for bill length mutate(preds = predict(lm(bill_length_mm ~ bill_depth_mm + species, data = .

## A Condensed Key for A Visual Guide to Targeted Maximum Likelihood Estimation (TMLE)

A condensed key for my corresponding TMLE tutorial blog post. Initial set up Estimand of interest: $ATE = \Psi = E_W[\mathrm{E}[Y|A=1,\mathbf{W}] - \mathrm{E}[Y|A=0,\mathbf{W}]]$ Step 1: Estimate the Outcome First, estimate the expected value of the outcome using treatment and confounders as predictors. $Q(A,\mathbf{W}) = \mathrm{E}[Y|A,\mathbf{W}]$ Then use that fit to obtain estimates of the expected outcome under varying three different treatment conditions:


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Now showing items 115-134 of 240

• #### New bounds for bilinear Calderón-Zygmund operators and applications

(2016-11-25)
In this work we extend Lacey’s domination theorem to prove the pointwise control of bilinear Calderón–Zygmund operators with Dini–continuous kernel by sparse operators. The precise bounds are carefully tracked following ...
• #### Nondegeneracy of heteroclinic orbits for a class of potentials on the plane

(2021)
In the scalar case, the nondegeneracy of heteroclinic orbits is a well-known property, commonly used in problems involving nonlinear elliptic, parabolic or hyperbolic P.D.E. On the other hand, Schatzman proved that in the ...
• #### Nonlocal discrete diffusion equations and the fractional discrete Laplacian, regularity and applications

(2018)
The analysis of nonlocal discrete equations driven by fractional powers of the discrete Laplacian on a mesh of size $h>0$ $(-\Delta_h)^su=f,$ for $u,f:\Z_h\to\R$, $0<s<1$, is performed. The pointwise nonlocal ...
• #### A note on generalized Fujii-Wilson conditions and BMO spaces

(2020-07-01)
In this note we generalize the definition of the Fujii-Wilson condition providing quantitative characterizations of some interesting classes of weights, such as A∞, A∞weak and Cp, in terms of BMO type spaces suited to them. ...
• #### A note on generalized Poincaré-type inequalities with applications to weighted improved Poincaré-type inequalities

(2020)
The main result of this paper supports a conjecture by C. P\'erez and E. Rela about the properties of the weight appearing in their recent self-improving result of generalized inequalities of Poincar\'e-type in the Euclidean ...
• #### A note on the off-diagonal Muckenhoupt-Wheeden conjecture

(2016-07-01)
We obtain the off-diagonal Muckenhoupt-Wheeden conjecture for Calderón-Zygmund operators. Namely, given $1 < p < q < \infty$ and a pair of weights $(u; v)$, if the Hardy-Littlewood maximal function satisfies the following ...
• #### Numerical approximation of the fractional Laplacian on R using orthogonal families

(2020-12-01)
In this paper, using well-known complex variable techniques, we compute explicitly, in terms of the F12 Gaussian hypergeometric function, the one-dimensional fractional Laplacian of the complex Higgins functions, the complex ...
• #### The observational limit of wave packets with noisy measurements

(2019)
The authors consider the problem of recovering an observable from certain measurements containing random errors. The observable is given by a pseudodifferential operator while the random errors are generated by a Gaussian ...
• #### On a hyperbolic system arising in liquid crystal modelling

(2017-11)
We consider a model of liquid crystals, based on a nonlinear hyperbolic system of differential equations, that represents an inviscid version of the model proposed by Qian and Sheng. A new concept of dissipative solution ...
• #### On a probabilistic model for martensitic avalanches incorporating mechanical compatibility

(2021-07-01)
Building on the work by Ball et al (2015 MATEC Web of Conf. 33 02008), Cesana and Hambly (2018 A probabilistic model for interfaces in a martensitic phase transition arXiv:1810.04380), Torrents et al (2017 Phys. Rev. E 95 ...
• #### On a sharp Poincare-type inequality on the 2-sphere and its application in micromagnetics

(2019-08-21)
The main aim of this note is to prove a sharp Poincaré-type inequality for vector-valued functions on $\mathbb{S}^2$ that naturally emerges in the context of micromagnetics of spherical thin films.
• #### On Bloom type estimates for iterated commutators of fractional integrals

(2018-04)
In this paper we provide quantitative Bloom type estimates for iterated commutators of fractional integrals improving and extending results from [15]. We give new proofs for those inequalities relying upon a new sparse ...
• #### On extension problem, trace hardy and Hardy’s inequalities for some fractional Laplacians

(2019-09)
We obtain generalised trace Hardy inequalities for fractional powers of general operators given by sums of squares of vector fields. Such inequalities are derived by means of particular solutions of an extended equation ...
• #### On pointwise and weighted estimates for commutators of Calderón-Zygmund operators

(2017)
In recent years, it has been well understood that a Calderón-Zygmund operator T is pointwise controlled by a finite number of dyadic operators of a very simple structure (called the sparse operators). We obtain a similar ...
• #### On sums involving Fourier coefficients of Maass forms for SL(3,Z)

(2016-09-10)
We derive a truncated Voronoi identity for rationally additively twisted sums of Fourier coefficients of Maass forms for SL(3,Z), and as an application obtain a pointwise estimate and a second moment estimate for the sums ...
• #### On the absolute divergence of Fourier series in the infinite dimensional torus

(2019-03-22)
• #### On the controllability of Partial Differential Equations involving non-local terms and singular potentials

(2016-12-12)
In this thesis, we investigate controllability and observability properties of Partial Differential Equations describing various phenomena appearing in several fields of the applied sciences such as elasticity theory, ...
• #### On the energy of critical solutions of the binormal flow

(2019-07-20)
The binormal flow is a model for the dynamics of a vortex filament in a 3-D inviscid incompressible fluid. The flow is also related with the classical continuous Heisen- berg model in ferromagnetism, and the 1-D cubic Schr ...
• #### On the energy of critical solutions of the binormal flow

(2020-07-02)
The binormal flow is a model for the dynamics of a vortex filament in a 3-D inviscid incompressible fluid. The flow is also related with the classical continuous Heisenberg model in ferromagnetism, and the 1-D cubic ...


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# Meaning of multiplication

### Problem

The desks are in 3 rows, with 8 desks in each row.
How can we find out how many desks there are?


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# Imposing the topology of open rays in $\Bbb R$

After having received Brian M. Scott's permission (see comments in the selected answer) I am integrating his suggestions with my own solutions to form a complete answer to the questions apperaing below.

Let $\mathscr{T}$ be the collection of subsets of $\Bbb R$ consisting of $\emptyset, \Bbb R$ and the rays of the form $(r, \infty)$, where $r \in \Bbb R$.

$(a)$ Exhibit that this, indeed, is a topology on $\Bbb R$.

Proof: Given any finite number of open sets of the form above, $\exists$ a maximal $r_n$. The set $(r_n, \infty)$ is the intersection of these finite sets. Let $\{U_i\}_{i \in I}$ be an arbitrary family of open sets. Let $r_{\min} = \inf\{r_i\}$. Then $(r_{\min}, \infty)$ is the union of the $U_i$ and clearly it is of the desired form to be an element of the topology. If $r_{\min} = -\infty$ then the union of the $U_i$ is the entire $\Bbb R$.

$(b)$ Show that it fails to be a topology if $r \in \Bbb Q$.

I don't have the full answer to this. I think that there should be a problem in the union of arbitrarily many open sets. Any help with a counterexample will be very helpful.

Answer the following questions. Is $(\Bbb R, \mathscr{T})$:

$(c)$ $T_1$?

No. Let $x_1 \neq x_2 \in \Bbb R$ and assume without loss of generality that $x_1 < x_2$. Then any open set containing $x_1$ will be of the form $x_1 - \epsilon, \infty$ for some $\epsilon > 0$ and will encessarily contain $x_2$.

$(d)$ Hausdorff

No. Being Hausdorff (or $T_2$) would imply that it is $T_1$, a contradiction to (a).

$(e)$ metrizable

No. If there was a metric, the metric space would have to be Hausdorff, contradicting (b).

$(f)$ second - countable

$(g)$ compact

No. There exists no finite cover of this space. Suppose we are given a finite cover of this space. Then there exists a minimal $r$ such that $(r_{\min}, \infty)$ covers the entire space. Of ,course, this is only possible if $(r_{\min}, \infty) = \Bbb R$.

$(h)$ locally compact

Yes. We need to exhibit that any point has a compact neighbourhood. To this end, fix $x \in \Bbb R$. Let $(r, \infty)$ be a neighbourhood of $x$ and let $\{U_i\}_{i \in I}$ be an open cover. Then, there must exist $q \in \Bbb R$ such that $q \ge r$ so that $(q, \infty)$ is a set of the open cover. Clearly, taking $(q, \infty)$ as the subcover completes the proof.

$(i)$ connected

In part $(j)$ we prove that $(\Bbb R, \mathcal{T})$ is path-wise connected, hence connected.

$(j)$ path-wise connected

$\Bbb R$ is convex. Given $x, y \in \Bbb R$, the path $f: [0,1] \to \Bbb R$ given by $f(t) = (1-t)x + ty$ is continuous and $f(0) = x$ and $f(1) = y$. Since path-wise connected implies connected we also answered $(i)$.

What is the closure of $\{1\}$ in $(\Bbb R, \mathcal{T})$?

Proof: Since the closure is the smallest closed set containing $\{1\}$, it is clear that it is $(-\infty, 1]$

Any suggestions, corrections, hints and any help, in general, will be tremendously appreciated! Any stylistic improvements in the formatting of the question are also greatly encouraged!

-

(b) Consider $\bigcup\{(r,\to):r\in\Bbb Q\text{ and }r>\sqrt2\}$.

(f) Consider the set $\mathscr{B}=\{(r,\to):r\in\Bbb Q\}$: is this a base for $\mathscr{T}$?

(g) A small correction: $\{\Bbb R\}$ is a finite cover of the space, as is any finite set of open sets that includes $\Bbb R$. However, $\{(r,\to):r\in\Bbb R\}$ is certainly an open cover with no finite subcover.

(h) Fix $r\in\Bbb R$, and let $\mathscr{U}=\{(q,\to):q>r\}$; $\mathscr{U}$ is an open cover of $(r,\to)$ with no finite subcover. Thus, no member of $\mathscr{T}$ is compact. However, if your definition of neighborhood does not require it to be open, then for any $x\in\Bbb R$ the set $[x-1,\to)$ is a nbhd (though not an open nbhd) of $x$ that is compact: if $x-1\in(r,\to)\in\mathscr{T}$, then $(r,\to)\supseteq[x-1,\to)$, so every open cover of $[x-1,\to)$ has a one-element subcover.

(j) You should say why $f$ is continuous, though I grant that it’s pretty easy.

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Thank you very much for your most helpful comments! – Orest Xherija Mar 15 '13 at 23:55
@Orest: You’re very welcome. (You’d already made a pretty good start.) – Brian M. Scott Mar 16 '13 at 0:00
Dear Brian, I would like to ask your permission to integrate your suggestions and hints into a complete answer in my original post. May I do so? – Orest Xherija Mar 24 '13 at 22:25
@Orest: Be my guest. – Brian M. Scott Mar 24 '13 at 22:32

Hint: For $(b)$, use the fact that $\Bbb Q$ is dense in $\Bbb R$. Maybe consider $\sqrt{2}+1/n\to\sqrt{2}$ as $n\to\infty$. For $(g)$, I'm not sure why you say there doesn't exist a finite cover; $\Bbb R$ itself is a perfectly good finite open cover. For $(j)$, I think you need more argument. It seems like you are jumping topologies, which is okay, but you have to argue that the function you've defined is actually continuous via the definition for continuous in topology.

I'm not sure about $(f)$, so I'll leave that for someone else, but hopefully this gets you started on the other parts. I think everything I didn't mention was okay, though.

-


text:
# What are 'Symmetric Agents' vs. 'Asymmetric Agents'?

Here is a contextual example from Landry et. al. 2006 :

We concentrate on an economy with n symmetric agents who derive utility from consuming a numeraire good, yi, a public good at level G, and (possibly) from their own contribution bi to the public good. Each agent faces a budget constraint yi bi w and derives ex post utility according to

(1) $$Ui = u(yi) + \Theta h(G) + \gamma f(bi)$$

• $\lambda$ yields $\lambda$, and $\Lambda$ gives you $\Lambda$ – caverac Nov 24 '18 at 19:59


text:
# Unique limit of a convergent filter

1. Feb 2, 2008

### andytoh

Question: Prove that if any convergent filter on a space X converges to a unique limit, then X is Hausdorff.

I think the solution in my textbook is faulty. It says "Suppose X is not Hausdorff. Let Nx and Ny be the collection of all neighbourhoods of x and y, respectively. Then Nx U Ny generates a filter that converges to both x and y, a contradiction."

The proof likes right except we must verify that Nx U Ny is actually a filter base, i.e. the intersection of two members of Nx U Ny must contain a member of Nx U Ny. If A belongs to Nx and B belongs to Ny, then indeed AnB is nonempty since by assumption X is not Hausdorff. But in order for AnB to contain a member of Nx U Ny, it must contain either x or y. But if neither x nor y belongs to AnB, where is the contradiction? What am I missing? I stated that if neither x nor y belongs to AnB, then int(A-intB) and int(B-intA) are disjoint open sets containing x and y, respectively, contradicting the assumption that X is not Hausdorff--but is that really so? What if they exist on the boundary of AnB?

By the way, I looked up the author's definition of neighbourhood of a point p. He defines it to be a set whose interior contains p. Ok, so a neighbourhood, by his defintion, does not have to be open, but I still don't see how that ensures that either x or y belongs to AnB.

Last edited: Feb 2, 2008
2. Feb 2, 2008

### andytoh

Ok, I fixed it. The textbook is wrong!

The correct filter base that does the job is Nx U Ny U W, where W is the collection of all sets of the form AUB, where A is from Nx and B is from Ny. I proved rigorously that this is indeed a filter base and generates a filter that converges to both x and y--the desired contradiction.

Mistakes like this in a textbook are unforgiveable. I lost at least 2 hours of wasted time over this mistake.

3. Aug 20, 2011

### Landau

No, the author was correct: he did not mean that Nx \cup Ny is a filter base, but a filter subbase.

A filter subbase is a collection subsets, such that every finite intersection is non-empty. Then the collection of those sets which contain such a finite intersection forms a filter, it is the filter generated by the subbase.

Said differently, starting from a filter subbase, the collection of its finite intersections forms a filter base. The filter generated by this base, equals the filter I just described, and both the subbase ande the base are said to generate this filter.

This is analoguous to topology: any collection serves as a subbase; the collection of its finite intersections forms a base, and the collection of all unions of the base is a topology; both the subbase and the base are said to generate this topology.

(But not every base arises in this way from a subbase.)

Returning to the subbase Nx \cup Ny, it is not so hard to see that the collection of its finite intersections consists precisely of things of the form Ux \cap Uy, with Ux resp. Uy a neighborhood of x resp. y. So this is a base (also easy to check directly) containing Nx \cup Ny, so the filter it generates converges to both x and y.

4. Aug 20, 2011

### micromass

Staff Emeritus
Indeed, so if you come across the sentence that "something generates a filter", then this only means that this something has the finite intersection property.
If something is a base, then this is usually specifically mentioned.


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# X = ?

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04 Nov 2009, 13:54
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I'm having some trouble with the formula...
Attachment:

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04 Nov 2009, 14:32
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Hi,

try the following approach:

1. Divide both sides by $$(2-sqrt5)$$.

Results in $$x=-\frac{1}{(2-sqrt5)}$$

Skim the answers. There is no any that match. So, modify further

2. Multiply two part of fraction by $$(2+sqrt5)$$ and get

$$x=-\frac{(2+sqrt5)}{(2^2-sqrt5^2)}=-\frac{(2+sqrt5)}{(4-5)}=-\frac{(2+sqrt5)}{(-1)}=2+sqrt5$$

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11 Nov 2009, 07:09
Any known roots formula that is useful in these types of questions?
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11 Nov 2009, 08:10
mendelay wrote:
I'm having some trouble with the formula...
Attachment:
untitled.JPG

Hi I dont know about the formula........but for a quick method just plugin the answer choice and see which answer choice result in the desired value i.e x= -1 in this case..

(2+root 5)*(2-root5)
= 2(2-root5)+root5(2-root5)
=4-2root5+2root5-5
= -1 (root5 gets canceled)

So option A is the correct answer choice
I think this is the quickest method that you can even apply during the real test.
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11 Nov 2009, 08:18
mendelay wrote:
Any known roots formula that is useful in these types of questions?

a tip, whenever i see the square root on the bottom always use the same numbers but change the sign

for example: 1/(2 - sqrt3)

multiply by the same numbers but different sign
(2 + sqrt 3) (1) / (2 - sqrt3)(2+sqrt 3)
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23 Oct 2011, 08:07
In case of surds, try to multiply with the conjugate of the given no. If a+sqrt(b) is given multiply with a-sqrt(b) and vice-versa.

That will lead to the formula (a^2)-(b^2) and it can be solved further from here.
As above, 2-sqrt5 is there, multiply by 2+sqrt5. (Just change the sign and multiply). This helps to eliminate the sqrt part.
Re: X = ?   [#permalink] 23 Oct 2011, 08:07
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text:
# Text with spaces in units using siunitx

I used to (a month or two ago) be able to make my units with some arbitrary text, with spaces, which helped a lot since I use a lot of different units and labels in my data, and I have no need to define special cases for each and every one.

For instance,

\SI{1}{liter soln}

would produce "1 liter soln"

However I recently updated all my packages and apparently this does not work anymore. I now get all the spaces in the units removed, so "1 litersoln".

Can someone confirm that siunitx changed recently to enforce always using properly declared units in \SI{}{}? And if so, is there some easy way to work around? To be clear I DO NOT want to define unique qualifiers and units for each case because I would literally have dozens in my file, each used once.

• You can define a common configuration file for siunitx that can contain all of those definitions, and gets loaded each time you use siunitx. See this comment thread for how to create such a file: tex.stackexchange.com/questions/27614/… (although I keep searching for this, maybe I should ask a question about it...) – darthbith Jan 28 '15 at 18:37
• That would be a waste of time, over all my documents there would literally be hundreds of different defintions, not to mention having to refactor all these documents. Nightmare. Is there some kind of "text mode" for units in siunitx? – William Kennerly Jan 28 '15 at 18:40
• There is, you can use \sisetup{mode=text} in your preamble to force siunitx to typeset everything in text mode instead of math mode. Beware if you have different fonts for your text and math though, it may not look quite right :-) Out of curiosity, why do you have so many different unit combinations? – darthbith Jan 28 '15 at 18:45
• Thanks darthbith, that worked perfectly. All the other official declared units work just fine too. I'm a chemist and include the chemical formula along with each quantity... so I have different "units" for "liters soln", "liters methane" "liters gas" "liters nitrogen" ...etc, etc. – William Kennerly Jan 28 '15 at 18:56

Unit input in siunitx should in general not include spaces as these are not part of the unit itself. If you really want a space, make an explicit one

\documentclass{article}
\usepackage{siunitx}

\begin{document}
\si{litre\ soln}
\end{document}

(Input should as far as possible not depend on math/text mode or should force the appropriate mode.)

That said, litre is not a unit symbol and siunitx is not intended for dealing the the names of units. As such, a more typical piece of input might read

\documentclass{article}
\usepackage{siunitx}
\sisetup{qualifier-mode = space}
\begin{document}
\si{\litre\of{soln}}
\end{document}

where it is not necessary to have a separate qualifier defined for each type. (Of course, you should not actually have qualifiers in units, as the BIPM make clear!) Done in this way you can define \litre to give 'litre' if you wish.

• Thanks for giving the "real" answer to this question. I only wonder: What do you mean by "Input... should force the appropriate mode"? Do you mean we should put \si{\text{liter soln}} and leave the mode setting alone? – darthbith Jan 28 '15 at 20:51
• @darthbith In general the input in the document body should work irrespective of math/text mode selection. That's why for example for anything that requires math mode I recommend using \ensuremath. The trick is to 'protect' as little input as possible (it can affect font choice): here, there's no reason 'litre' or 'soln' to be forced into text mode, so I'd prefer a solution which avoids using \text. – Joseph Wright Jan 28 '15 at 20:54
• Thanks! Two more follow ups: 1) Does setting mode=text set the input into a \text{}, equivalent to the user using that macro in the input (should I ask this as a new question)? 2) As I found out, using \text{} may also affect line breaking, most especially in the range and list macros (although in the case of this question, one may not want a line break between the unit and qualifier) – darthbith Jan 28 '15 at 21:28
• @darthbith On the line breaking, it's a technical limitation of the way I've implemented the list/range stuff. I have a plan for v3, but in any case it doesn't affect the 'usual' cases. mode=text does indeed (broadly) put everything inside \text. – Joseph Wright Jan 28 '15 at 21:35

You can use the mode=text option to get the spaces to be respected.

\documentclass{article}
\usepackage{siunitx}
\begin{document}
Default: \SI{1}{liter soln}

\sisetup{mode=text}
\verb|mode=text|:  \SI{1}{liter soln}
\end{document}