SuicideFuel Math thread problem (official)

[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

 

[Ahnfeltia]

There is no creativity involved in the calculation of values.

I beg to differ. Regarding the problem whose solution you quoted, it's anything but obvious that any n > 32 cannot yield a valid solution, yet all it took to prove this highly nonobvious claim was high school-level math. Moreover, one cannot simply exhaust all those infinitely many options, so some sort of trick is required. How is finding a way to make the seemingly impossible possible with only simple tools not creative?

 

[Ahnfeltia]

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

Your solution is unfortunately flawed. Here are a couple of things you overlooked:

• Just because only one of a, b, and c can have 2 as a prime divisor doesn't necessarily mean that one of them does. They could potentially all be odd. Similarly for 3 and 5. This oversight is the main flaw of your argument. It's incidentally also the bulk of the work.
• Technically not every positive integer has a prime divisor. 1 does not have any.
• Once one has properly deduced that WLOG 2 divides a, 3 divides b, and 5 divides c, then one merely knows that 30 divides abc, whence rad(abc) is at least 30. Your equal signs are unwarranted. In fact, there's no Pythagorean triple wherefor abc = 30.

 

[Ahnfeltia]

you could provide a method of proving 2+2=5, but that does not prove that 2+2≠4.

Please do. I'm skeptical.

If we eliminate Euclid's fifth postulate one might hastily conclude that they have stumbled upon impossible shapes, yet these shapes were never impossible, they simply remained undiscovered by man for a great deal of time.

Who is the "one" you speak of here? I suppose hypothetically one could, but could hypothetically also stick a rod up his ass. I fail to see the relevance tbh.

Is there an introduction for complex analysis, calculus, and the varying other forms of mathematics that would at least be comprehensive to someone who only has a high-school education? I have books on complex analysis, problem solving, and other topics but they are all too advanced despite having been advertised for beginners.

Calculus is meant to follow up on high school math, so most introductory calculus textbooks should suffice. Complex analysis is quite a bit more advanced and requires more mathematical maturity. Most complex analysis books presume familiarity with real analysis at the very least.

What does that even mean?

Is every number of n{1,2,3,4,...} being divided by a different number?

Sorry not sure what you're asking here.

Are the three integers representing each side of a supposed right triangle prime numbers? Is that why it is called a "primitive" Pythagorean triple?

No, in a primitive Pythagorean triple no two of the three integral side lenghts are to share any prime factors. That way any Pythagorean triple which is not primitive has all three sides sharing a common factor and can therefore be downscaled (as a triangle) so as to obtain a primitive one. Primitive Pythagorean triples cannot be downscaled, however, making them somehow fundamental.

 

[Ahnfeltia]

https://www.quora.com/How-can-we-make-2+2-5-1
There was an attempt.

2 + 2 ≠ 2 + 2 if 2 + 2 = 4

Yeah, I'm still not convinced.

What do you mean by "downscaled"? Decreased in value(subtracted) or factored?

Sorry for being vague. I meant that every triangle corresponding to a Pythagorean triple which is not primitive is similar to a triangle corresponding to a primitive Pythagorean triple. In this case the triangle corresponding to the primitive Pythagorean triple will always be smaller, meaning it's like a downscaled version. No two different primitive Pythagorean triples have similar corresponding triangles.

 

[Ahnfeltia]

I wasn't necessarily trying to say that one of them necessarily had 2 as a prime divisor, I'm saying the smallest product of 3 distinct prime numbers is 30 which means that the radical of abc would have to be at least 30. Since a, b, and c are coprime then each would have to have a distinct prime factor. If you try another primitive Pythagorean triple then you would have to change at least one of those prime factors to be a number greater than 2, 3, or 5, which has to have a product greater than 30.

Right. You're right. My bad. I had actually posted a corrected version of the question in which I ask to prove the statement for any Pythagorean triple (not just primitive ones). As you've correctly identified, the claim's pretty uninteresting for primitive Pythagorean triples.

 

[trying to ascend]

Given the following polynomial p(x) = x^4 - 3x³ + 2x² - x + 4, with a, b, c and d as its roots, find the value of a^5 + b^5 + c^5 + d^5

 

[Ahnfeltia]

Given the following polynomial p(x) = x^4 - 3x³ + 2x² - x + 4, with a, b, c and d as its roots, find the value of a^5 + b^5 + c^5 + d^5

Do you perchance mean a^4 + b^4 + c^4 + d^4?

 

[trying to ascend]

Do you perchance mean a^4 + b^4 + c^4 + d^4?

No

 

[Ahnfeltia]

Given the following polynomial p(x) = x^4 - 3x³ + 2x² - x + 4, with a, b, c and d as its roots, find the value of a^5 + b^5 + c^5 + d^5

We'll find the values of a^n + b^n + c^n + d^n for n = 1 thru 5 in order. But first, some notation. Let e1 = a + b + c + d, e2 = ab + ac + ad + bc + bd + cd, e3 = abc + abd + acd + bcd, and e4 = abcd. By Vieta's formulae, e1 = 3, e2 = 2, e3 = 1, and e4 = 4. Now, a + b + c + d = e1 = 3. Next,

• a^2 + b^2 + c^2 + d^2 = e1^2 - 2e2 = 3^2 - 2*2 = 5
• a^3 + b^3 + c^3 + d^3 = e1^3 - 3e1e2 + 3e3 = 3^3 - 3*3*2 + 3*1 = 27 - 18 + 3 = 12
• a^4 + b^4 + c^4 + d^4 = e1^4 - 4e1^2e2 + 2e2^2 + 4e1e3 - 4e4 = 3^4 - 4*3^2*2 + 2*2^2 + 4*3*1 - 4*4 = 81 - 72 + 8 + 12 - 16 = 13

As for a^5 + b^5 + c^5 + d^5, we know that a^5 = a*a^4 = a(3a^3 - 2a^2 + a - 4) = 3a^4 - 2a^3 + a^2 - 4a as a is zero of p. Similarly for b, c, and d. Ergo, a^5 + b^5 + c^5 + d^5 = 3(a^4 + b^4 + c^4 + d^4) - 2(a^3 + b^3 + c^3 + d^3) + (a^2 + b^2 + c^2 + d^2) - 4(a + b + c + d) = 3*13 - 2*12 + 5 - 3 = 39 - 24 + 2 = 17.

In hindsight using Newton's identities would've been faster...

 

[trying to ascend]

We'll find the values of a^n + b^n + c^n + d^n for n = 1 thru 5 in order. But first, some notation. Let e1 = a + b + c + d, e2 = ab + ac + ad + bc + bd + cd, e3 = abc + abd + acd + bcd, and e4 = abcd. By Vieta's formulae, e1 = 3, e2 = 2, e3 = 1, and e4 = 4. Now, a + b + c + d = e1 = 3. Next,

• a^2 + b^2 + c^2 + d^2 = e1^2 - 2e2 = 3^2 - 2*2 = 5
• a^3 + b^3 + c^3 + d^3 = e1^3 - 3e1e2 + 3e3 = 3^3 - 3*3*2 + 3*1 = 27 - 18 + 3 = 12
• a^4 + b^4 + c^4 + d^4 = e1^4 - 4e1^2e2 + 2e2^2 + 4e1e3 - 4e4 = 3^4 - 4*3^2*2 + 2*2^2 + 4*3*1 - 4*4 = 81 - 72 + 8 + 12 - 16 = 13

As for a^5 + b^5 + c^5 + d^5, we know that a^5 = a*a^4 = a(3a^3 - 2a^2 + a - 4) = 3a^4 - 2a^3 + a^2 - 4a as a is zero of p. Similarly for b, c, and d. Ergo, a^5 + b^5 + c^5 + d^5 = 3(a^4 + b^4 + c^4 + d^4) - 2(a^3 + b^3 + c^3 + d^3) + (a^2 + b^2 + c^2 + d^2) - 4(a + b + c + d) = 3*13 - 2*12 + 5 - 3 = 39 - 24 + 2 = 17.

In hindsight using Newton's identities would've been faster...

Correct

Indeed, the question was meant to be solved using them.