I’m waiting on my flight at the airport, and I love math.

Count Calculus·Jan 1, 2019·12 votes·10,972 views

Ask me any (math) problem! Not the millenium problems, though, if I had the solutions to those I wouldn’t be waiting at an airport.

63 Comments

Bob the ToastrJan 1, 20193 votes

Is mayonnaise an instrument?

Count CalculusJan 1, 20197 votes

[{quoted}](name=Bob the Toastr,realm=NA,application-id=yrc23zHg,discussion-id=8nfTLEQi,comment-id=0005,timestamp=2019-01-01T23:00:29.624+0000)

Is mayonnaise an instrument?

No, Patrick, mayonnaise is not an instrument.

...

Horseradish isn’t an instrument either.

Count CalculusJan 1, 20192 votes

[{quoted}](name=Count Calculus,realm=NA,application-id=yrc23zHg,discussion-id=8nfTLEQi,comment-id=,timestamp=2019-01-01T21:22:26.676+0000)

Ask me any (math) problem! Not the millenium problems, though, if I had the solutions to those I wouldn’t be waiting at an airport.

Edit: this is not the person I meant to quote xD

KythersJan 2, 20192 votes

how do i get a gf

Count CalculusJan 2, 20191 votes

{quoted}

how do i get a gf

With the power of math!

5 VJan 1, 20191 votes

Quote my comment @anyone. I want to get notifications when this thread gets a comment be it a Question or Answer. I also like this poster a lot.

S V Jan 1, 20191 votes

Forgot to ask, how did you best learn math? Teacher? Books? Videos? Brute Force in working through a mass amount of problems? (Also don't remember if I asked this) What's the use of quadruple integrals? I understand the other integrals are similar to 1d 2d 3d but 4 integrals? That seems like a bit much.

[{quoted}](name=5 V,realm=NA,application-id=yrc23zHg,discussion-id=8nfTLEQi,comment-id=0000,timestamp=2019-01-01T21:25:43.415+0000)

Quote

Count CalculusJan 1, 20193 votes

[{quoted}](name=S V ,realm=NA,application-id=yrc23zHg,discussion-id=8nfTLEQi,comment-id=00000000,timestamp=2019-01-01T21:32:06.912+0000)

Forgot to ask, how did you best learn math? Teacher? Books? Videos? Brute Force in working through a mass amount of problems? (Also don't remember if I asked this) What's the use of quadruple integrals? I understand the other integrals are similar to 1d 2d 3d but 4 integrals? That seems like a bit much.

I’ve learned math through a number of methods, but the one that’s worked best for me is application on homework and exams. I learn best through examples since I’m really good at breaking apart reasoning and putting it back together (it’s how I learned to program in python).

As for quadruple integrals, you may not have seen one explicitly, but consider applications in physics. An important note to consider is that, if your function has some dimension (for instance, space, time, mass, etc.), then integrating with respect to that dimension increases the power of that dimension. If you integrate, say, position with respect to space, you now have two dimensions of space, whereas if you integrate with respect to time, you now have dimensions of space times time. So, while a quadruple space integral might seem useless in our 3D world, a triple space integral followed by a time integral might be (I don’t think it is, but there might be other quantities that are).

Count CalculusJan 1, 20192 votes

[{quoted}](name=5 V,realm=NA,application-id=yrc23zHg,discussion-id=8nfTLEQi,comment-id=0000,timestamp=2019-01-01T21:25:43.415+0000)

Quote my comment @anyone. I want to get notifications when this thread gets a comment be it a Question or Answer. I also like this poster a lot.

Responded to SV (at least I think you’re different people).

Count CalculusJan 1, 20192 votes

[{quoted}](name=5 V,realm=NA,application-id=yrc23zHg,discussion-id=8nfTLEQi,comment-id=0000,timestamp=2019-01-01T21:25:43.415+0000)

You’ve just been quoted ;)

Count CalculusJan 1, 20193 votes

[{quoted}](name=5 V,realm=NA,application-id=yrc23zHg,discussion-id=8nfTLEQi,comment-id=0000,timestamp=2019-01-01T21:25:43.415+0000)

Quote my comment @anyone. I want to get notifications when this thread gets a comment be it a Question or Answer. I also like this poster a lot.

Someone asked a math question, so here’s your ping.

MonstermushmushJan 2, 20191 votes

[{quoted}](name=5 V,realm=NA,application-id=yrc23zHg,discussion-id=8nfTLEQi,comment-id=0000,timestamp=2019-01-01T21:25:43.415+0000)

Idk how to quote. Anyways, what is the point of the imaginary number (i)? It doesn't exist in real life.

Count CalculusJan 2, 20191 votes

{quoted}

Idk how to quote. Anyways, what is the point of the imaginary number (i)? It doesn't exist in real life.

Hoooo boy. Well, there's a LOT of interesting things that involve i. Quantum mechanics is a pretty big one, since the time evolution operator in the Schrodinger picture looks like e^{-iEt/hbar}, where E is the energy of an energy eigenstate and hbar is Planck's constant divided by 2*pi.

Another cool thing to note is that a + b*i is a vector in the complex plane, meaning the "number" i can be used to define a vector space. However, I'm going to choose my personal favorite application:

Claim: e^{i*pi} = -1

Proof: First, recall the taylor expansion of e^x as SUM{n=0,infty} (x^n)/(n!). Now recall the taylor expansions of sine and cosine:

sin(x) = SUM{n=0,infty} (-1)^(n)*x^(2n+1)/(2n+1)! cos(x) = SUM{n=0,infty} (-1)^(n)*x^(2n)/(2n)!

If you don't know what a Taylor Series is, it's basically an infinite sum that exactly equals the function in question, though it can be truncated to estimate the function to arbitrary accuracy

Note that, if we add the Taylor expansion of cos(x) and sin(x) together, we ALMOST get the Taylor Series for e^x. However, we're off by some minus signs. Using the "number" i, it's pretty easy to check that:

e^(ix) = cos(x) + i*sin(x) (sub in ix for x in the Taylor expansion of e^x and see what happens).

Therefore, e^(i*pi) = cos(pi) + i*sin(pi) = -1 + 0 = -1.

Corollary: ln(-1) = i*pi

MonstermushmushJan 2, 20191 votes

[{quoted}](name=Count Calculus,realm=NA,application-id=yrc23zHg,discussion-id=8nfTLEQi,comment-id=000000050000,timestamp=2019-01-02T20:10:12.735+0000)

Hoooo boy. Well, there's a LOT of interesting things that involve i. Quantum mechanics is a pretty big one, since the time evolution operator in the Schrodinger picture looks like e^{-iEt/hbar}, where E is the energy of an energy eigenstate and hbar is Planck's constant divided by 2*pi.

Another cool thing to note is that a + b*i is a vector in the complex plane, meaning the "number" i can be used to define a vector space. However, I'm going to choose my personal favorite application:

Claim: e^{i*pi} = -1

Proof: First, recall the taylor expansion of e^x as SUM{n=0,infty} (x^n)/(n!). Now recall the taylor expansions of sine and cosine:

sin(x) = SUM{n=0,infty} (-1)^(n)*x^(2n+1)/(2n+1)! cos(x) = SUM{n=0,infty} (-1)^(n)*x^(2n)/(2n)!

If you don't know what a Taylor Series is, it's basically an infinite sum that exactly equals the function in question, though it can be truncated to estimate the function to arbitrary accuracy

Note that, if we add the Taylor expansion of cos(x) and sin(x) together, we ALMOST get the Taylor Series for e^x. However, we're off by some minus signs. Using the "number" i, it's pretty easy to check that:

e^(ix) = cos(x) + i*sin(x) (sub in ix for x in the Taylor expansion of e^x and see what happens).

Therefore, e^(i*pi) = cos(pi) + i*sin(pi) = -1 + 0 = -1.

Corollary: ln(-1) = i*pi

I don't get any of this. I'm 15 and I'm learning Algebra 2.

Count CalculusJan 2, 20191 votes

{quoted}

I don't get any of this. I'm 15 and I'm learning Algebra 2.

Well, the only thing we need to understand here is a Taylor Series. Consider this series:

1 + 2 + 3 + 4 + 5 + ...

This is an infinite series that doesn't converge to any value (it has no solution). It can be expressed as:

S = SUM{n=1,infty} n

In other words, start at n = 1 and add n + (n+1) + (n+2) + ... A TAYLOR series uses calculus to find a polynomial that represents a function. The exponential function can be represented by:

e^x = 1 + x + (x^2)/2 + (x^3)/6 + ... Here's a plot to compare them: https://gyazo.com/828205b36a0ba249a8680aefd36dbe29

And so on. If you're still with me, try writing out a few terms of the sine and cosine series representations I gave you. Then, multiply the sine series by i and add it to the cosine series. Next, take the terms of the e^x series above, sub in (ix) for x, and compare :).

MonstermushmushJan 3, 20191 votes

[{quoted}](name=Count Calculus,realm=NA,application-id=yrc23zHg,discussion-id=8nfTLEQi,comment-id=00000005000000000000,timestamp=2019-01-02T23:25:04.559+0000)

Well, the only thing we need to understand here is a Taylor Series. Consider this series:

1 + 2 + 3 + 4 + 5 + ...

This is an infinite series that doesn't converge to any value (it has no solution). It can be expressed as:

S = SUM{n=1,infty} n

In other words, start at n = 1 and add n + (n+1) + (n+2) + ... A TAYLOR series uses calculus to find a polynomial that represents a function. The exponential function can be represented by:

e^x = 1 + x + (x^2)/2 + (x^3)/6 + ... Here's a plot to compare them: https://gyazo.com/828205b36a0ba249a8680aefd36dbe29

And so on. If you're still with me, try writing out a few terms of the sine and cosine series representations I gave you. Then, multiply the sine series by i and add it to the cosine series. Next, take the terms of the e^x series above, sub in (ix) for x, and compare :).

No. me brane hurt. Me dum. Is y i play dis game.

KFCeytronJan 2, 20191 votes

[{quoted}](name=5 V,realm=NA,application-id=yrc23zHg,discussion-id=8nfTLEQi,comment-id=0000,timestamp=2019-01-01T21:25:43.415+0000)

Quote my comment @anyone. I want to get notifications when this thread gets a comment be it a Question or Answer. I also like this poster a lot.

(ping)

Count CalculusJan 2, 20191 votes

[{quoted}](name=5 V,realm=NA,application-id=yrc23zHg,discussion-id=8nfTLEQi,comment-id=0000,timestamp=2019-01-01T21:25:43.415+0000)

Quote my comment @anyone. I want to get notifications when this thread gets a comment be it a Question or Answer. I also like this poster a lot.

SURPRISE! My phone couldn't access the second page yesterday, so here's more responses.

Count CalculusJan 2, 20191 votes

[{quoted}](name=5 V,realm=NA,application-id=yrc23zHg,discussion-id=8nfTLEQi,comment-id=0000,timestamp=2019-01-01T21:25:43.415+0000)

Quote my comment @anyone. I want to get notifications when this thread gets a comment be it a Question or Answer. I also like this poster a lot.

Boop

Pile Of PillowsJan 1, 20191 votes

You might have seen this one, but it is one of my favorites :D Suppose you have a sphere of any volume but 0, and create 4 randomly placed points on the surface of the sphere. If you draw lines from the points to the inside of the sphere creating a tetrahedron, what are the exact chances that the tetrahedron made would encapsulate the center of the sphere? From the Putnam, took me about 3 weeks to solve for myself, but I'm not some savant lol. If you can get a piece of paper, drawing this out really helps. Good luck!

Pile Of PillowsJan 1, 20192 votes

[{quoted}](name=5 V,realm=NA,application-id=yrc23zHg,discussion-id=8nfTLEQi,comment-id=0000,timestamp=2019-01-01T21:25:43.415+0000)

Quote my comment @anyone. I want to get notifications when this thread gets a comment be it a Question or Answer. I also like this poster a lot.

Count CalculusJan 1, 20191 votes

[{quoted}](name=Pile Of Pillows,realm=NA,application-id=yrc23zHg,discussion-id=8nfTLEQi,comment-id=0003,timestamp=2019-01-01T22:47:22.129+0000)

You might have seen this one, but it is one of my favorites :D Suppose you have a sphere of any volume but 0, and create 4 randomly placed points on the surface of the sphere. If you draw lines from the points to the inside of the sphere creating a tetrahedron, what are the exact chances that the tetrahedron made would encapsulate the center of the sphere? From the Putnam, took me about 3 weeks to solve for myself, but I'm not some savant lol. If you can get a piece of paper, drawing this out really helps. Good luck!

I don’t have paper and I don’t think I’ll be able to answer your question, but I do have a fun Putnam question of my own! Choose any odd number and double it. Make a list of 1 through that number (ex: choose 3, take list 1, 2, 3, 4, 5, 6). Choose any two numbers from the list and take the absolute value of their difference. Remove the two numbers you chose from the list and replace them with the difference you calculated. Repeat this until you only have one number remaining.

Claim: the number will always be odd. Provide the proof :)

Demonstration: 1, 2, 3, 4, 5, 6 Choose 1,2: 2-1 = 1, new list is 1, 3, 4, 5, 6 Choose 5,6: 6-5 = 1, new list is 1, 1, 3, 4 Choose 1,1: 1-1 = 0, new list is 0, 3, 4 Choose 0,3: 3-0 = 3, new list is 3, 4 Choose 3,4: 4-3 = 1, remaining number is 1.

Pile Of PillowsJan 1, 20191 votes

I sat down and tried to model the equation algebraically, but I didn't get very far before I got stumped. One thing I find interesting is that with the nature of odd numbers, the list you give will (if I'm not mistaken) give an odd number if every number in the sequence is added up. What I'm still a bit stumped about is how you can choose any number and make the ABSOLUTE VALUE of the difference, making it always be positive. I'm going to model a few more and see what other properties I can find. Certainly an interesting problem. Regarding my problem, 3blue 1brown has a great video on the problem, and explains it better than I ever could, and gives how one would find the solution. I would recommend watching it :D https://www.youtube.com/watch?v=OkmNXy7er84

Count CalculusJan 2, 20191 votes

[{quoted}](name=Pile Of Pillows,realm=NA,application-id=yrc23zHg,discussion-id=8nfTLEQi,comment-id=000300010000,timestamp=2019-01-01T23:38:48.187+0000)

I sat down and tried to model the equation algebraically, but I didn't get very far before I got stumped. One thing I find interesting is that with the nature of odd numbers, the list you give will (if I'm not mistaken) give an odd number if every number in the sequence is added up. What I'm still a bit stumped about is how you can choose any number and make the ABSOLUTE VALUE of the difference, making it always be positive. I'm going to model a few more and see what other properties I can find. Certainly an interesting problem. Regarding my problem, 3blue 1brown has a great video on the problem, and explains it better than I ever could, and gives how one would find the solution. I would recommend watching it :D https://www.youtube.com/watch?v=OkmNXy7er84

I'm not sure what you mean by "What I'm still a bit stumped about is how you can choose any number and make the ABSOLUTE VALUE of the difference, making it always be positive", but if I'm understanding your question correctly: we take the absolute value of the difference to guarantee we add a positive number to the list. I'm actually not sure if that's even necessary, based on the solution I found, but essentially, what we do is:

Take two numbers
Subtract one from the other
If the resulting value is negative, multiply by negative 1. If it's not, leave it alone

That's how we take the absolute value.

Also, HINT: Consider the number of odd numbers in the list :)

Ćount KledulaJan 1, 20191 votes

what's nine plus ten?

Count CalculusJan 1, 20194 votes

[{quoted}](name=Ćount Kledula,realm=EUW,application-id=yrc23zHg,discussion-id=8nfTLEQi,comment-id=0004,timestamp=2019-01-01T22:53:05.907+0000)

what's nine plus ten?

We can make this problem more complicated by converting to base 2. In base 2, 9 = 2^0 + 2^3, written in binary as 00001001. Ten is given by 00001010. Added together, this is 00010011. This is 2^0 + 2^1 + 2^4 = 16 + 2 + 1 = 19. This is how your computer would have calculated it.

Ćount KledulaJan 2, 20191 votes

[{quoted}](name=Count Calculus,realm=NA,application-id=yrc23zHg,discussion-id=8nfTLEQi,comment-id=00040000,timestamp=2019-01-01T23:03:08.772+0000)

We can make this problem more complicated by converting to base 2. In base 2, 9 = 2^0 + 2^3, written in binary as 00001001. Ten is given by 00001010. Added together, this is 00010011. This is 2^0 + 2^1 + 2^4 = 16 + 2 + 1 = 19. This is how your computer would have calculated it.

fucking right, count.

######im a count too, how's your day, brother?

Count CalculusJan 2, 20191 votes

[{quoted}](name=Ćount Kledula,realm=EUW,application-id=yrc23zHg,discussion-id=8nfTLEQi,comment-id=000400000000,timestamp=2019-01-02T00:27:27.393+0000)

fucking right, count.

######im a count too, how's your day, brother?

It was pretty good. I actually couldn't access this page of responses on my phone for some reason, so I'm responding to them now.

KFCeytronJan 2, 20191 votes

[{quoted}](name=Count Calculus,realm=NA,application-id=yrc23zHg,discussion-id=8nfTLEQi,comment-id=000000000000,timestamp=2019-01-01T21:43:15.060+0000)

(it’s how I learned to program in python)

GitHub/etc. link?

[{quoted}](name=Count Calculus,realm=NA,application-id=yrc23zHg,discussion-id=8nfTLEQi,comment-id=00040000,timestamp=2019-01-01T23:03:08.772+0000)

We can make this problem more complicated by converting to base 2. In base 2, 9 = 2^0 + 2^3, written in binary as 00001001. Ten is given by 00001010. Added together, this is 00010011. This is 2^0 + 2^1 + 2^4 = 16 + 2 + 1 = 19. This is how your computer would have calculated it.

How about 0.1 + 0.2? :^)

Count CalculusJan 2, 20191 votes

{quoted}

GitHub/etc. link?

How about 0.1 + 0.2? :^)

DAMMIT, I FORGOT HOW TO DO FLOATING POINT CALCULATIONS! AAAAAAGH!

Also, I don't really have a GitHub link, if that's what you're asking for here. I took a 1 hour "programming for physicists" course at my university, where we were taught basic python. In the class, we had assignments to turn in, so I would look up the packages I needed to understand and look at examples they had. I basically copied enough examples to learn the language, and then started winging everything else xD.

KFCeytronJan 2, 20191 votes

[{quoted}](name=Count Calculus,realm=NA,application-id=yrc23zHg,discussion-id=8nfTLEQi,comment-id=0004000000010000,timestamp=2019-01-02T20:11:29.466+0000)

DAMMIT, I FORGOT HOW TO DO FLOATING POINT CALCULATIONS! AAAAAAGH!

Also, I don't really have a GitHub link, if that's what you're asking for here. I took a 1 hour "programming for physicists" course at my university, where we were taught basic python. In the class, we had assignments to turn in, so I would look up the packages I needed to understand and look at examples they had. I basically copied enough examples to learn the language, and then started winging everything else xD.

Just tell me whether it's 0.3 or not. :^)

I meant some sample of programs you've written, that sort of thing. Do you only write specific, one-off scripts for whatever physics problem you're solving at the moment? Not that there's anything wrong with that, of course. I mean, my dad programs all day for work, and yet has no GitHub or other outside programming presence; it's not a hobby for him. He actually just learned git a few weeks ago so he can do some basic version control on his local machine and hopefully never again have to debug obfuscated source code after his boss accidentally saves over a release version of his code. :)

Count CalculusJan 2, 20191 votes

{quoted}

Just tell me whether it's 0.3 or not. :^)

I meant some sample of programs you've written, that sort of thing. Do you only write specific, one-off scripts for whatever physics problem you're solving at the moment? Not that there's anything wrong with that, of course. I mean, my dad programs all day for work, and yet has no GitHub or other outside programming presence; it's not a hobby for him. He actually just learned git a few weeks ago so he can do some basic version control on his local machine and hopefully never again have to debug obfuscated source code after his boss accidentally saves over a release version of his code. :)

Yeah, it's 0.3. If you PM me your e-mail, I can send you some sample codes I've written. Be warned: none of them are commented (except maybe a bit).

EDIT: I found something to remind me of how to convert to binary:

0.1*2 = 0.2: 0 0.2*2 = 0.4: 0 0.4*2 = 0.8: 0 0.8*2 = 1.6: 1 0.6*2 = 1.2: 1 0.2*2 = 0.4: 0 0.4*2 = 0.8: 0 0.8*2 = 1.6: 1 (we've reached 8 bits, but this could continue infinitely, suggesting that 0.1 can't be exactly represented in binary).

0.2*2 = 0.4: 0 0.4*2 = 0.8: 0 0.8*2 = 1.6: 1 0.6*2 = 1.2: 1 0.2*2 = 0.4: 0 0.4*2 = 0.8: 0 0.8*2 = 1.6: 1 0.6*2 = 1.2: 1

Therefore, in binary, 0.1 + 0.2 is given by 0.00011001 + 0.00110011 = 0.01001100 = 2^{-2} + 2^{-5} + 2^{-6} = 0.296875, which is almost 0.3, but that's because 0.1 and 0.2 have infinite representations in binary.

KFCeytronJan 3, 20192 votes

[{quoted}](name=Count Calculus,realm=NA,application-id=yrc23zHg,discussion-id=8nfTLEQi,comment-id=000400000001000000000000,timestamp=2019-01-02T21:22:55.468+0000)

Yeah, it's 0.3. If you PM me your e-mail, I can send you some sample codes I've written. Be warned: none of them are commented (except maybe a bit).

EDIT: I found something to remind me of how to convert to binary:

0.1*2 = 0.2: 0 0.2*2 = 0.4: 0 0.4*2 = 0.8: 0 0.8*2 = 1.6: 1 0.6*2 = 1.2: 1 0.2*2 = 0.4: 0 0.4*2 = 0.8: 0 0.8*2 = 1.6: 1 (we've reached 8 bits, but this could continue infinitely, suggesting that 0.1 can't be exactly represented in binary).

0.2*2 = 0.4: 0 0.4*2 = 0.8: 0 0.8*2 = 1.6: 1 0.6*2 = 1.2: 1 0.2*2 = 0.4: 0 0.4*2 = 0.8: 0 0.8*2 = 1.6: 1 0.6*2 = 1.2: 1

Therefore, in binary, 0.1 + 0.2 is given by 0.00011001 + 0.00110011 = 0.01001100 = 2^{-2} + 2^{-5} + 2^{-6} = 0.296875, which is almost 0.3, but that's because 0.1 and 0.2 have infinite representations in binary.

That's not actually how your computer does it. :) There's a fast, inaccurate way and a slow, accurate way, and the fast way is used most often, due to being much faster (in my small tests, about 3x faster for simple addition of two saved numbers and 40x faster for creating and adding two new values).

Also, obligatory SMBC:

http://www.smbc-comics.com/comics/20130605.png

LigseoJan 2, 20191 votes

If you still want some: here is mine:

You have 9 pennies Each are visually identical One is a fake The fake one is either lighter or heavier

You have a Balance scale to check them

Can you identify the fake one with only 3 uses of the balance?

Pile Of PillowsJan 2, 20191 votes

You also have a marker :D

Count CalculusJan 2, 20191 votes

{quoted}

If you still want some: here is mine:

You have 9 pennies Each are visually identical One is a fake The fake one is either lighter or heavier

You have a Balance scale to check them

Can you identify the fake one with only 3 uses of the balance?

UUUUUGH I just spent an hour and a half trying to find the answer. I remember solving this once, but I forgot what the solution was.

BladeSingularityJan 2, 20191 votes

{quoted}

If you still want some: here is mine:

You have 9 pennies Each are visually identical One is a fake The fake one is either lighter or heavier

You have a Balance scale to check them

Can you identify the fake one with only 3 uses of the balance?

Here's my answer:

Divide the pennies into 3 groups of 3.

First use: weigh one group of 3 against another group of three. If the two groups have the same weight, put them aside, the fake is in the third group. Second use (first result is equal): weigh one penny from the third group against another one from that group. If their weights are equal, it's the last one. If not, mark the two pennies based on which one was heavier, and then ... Third use: weigh one of the two pennies you just weighed against the last one. If they're equal, it's the last one. If not, check the marking. If one of the pennies was lighter both times or heavier both times, it's that one.

Now, if with your first pass, the weights weren't equal, you need to mark the groups based on which one was lighter or heavier. Second use (first result is unbalanced): Take one of the pennies from the heavier group, two of the pennies from the lighter group, and weigh them against the last penny in the lighter group with two from the third group (which we know are all normal weight). If this result is not balanced (the side with that one heavier coin is heavier), then ... Third use (unbalanced): Weigh either the lighter coin or the heavier coin against a coin from the third group. If the results are consistent (still lighter/heavier than the neutral), that's the one. If those two weigh the same, it's the other one, which you didn't weigh this time. Now, if the second use resulted in the side with the heavier coin being lighter, then one of the lighter coins is the fake. Third use: Weigh those lighter coins against each other. The fake is whichever one is lighter. If the second use was balanced ... Third use (balanced): Weigh the two coins you set aside from the heavier group against each other. The heavier one of the two is the fake.

Whew ... I think I got that right.

LigseoJan 9, 20191 votes

{quoted}

Forgot to answer, well played

Yordle XayahJan 2, 20191 votes

Not really a question but a math video I found really interesting:

https://youtu.be/DGpwkWhnWAI

Thought I'd share it.

Count CalculusJan 2, 20191 votes

{quoted}

Not really a question but a math video I found really interesting:

https://youtu.be/DGpwkWhnWAI

Thought I'd share it.

I, too, have a math video I love. Thanks for sharing:

https://www.youtube.com/watch?v=_s5RFgd59ao

MiraculumJan 2, 20191 votes

i dunno if youre still answering but like math problem an airplane travels west at 180 km/h and returns east with the jet stream at 300 km/h. what was the average speed in km/h for the whole trip? math homework over break woO

Count CalculusJan 2, 20191 votes

{quoted}

i dunno if youre still answering but like math problem an airplane travels west at 180 km/h and returns east with the jet stream at 300 km/h. what was the average speed in km/h for the whole trip? math homework over break woO

There's a nice and easy formula for doing this, but let's derive it! A quick google search will result in this image:

https://www.onlinemath4all.com/images/averagetspeed1.png

Let's prove it:

Proof: First note that average speed is defined as total distance divided by total time. Suppose the aircraft travels a distance d in both directions (which I assume we are able to do since it, if it doesn't start and end in the same place, this problem becomes much more complicated, and I'm not even sure if you can solve it). Anyways, we have:

S_{avg} = 2d/T

Where T = t_1 + t_2, t_1 is the amount of time spent flying at speed x, and t_2 the amount of time spent flying at speed y. Therefore, t_1 = d/x, and t_2 = d/y. So:

S_{avg} = 2d/(d/x + d/y) = 2/(1/x + 1/y) = 2/((x+y)/xy) = 2xy/(x+y).

the shy topJan 2, 20191 votes

If 3x−y=12, what is the value of 8x2y?

A) 212 B) 44 C) 82 D) The value cannot be determined from the information given.

Count CalculusJan 2, 20191 votes

[{quoted}](name=the shy top,realm=OCE,application-id=yrc23zHg,discussion-id=8nfTLEQi,comment-id=000c,timestamp=2019-01-02T10:45:03.347+0000)

If 3x−y=12, what is the value of 8x2y?

A) 212 B) 44 C) 82 D) The value cannot be determined from the information given.

D. You have one equation and two unknowns. 3x - y = 12 gives you information about y in terms of x, but to solve 8 * 2y (I'm assuming your x means multiplication and is not the variable x), you would need to have a second equation that can be used to find a value for x (either as a number, or in terms of y). Then, you would be able to solve for a numerical value of y.

DrCyanideJan 2, 20191 votes

A bit of an odd ball problem to throw at you, but I figure you might enjoy it.

As you probably know, going from a cube with sides of X to a cube with sides of 2*X, the mass doesn't increase 2 times, but rather 8 times. Similarly, going to sides of 0.5*X makes the mass 1/8th.

Lets say you had 2 cubes, one with sides of length X, the other length Y. If you decrease the side of the first cube by N and added that mass to the other cube, what formula would describe how much longer the other cube's sides are now (M)? Can this formula be adapted to two rectangular prisms, assuming the prisms are proportional in their dimensions?

Count CalculusJan 2, 20191 votes

{quoted}

A bit of an odd ball problem to throw at you, but I figure you might enjoy it.

As you probably know, going from a cube with sides of X to a cube with sides of 2*X, the mass doesn't increase 2 times, but rather 8 times. Similarly, going to sides of 0.5*X makes the mass 1/8th.

Lets say you had 2 cubes, one with sides of length X, the other length Y. If you decrease the side of the first cube by N and added that mass to the other cube, what formula would describe how much longer the other cube's sides are now (M)? Can this formula be adapted to two rectangular prisms, assuming the prisms are proportional in their dimensions?

Assume both cubes have the same density (mass per volume). The initial mass of the system is:

rho*(X^3 + Y^3)

Where rho is the density mentioned above. Suppose the new side length of the second cube is Y + M. We seek M:

rho*(Y+M)^3 + rho*(X-N)^3 = rho*(X^3 + Y^3) (total mass doesn't change)

=> M = [X^3 + Y^3 - (X-N)^3]^{1/3} - Y

This formula assumes that the volume of the solid in question can be exactly written as the cube of one of its sides (in other words, it assumes the solid is a cube). A rectangular prism's volume cannot be written in this way, so THIS formula definitely does not apply.

Question: Are the ratios of the sides of the rectangular prism ALWAYS the same, even after adding the mass? While we were forced to add the same length to each side of the cube to keep it a cube, we're able to change the sides of a rectangular prism any way we want (length wise, at least) and it would still be a rectangular prism. A similar technique should be applicable in the case of the prism, but we would need to know that the dimensions of the prisms are kept proportional after adding mass.

DrCyanideJan 2, 20191 votes

[{quoted}](name=Count Calculus,realm=NA,application-id=yrc23zHg,discussion-id=8nfTLEQi,comment-id=00110000,timestamp=2019-01-02T21:46:33.120+0000)

Assume both cubes have the same density (mass per volume). The initial mass of the system is:

rho*(X^3 + Y^3)

Where rho is the density mentioned above. Suppose the new side length of the second cube is Y + M. We seek M:

rho*(Y+M)^3 + rho*(X-N)^3 = rho*(X^3 + Y^3) (total mass doesn't change)

=> M = [X^3 + Y^3 - (X-N)^3]^{1/3} - Y

This formula assumes that the volume of the solid in question can be exactly written as the cube of one of its sides (in other words, it assumes the solid is a cube). A rectangular prism's volume cannot be written in this way, so THIS formula definitely does not apply.

Question: Are the ratios of the sides of the rectangular prism ALWAYS the same, even after adding the mass? While we were forced to add the same length to each side of the cube to keep it a cube, we're able to change the sides of a rectangular prism any way we want (length wise, at least) and it would still be a rectangular prism. A similar technique should be applicable in the case of the prism, but we would need to know that the dimensions of the prisms are kept proportional after adding mass.

For what I was thinking of, the proportions would be roughly humanoid - so about 7x tall, 3x wide, 1x deep. If I downscaled one prism so that it was X inches shorter, how much taller would I be able to make the other prism.

I would want to keep the proportions the same, so if one prism started out 140cm tall, 60cm wide, 20cm deep and they became 70cm shorter I'd expect their dimensions to be 70cm, 30cm, 10cm. What I don't know is if I was to take that missing mass and add it to the other prism what it's new dimensions would be.

Count CalculusJan 2, 20191 votes

{quoted}

For what I was thinking of, the proportions would be roughly humanoid - so about 7x tall, 3x wide, 1x deep. If I downscaled one prism so that it was X inches shorter, how much taller would I be able to make the other prism.

I would want to keep the proportions the same, so if one prism started out 140cm tall, 60cm wide, 20cm deep and they became 70cm shorter I'd expect their dimensions to be 70cm, 30cm, 10cm. What I don't know is if I was to take that missing mass and add it to the other prism what it's new dimensions would be.

If all you did was make the prism 70cm shorter, how are you changing the dimensions of the width and depth? Are you saying to halve every dimensions? I was think you could have dimensions A x B x C and remove N, M, and P, respectively. If the ratios of the prisms are kept the same (in other words, the ratio of their heights is X:Y), and the ratio is kept the same after adding the mass to the other one? Actually, I don't think that would be possible if we did it for all three sides. Actually, if we can only add mass to the second, then the ratios can't be kept constant for any side. At best, we could flip the ratio to be Y:X. The biggest questions that need to be addressed here are:

HOW are we taking mass off the first prism?
WHERE are we putting this mass on the second prism?

In the cube question, the answers were easy:

Same amount taken off each side.
Same amount put on each side.

The question doesn't exactly give us enough information to work with, here.

DrCyanideJan 2, 20191 votes

[{quoted}](name=Count Calculus,realm=NA,application-id=yrc23zHg,discussion-id=8nfTLEQi,comment-id=0011000000000000,timestamp=2019-01-02T23:21:11.795+0000)

If all you did was make the prism 70cm shorter, how are you changing the dimensions of the width and depth? Are you saying to halve every dimensions? I was think you could have dimensions A x B x C and remove N, M, and P, respectively. If the ratios of the prisms are kept the same (in other words, the ratio of their heights is X:Y), and the ratio is kept the same after adding the mass to the other one? Actually, I don't think that would be possible if we did it for all three sides. Actually, if we can only add mass to the second, then the ratios can't be kept constant for any side. At best, we could flip the ratio to be Y:X. The biggest questions that need to be addressed here are:

HOW are we taking mass off the first prism?

WHERE are we putting this mass on the second prism?

In the cube question, the answers were easy:

Same amount taken off each side.

Same amount put on each side.

The question doesn't exactly give us enough information to work with, here.

The idea is to scale down one, while scaling up the other. I thought of it more as a fantasy game idea so it doesn't have a real world equivalent per say, but if you need a visual you can think about 3D printing two figurines to sell. They'll keep the same proportions regardless of if you scale them up or down, but you've got a set amount of filament (mass) to work with, so if you scale one up the other has to be scaled down. Much like selling figurines, the hight is the thing the customer mostly cares about, but if the other dimensions are needed they can be found by applying the known ratio.

I know if I were to make one figurine half as tall I wouldn't have the mass to make the other figuring twice as tall, since I'd have 7/8 starting mass available (1/8 needed for the shrunken figurine) and I'd need 7 times starting mass available to add to the other figurine to double it's size. But I don't know how much taller than original a figure with 15/8ths mass would be.

Of course it might not be shrunk by half, it could be any arbitrary amount. The goal is to know how much the other figurine would be changed.

So HOW: Magic! WHERE: Everywhere, but proportionally

Edit: the figurines scale like Zac does, uniformly.

DrCyanideJan 3, 20191 votes

Posting a potential answer to my own question.

The prism dimensions don't change the principle of doubling/halfing. A 7*3*1 prism has a volume of 21 units, which cubed gets a volume of 168 units. 168 = 7X*3X*1X 168 = (7*3*1)*X^3 168 = 21*X^3 8 = X^3 2 = X, Same as with the ordinary cube. (You can check, 168 = 14*6*2, which is what I originally did.)

Now, knowing that, I can apply the principles to two figurines - both originally 7*3*1

One becomes 4.5*1.5*0.5, or a volume of 3.375. This means it lost 21 - 3.375 = 17.625, which goes over to the other figurine, for a total of 21 + 17.625 = 38.625.

38.625 = 21*X^3 1.8392857142857142 = X^3 1.2252265501940295 = X

So the second figurine can be 1.225*7 units tall, or 8.575 units tall.