SuicideFuel Math thread problem (official)

[Ahnfeltia]

I've read some parts of the book with the aid of Michael Penn's videos, funny twist of fate huh, I unsubscribed and now I'm watching his real analysis videos for help.

Which book if I may ask?

I've recently unsubscribed as most of his content is too advanced for me at this point of time.

Since you're watching Michael Penn's videos now, I guess that means you matured mathematically. Congrats

 

[Fallenleaves]

Which book if I may ask?

Since you're watching Michael Penn's videos now, I guess that means you matured mathematically. Congrats

Introduction to Real Analysis
Textbook by Robert G. Bartle

It was recommended to me by the Math Sorcerer, I still watch the Math Sorcerer though. I watch whoever helps me learn, but thanks for the compliment, I appreciate it.

 

[Fallenleaves]

@trying to ascend Turns out real analysis and calculus go hand in hand jfl

 

[Fallenleaves]

@Ahnfeltia His real analysis videos are great, it's helping me learn about proofs along with the "proofs" part of the book, love him.

 

[Ahnfeltia]

@Ahnfeltia His real analysis videos are great, it's helping me learn about proofs along with the "proofs" part of the book, love him.

Michael Penn you mean?

 

[Fallenleaves]

Michael Penn you mean?

That guy

 

[Ahnfeltia]

That guy

I've never watched his course videos, but the videos on his main channel are pretty OK. How do you like epsilon-delta arguments? Most students hate them.

 

[Fallenleaves]

I've never watched his course videos, but the videos on his main channel are pretty OK. How do you like epsilon-delta arguments? Most students hate them.

Ah I see.

It was hard for me when I first read it. I stopped thinking about it for almost half a year, re-read the book and understood maybe 80% of it. Then I went to watch his videos to hear a verbal explanation of what's going on as well as to see how he constructs the proof.

 

[Ahnfeltia]

It was hard for me when I first read it. I stopped thinking about it for almost half a year, re-read the book and understood maybe 80% of it. Then I went to watch his videos to hear a verbal explanation of what's going on as well as to see how he constructs the proof.

You still haven't said whether you enjoy epsilon-delta arguments

 

[Fallenleaves]

You still haven't said whether you enjoy epsilon-delta arguments

I'm not sure if I enjoy them but overall I feel lighter after some time of self studying math, I will take a massive detour in life soon (don't ask, I'll be offline for a few months) so I'm treasuring the time I get to study on my own accord.

 

[Ahnfeltia]

I will take a massive detour in life soon

Good luck with whatever that entails

 

[Caesercel]

 

[Fallenleaves]

@Ahnfeltia @trying to ascend Should I start insanemaxxing by studying math for up to 10 hours a day? I'm finally starting to fix my sleep schedule.

 

[Ahnfeltia]

@Ahnfeltia @trying to ascend Should I start insanemaxxing by studying math for up to 10 hours a day? I'm finally starting to fix my sleep schedule.

Considering this doesn't sound insane to me, I'm sure you can guess my answer.

 

[Fallenleaves]

Considering this doesn't sound insane to me, I'm sure you can guess my answer.

Took a break today, my head hurts jfl

 

[Ahnfeltia]

Took a break today, my head hurts jfl

Do take breaks. If you try to cram as much math as fast as possible, most of it ain't gonna stick and your efforts will be for naught.

 

[Fallenleaves]

Do take breaks. If you try to cram as much math as fast as possible, most of it ain't gonna stick and your efforts will be for naught.

 

[Ahnfeltia]

I also recommend doing plenty of exercises as opposed to only cramming theory. Actively engage with the material and all that. I'm sure you've heard that spiel before.

 

[trying to ascend]

Find all values of n (n belonging to the natural set), such that sqrt(n) + sqrt(n + 2005) is a natural number

 

[Ahnfeltia]

Find all values of n (n belonging to the natural set), such that sqrt(n) + sqrt(n + 2005) is a natural number

198^2 and 1002^2

 

[GriffithIsInnocent]

Just read Michael Spivak's Calculus book and you'll be fine.

 

[Ahnfeltia]

Find all values of n (n belonging to the natural set), such that sqrt(n) + sqrt(n + 2005) is a natural number

@Fallenleaves To solve this exercise, you need the following lemma:

Let a and b be positive integers. Then sqrt(a) + sqrt(b) is a positive integer if and only if both a and b are squares.

It might be a fun exercise to try and prove this fact.

 

[Fallenleaves]

@Fallenleaves To solve this exercise, you need the following lemma:

Let a and b be positive integers. Then sqrt(a) + sqrt(b) is a positive integer if and only if both a and b are squares.

It might be a fun exercise to try and prove this fact.

I'll pass, my self studying progress has been significantly slowed, I'm nowhere close to actually proving statements.

 

[Ahnfeltia]

I'll pass, my self studying progress has been significantly slowed, I'm nowhere close to actually proving statements.

fair enough

 

[trying to ascend]

198^2 and 1002^2

Correct

 

[trying to ascend]

A certain operation consists of connecting n number of dots in a circumference, such that no intersection between two lines connecting dots is not intercepted by others.

For example: When there are three dots, we have 4 divisions, when we have 4 dots, we have 8 divisions and so on.

Given that, find the number of parts the circle is divided in when there are 20 dots in its circumference.

 

[Ahnfeltia]

such that no intersection between two lines connecting dots is not intercepted by others.

I don't understand what you mean by this

 

[trying to ascend]

I don't understand what you mean by this

There can't be any point, such that three or more lines contain it (as it happens with polygonals with an even number of sides).

You don't have to consider those lines in this problem

 

[Fallenleaves]

There can't be any point, such that three or more lines contain it (as it happens with polygonals with an even number of sides).

You don't have to consider those lines in this problem

Are we still doing this?

 

[trying to ascend]

Are we still doing this?

Yes

 

[Fallenleaves]

Yes

Could you post a question on Calc 2 or Calc 3 (but only including the introductory chapter on vectors)?

 

[Fallenleaves]

Don't post a question on cylinders and quadric surfaces.

 

[Caesercel]

A certain operation consists of connecting n number of dots in a circumference, such that no intersection between two lines connecting dots is not intercepted by others.

For example: When there are three dots, we have 4 divisions, when we have 4 dots, we have 8 divisions and so on.

Given that, find the number of parts the circle is divided in when there are 20 dots in its circumference.

5036?

 

[trying to ascend]

5036?

Correct

 

[Caesercel]

Correct

I googled the formula lol

 

[trying to ascend]

I googled the formula lol

Is this the only problem here you couldn't solve?

 

[Caesercel]

Is this the only problem here you couldn't solve?

Nah there are many I don't even bother trying. This one though, the insight on the regions created by intersection of diagonals. I could never have guessed. I did end up drawing a bunch of diagrams and summoning Satan though

 

[Ahnfeltia]

Nah there are many I don't even bother trying. This one though, the insight on the regions created by intersection of diagonals. I could never have guessed. I did end up drawing a bunch of diagrams and summoning Satan though

For those who are curious, here's an excellent exposition

I remembered seeing this video a long time ago (I just rewatched it)

 

[ShortestManlet_2]

Everyone here solving hard ass maths and my low IQ failure here is sitting

 

[Ahnfeltia]

Everyone here solving hard ass maths and my low IQ failure here is sitting

@Caesercel and I both literally looked up the answer to the latest problem.

 

[Caesercel]

For those who are curious, here's an excellent exposition

I remembered seeing this video a long time ago (I just rewatched it)

I think 3b1b is the biggest math youtuber at this point and its deserved

 

[Ahnfeltia]

I think 3b1b is the biggest math youtuber at this point and its deserved

He's the biggest math youtuber by quite a margin AFAIK. His videos are indeed quite good.

 

[Ahnfeltia]

Let (a,b,c) be a primitive Pythagorean triple. Prove that rad(abc) is at least 30, where rad(n) denotes the radical of the positive integer n -- i.e., the product of the distinct prime divisors of n (with rad(1) = 1).

Bonus (difficult): prove that the only primitive Pythagorean triple for which rad(abc) = 30 is (a,b,c) = (2,3,5).

 

[Ahnfeltia]

Let (a,b,c) be a primitive Pythagorean triple. Prove that rad(abc) is at least 30, where rad(n) denotes the radical of the positive integer n -- i.e., the product of the distinct prime divisors of n (with rad(1) = 1).

Bonus (difficult): prove that the only primitive Pythagorean triple for which rad(abc) = 30 is (a,b,c) = (2,3,5) (3,4,5).

corrected

 

[trying to ascend]

Consider the set of integers, such that A = (1, 2, 3, ... n), with n elements. If we take away a number P from set A, the arithmetic mean becomes 16,4.

Given that, find the absolute value of p - n

 

[Ahnfeltia]

The (arithmetic) mean of A without p is (n(n+1)/2 - p)/(n - 1) = 16.4 = 164/10. Cross multiplying yields that

5n(n+1) - 10p = 164(n - 1) -- i.e., 10p - 10n = 5n^2 - 169n + 164 = (n - 1)(5n - 164)

so 10(n - p) = (n - 1)(164 - 5n). Since p is an element of A, it cannot exceed n, so |p - n| = n - p is nonnegative. Since n is a positive integer, n - 1 is nonnegative, so 164 - 5n must be nonnegative as well -- i.e., n ≤ 164/5 = 32.8. Since n is an integer, n ≤ 32. Moreover, since the mean of the list (2, 3, ..., 30) is 16 < 16.4, n > 30. So either n = 31 or n = 32.

Going back to 10(n - p) = (n - 1)(164 - 5n), we see that 10 must divide (n - 1)(164 - 5n), which it does when n = 31, but not when n = 32, so n = 31, whence |p - n| = n - p = (31 - 1)(164 - 5*31)/10 = 3*(164 - 155) = 3*9 = 27 (meaning p = n - 27 = 31 - 27 = 4).

 

[trying to ascend]

The (arithmetic) mean of A without p is (n(n+1)/2 - p)/(n - 1) = 16.4 = 164/10. Cross multiplying yields that

5n(n+1) - 10p = 164(n - 1) -- i.e., 10p - 10n = 5n^2 - 169n + 164 = (n - 1)(5n - 164)

so 10(n - p) = (n - 1)(164 - 5n). Since p is an element of A, it cannot exceed n, so |p - n| = n - p is nonnegative. Since n is a positive integer, n - 1 is nonnegative, so 164 - 5n must be nonnegative as well -- i.e., n ≤ 164/5 = 32.8. Since n is an integer, n ≤ 32. Moreover, since the mean of the list (2, 3, ..., 30) is 16 < 16.4, n > 30. So either n = 31 or n = 32.

Going back to 10(n - p) = (n - 1)(164 - 5n), we see that 10 must divide (n - 1)(164 - 5n), which it does when n = 31, but not when n = 32, so n = 31, whence |p - n| = n - p = (31 - 1)(164 - 5*31)/10 = 3*(164 - 155) = 3*9 = 27 (meaning p = n - 27 = 31 - 27 = 4).

Correct

 

[incelerated]

avi thief

 

[Orzmund]

The (arithmetic) mean of A without p is (n(n+1)/2 - p)/(n - 1) = 16.4 = 164/10. Cross multiplying yields that

5n(n+1) - 10p = 164(n - 1) -- i.e., 10p - 10n = 5n^2 - 169n + 164 = (n - 1)(5n - 164)

so 10(n - p) = (n - 1)(164 - 5n). Since p is an element of A, it cannot exceed n, so |p - n| = n - p is nonnegative. Since n is a positive integer, n - 1 is nonnegative, so 164 - 5n must be nonnegative as well -- i.e., n ≤ 164/5 = 32.8. Since n is an integer, n ≤ 32. Moreover, since the mean of the list (2, 3, ..., 30) is 16 < 16.4, n > 30. So either n = 31 or n = 32.

Going back to 10(n - p) = (n - 1)(164 - 5n), we see that 10 must divide (n - 1)(164 - 5n), which it does when n = 31, but not when n = 32, so n = 31, whence |p - n| = n - p = (31 - 1)(164 - 5*31)/10 = 3*(164 - 155) = 3*9 = 27 (meaning p = n - 27 = 31 - 27 = 4).

There is no creativity involved in the calculation of values.

 

[Indracel]

Let (a,b,c) be a primitive Pythagorean triple. Prove that rad(abc) is at least 30, where rad(n) denotes the radical of the positive integer n -- i.e., the product of the distinct prime divisors of n (with rad(1) = 1).

Bonus (difficult): prove that the only primitive Pythagorean triple for which rad(abc) = 30 is (a,b,c) = (2,3,5).

Since a, b, and c are coprime by the definition of a primitive Pythagorean triple, WLOG at the very least the only prime divisor of a is 2, b is 3, and c is 5. Every number has a prime divisor btw. Therefore a*b*c=2*3*5=30. Hence rad(abc)=30

Did I get it right