SuicideFuel Math thread problem (official)

[mindlessselfindulge]

sorry but I don't follow

I was using this thing: wikipedia.org/wiki/Fundamental_theorem_of_Galois_theory

It basically describes the rotations and transformations for the roots of a polynomial in a field extension. sqrt(p) for any p corresponds to the 2 element group because it has a degree 2 polynomial. In general the polynomial (x^2 - p_1)(x^2 - p_2) ... (x^2 - p_n) has the galois group (C_2)^n.

 

[Ahnfeltia]

I was using this thing: wikipedia.org/wiki/Fundamental_theorem_of_Galois_theory

It basically describes the rotations and transformations for the roots of a polynomial in a field extension. sqrt(p) for any p corresponds to the 2 element group because it has a degree 2 polynomial. In general the polynomial (x^2 - p_1)(x^2 - p_2) ... (x^2 - p_n) has the galois group (C_2)^n.

Yeah sure, I got that. But a) proving that the Galois group of Q(sqrt(p_1), ... , sqrt(p_m)) is (C_2)^m uses that sqrt(p_1) thru sqrt(p_m) are linearly independent over Q to begin with, which is part of what you're trying to prove, so induction is necessary. And b) how does this yield that "none of these subfields contain Q(sqrt(p_m+1))"?

 

[mindlessselfindulge]

Yeah sure, I got that. But a) proving that the Galois group of Q(sqrt(p_1), ... , sqrt(p_m)) is (C_2)^m uses that sqrt(p_1) thru sqrt(p_m) are linearly independent over Q to begin with, which is part of what you're trying to prove, so induction is necessary. And b) how does this yield that "none of these subfields contain Q(sqrt(p_m+1))"?

I will edit this post with the inductive step later, as for part b, it is basically proving that Q(p_m+1) is not a subfield of any of the subfields, and this goes down to the base level of Q(p_i) =/= Q(p_j)

 

[Ahnfeltia]

as for part b, it is basically proving that Q(p_m+1) is not a subfield of any of the subfields, and this goes down to the base level of Q(p_i) =/= Q(p_j)

Again, I don't think it's that simple. Q(√2) ≠ Q(√6) and Q(√3) ≠ Q(√6) yet Q(√6) is a subfield of Q(√2,√3). Also, what do you even need Galois groups for then?

 

[mindlessselfindulge]

Again, I don't think it's that simple. Q(√2) ≠ Q(√6) and Q(√3) ≠ Q(√6) yet Q(√6) is a subfield of Q(√2,√3). Also, what do you even need Galois groups for then?

In any linear combination of square roots of square free numbers involving p_1 through p_m+1 which equals 0, the value sqrt(p_m+1) can be factored out and written as a fraction of linear combinations of square free numbers involving the other primes.

Example: a * sqrt(2) + b * sqrt(3) + c * sqrt(6) = 0 => sqrt(3) = -a * sqrt(2) / (b + c * sqrt(2))

 

[mindlessselfindulge]

In any linear combination of square roots of square free numbers involving p_1 through p_m+1 which equals 0, the value sqrt(p_m+1) can be factored out and written as a fraction of linear combinations of square free numbers involving the other primes.

Example: a * sqrt(2) + b * sqrt(3) + c * sqrt(6) = 0 => sqrt(3) = -a * sqrt(2) / (b + c * sqrt(2))

Proof of (x^2 - p_1)...(x^2 - p_m) having Galois group (C_2)^m:

The group action corresponding to each p_i is tau_i(sqrt(p_i)) -> -sqrt(p_i). This is C_2, and combining all of these gives (C_2)^m.

I havent studied this stuff formally so sorry if this is not rigorous enough.

 

[Grim_Reaper]

Let f : R^2 → R be continuous and recall that the graph of f is given by the subspace Γ_f := {(x, f (x)) | x ∈ R^2} ⊂ R^3. Prove that Γ_f ⊂ R^3 is closed

 

[mindlessselfindulge]

Let f : R^2 → R be continuous and recall that the graph of f is given by the subspace Γ_f := {(x, f (x)) | x ∈ R^2} ⊂ R^3. Prove that Γ_f ⊂ R^3 is closed

Closed graph theorem - Wikipedia

en.wikipedia.org

 

[Grim_Reaper]

Closed graph theorem - Wikipedia

en.wikipedia.org

Ok now prove it's path-connected

 

[Ahnfeltia]

In any linear combination of square roots of square free numbers involving p_1 through p_m+1 which equals 0, the value sqrt(p_m+1) can be factored out and written as a fraction of linear combinations of square free numbers involving the other primes.

Example: a * sqrt(2) + b * sqrt(3) + c * sqrt(6) = 0 => sqrt(3) = -a * sqrt(2) / (b + c * sqrt(2))

Sure, but again, why does this help you? I don't think this is helpful. I could be wrong but as yet I'm not convinced.

 

[mindlessselfindulge]

Sure, but again, why does this help you? I don't think this is helpful. I could be wrong but as yet I'm not convinced.

It implies that sqrt(3) is in Q(sqrt(2)), or more generally sqrt(p_m+1) is in Q(sqrt(p_1), ... , sqrt(p_m))

 

[Ahnfeltia]

It implies that sqrt(3) is in Q(sqrt(2)), or more generally sqrt(p_m+1) is in Q(sqrt(p_1), ... , sqrt(p_m))

While I don't much love your approach, I reconsidered and it's workable, so I retract my previous statement of it not being helpful. You technically need to split off a case to avoid dividing by 0, but other than that your approach can work if applied inductively. What you now have to prove is that √p_(m+1) ∉ Q(√p_1, ..., √p_m). I really don't think Galois theory is the answer.

 

[Ahnfeltia]

Ok now prove it's path-connected

Let p & q be two points on the graph. Then p = (x, f(x)) and q = (y, f(y)) for some x, y ∈ R^2. Now consider the function
g : [0, 1] ∋ t ↦ ( tx + (1 − t)y, f(tx + (1 − t)y) ).
Since f is continuous, so is g. Since g(0) = p & g(1) = q we have our continuous path.

 

[Ahnfeltia]

Let X be a probability distribution of bounded vectors in R^d, and let X_1, X_2, and X_3 be 3 random independent vectors from X, and r is an arbitrary nonnegative real. Prove 2 * probability(norm(X_1 + X_2) >= r) >= (probability(norm(X_3) >= r))^2

• If r = 0, then the RHS = 0, so there's nothing to prove.
  If r > 0, then we can rescale the probability distribution so that we may WLOG assume that r = 1.
• If P( |X3| ⩾ 1 ) = 0, then the RHS = 0 again, so we'll assume that P( |X3| ⩾ 1 ) > 0. The LHS can now be rewritten as
  2 * P( |X1 + X2| ⩾ 1 ) = 2 * P( |X1 + X2| ⩾ 1 | |X1| ⩾ 1, |X2| ⩾ 1 ) * P( |X1| ⩾ 1, |X2| ⩾ 1 ).
  Since P( |X1| ⩾ 1, |X2| ⩾ 1 ) = P( |X3| ⩾ 1 ) ^ 2 > 0 by independence, all that needs to be proven is that
  P( |X1 + X2| ⩾ 1 | |X1| ⩾ 1, |X2| ⩾ 1 ) ⩾ ½ for all probability distributions X on R^d.
  To that end, we will minimize P( |X1 + X2| ⩾ 1 | |X1| ⩾ 1, |X2| ⩾ 1 ) w.r.t. X.
• |X1 + X2| ⩾ 1 iff
  1 ⩽ |X1 + X2|^2 = |X1|^2 + |X2|^2 + 2<X1, X2> = |X1|^2 + |X2|^2 + 2 * |X1| * |X2| * cos(angle) iff
  cos(angle) ⩾ ( 1 − |X1|^2 − |X2|^2 ) / ( 2 * |X1| * |X2| ).
  Since we are conditioning on the event that |X1| ⩾ 1 and |X2| ⩾ 1, we avoid division by 0.
  If we want to minimize
  P( |X1 + X2| ⩾ 1 | |X1| ⩾ 1, |X2| ⩾ 1 ) = P( cos(angle) ⩾ ( 1 − |X1|^2 − |X2|^2 ) / ( 2 * |X1| * |X2| ) | |X1| ⩾ 1, |X2| ⩾ 1 )
  then we want ( 1 − |X1|^2 − |X2|^2 ) / ( 2 * |X1| * |X2| ) to be as large as possible subject to |X1| ⩾ 1 and |X2| ⩾ 1.
  This is because norms and "angles" are "independent".
  I'll skip the proof, but ( 1 − |X1|^2 − |X2|^2 ) / ( 2 * |X1| * |X2| ) is maximal with value −½ when |X1| = |X2| = 1.
  As such, we need only minimize P( <Z1, Z2> ⩾ −½) for Z1, Z2 ~ Z, where Z is any probability distribution on S^(d−1).
  This is because cos(angle) = <Z1, Z2> when Z1 and Z2 lie on S^(d−1).

That was the easy part. Now onto the the hard part -- minimizing P( <Z1, Z2> ⩾ −½) w.r.t. Z.

• If d = 1, then S^(d−1) = S^0 = {±1}.
  It's easy to verify that the unique minimizer is the Rademacher distribution, which achieves a probability of ½ as expected.
• If d = 2, I don't know how to proceed.

 

[Grim_Reaper]

Let p & q be two points on the graph. Then p = (x, f(x)) and q = (y, f(y)) for some x, y ∈ R^2. Now consider the function
g : [0, 1] ∋ t ↦ ( tx + (1 − t)y, f(tx + (1 − t)y) ).
Since f is continuous, so is g. Since g(0) = p & g(1) = q we have our continuous path.

g(0) here is q while g(1) = p and how does it show that R^2 x {0} is path-connected

 

[Ahnfeltia]

g(0) here is q while g(1) = p

Oh yeah. Oops my bad.

 

[mindlessselfindulge]

While I don't much love your approach, I reconsidered and it's workable, so I retract my previous statement of it not being helpful. You technically need to split off a case to avoid dividing by 0, but other than that your approach can work if applied inductively. What you now have to prove is that √p_(m+1) ∉ Q(√p_1, ..., √p_m). I really don't think Galois theory is the answer.

What is your method? When I saw the problem it immediately reminded me of galois theory.

• If r = 0, then the RHS = 0, so there's nothing to prove.
  If r > 0, then we can rescale the probability distribution so that we may WLOG assume that r = 1.
• If P( |X3| ⩾ 1 ) = 0, then the RHS = 0 again, so we'll assume that P( |X3| ⩾ 1 ) > 0. The LHS can now be rewritten as
  2 * P( |X1 + X2| ⩾ 1 ) = 2 * P( |X1 + X2| ⩾ 1 | |X1| ⩾ 1, |X2| ⩾ 1 ) * P( |X1| ⩾ 1, |X2| ⩾ 1 ).
  Since P( |X1| ⩾ 1, |X2| ⩾ 1 ) = P( |X3| ⩾ 1 ) ^ 2 > 0 by independence, all that needs to be proven is that
  P( |X1 + X2| ⩾ 1 | |X1| ⩾ 1, |X2| ⩾ 1 ) ⩾ ½ for all probability distributions X on R^d.
  To that end, we will minimize P( |X1 + X2| ⩾ 1 | |X1| ⩾ 1, |X2| ⩾ 1 ) w.r.t. X.
• |X1 + X2| ⩾ 1 iff
  1 ⩽ |X1 + X2|^2 = |X1|^2 + |X2|^2 + 2<X1, X2> = |X1|^2 + |X2|^2 + 2 * |X1| * |X2| * cos(angle) iff
  cos(angle) ⩾ ( 1 − |X1|^2 − |X2|^2 ) / ( 2 * |X1| * |X2| ).
  Since we are conditioning on the event that |X1| ⩾ 1 and |X2| ⩾ 1, we avoid division by 0.
  If we want to minimize
  P( |X1 + X2| ⩾ 1 | |X1| ⩾ 1, |X2| ⩾ 1 ) = P( cos(angle) ⩾ ( 1 − |X1|^2 − |X2|^2 ) / ( 2 * |X1| * |X2| ) | |X1| ⩾ 1, |X2| ⩾ 1 )
  then we want ( 1 − |X1|^2 − |X2|^2 ) / ( 2 * |X1| * |X2| ) to be as large as possible subject to |X1| ⩾ 1 and |X2| ⩾ 1.
  This is because norms and "angles" are "independent".
  I'll skip the proof, but ( 1 − |X1|^2 − |X2|^2 ) / ( 2 * |X1| * |X2| ) is maximal with value −½ when |X1| = |X2| = 1.
  As such, we need only minimize P( <Z1, Z2> ⩾ −½) for Z1, Z2 ~ Z, where Z is any probability distribution on S^(d−1).
  This is because cos(angle) = <Z1, Z2> when Z1 and Z2 lie on S^(d−1).

That was the easy part. Now onto the the hard part -- minimizing P( <Z1, Z2> ⩾ −½) w.r.t. Z.

• If d = 1, then S^(d−1) = S^0 = {±1}.
  It's easy to verify that the unique minimizer is the Rademacher distribution, which achieves a probability of ½ as expected.
• If d = 2, I don't know how to proceed.

I will write up a thing in latex for this soon.

 

[Ahnfeltia]

What is your method? When I saw the problem it immediately reminded me of galois theory.

Upon closer inspection my method doesn't work... I feel so dumb right now. Sorry for talking you down while blowing smoke myself.
Anyways, here's a nice write-up regarding the problem:

https://abel.math.harvard.edu/hcmr/issues/2.pdf

 

[mindlessselfindulge]

The point P = (x_1, ... , x_n) is uniformly distributed on the hyperplane x_1 + ... + x_n = 1 with x_i >= 0 for 1 <= i <= n, with n >= 1. Find E(max(x_i)) and E(min(x_i)) in terms of n. For the more general case, find the expected value of the k^th max/min element.

 

[Ahnfeltia]

The point P = (x_1, ... , x_n) is uniformly distributed on the hyperplane x_1 + ... + x_n = 1 with x_i >= 0 for 1 <= i <= n, with n >= 1. Find E(max(x_i)) and E(min(x_i)) in terms of n. For the more general case, find the expected value of the k^th max/min element.

I tried for a bit, but no dice. Is this is what you had in mind?

 

[yung_w3rth3r]

math is gay xD

 

[mindlessselfindulge]

Functional power mean inequality:

https://ibb.co/4KtXF93

 

[Ahnfeltia]

Let (a,b,c) be a Pythagorean triple. Prove that 60 divides abc.

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

 

[mindlessselfindulge]

Let (a,b,c) be a Pythagorean triple. Prove that 60 divides abc.

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

All primitive triples can be represented as (m^2 - n^2, 2mn, m^2 + n^2) by https://en.wikipedia.org/wiki/Pythagorean_triple#Proof_of_Euclid's_formula, with m and n relatively prime, and one of them being even.

This means that mn(m - n)(m + n)(m^2 + n^2) must be divisible by 30. One of m, n is even, so the factor of 2 is taken care of. All possible pairs of values of m and n mod 3 produce something that is 0 mod 3 in m, n, m - n, m + n. In the case of mod 5, the only pairs which dont produce something that is 0 mod 5 in m, n, m - n, m + n, are (1, 2), (1, 3), (2, 4), (3, 4), which all produce 0 mod 5 in m^2 + n^2.

---------------------------------------------------------------------------------------------------------------------------------------------------

Bonus

m and n are relatively prime, and in order to satisfy the conditions, both of them can only have prime factors in {2, 3, 5}. By the Euclidean algorithm, {m, n, m - n, m + n} are pairwise relatively prime (using the condition that exactly 1 of {m, n} is even). In order to distribute the prime factors 2, 3, 5 among these, either n = 1, or m - n = 1, or both, since the 3 prime factors can not be distributed in 4 numbers.

If n = m - n = 1, then m = 2, creating the triple 3, 4, 5.

If n = 1, then the set {m - 1, m, m + 1} must contain 1 power of 2, 1 power of 3, and 1 power of 5. m has to be the power of 2. Since m^2 + 1 is represented as 4^k + 1 for some k, it can not be divisible by 2 or 3, so it must be a power of 5. However, since m^2 - 1 = (m - 1)(m + 1) is divisible by 5, m^2 + 1 is 2 mod 5, which is a contradiction.

If m - n = 1, then the set {m, n, m + n} must contain 1 power of 2, 1 power of 3, and 1 power of 5. This set can be rewritten as {m - 1, m, 2m - 1}. The value m^2 + n^2 can be rewritten as 2m^2 - 2m + 1. Since 2m^2 - 2m + 1 is 1 mod m - 1, 1 mod m, and m mod 2m - 1, it is not divisible by any of 2, 3, 5, so it must be divisible by different prime number(s).

 

[Ahnfeltia]

All primitive triples can be represented as (m^2 - n^2, 2mn, m^2 + n^2) by https://en.wikipedia.org/wiki/Pythagorean_triple#Proof_of_Euclid's_formula, with m and n relatively prime, and one of them being even.

Overkill, but yes, that works.

m and n are relatively prime, and in order to satisfy the conditions, both of them can only have prime factors in {2, 3, 5}. By the Euclidean algorithm, {m, n, m - n, m + n} are pairwise relatively prime (using the condition that exactly 1 of {m, n} is even). In order to distribute the prime factors 2, 3, 5 among these, either n = 1, or m - n = 1, or both, since the 3 prime factors can not be distributed in 4 numbers.

If n = m - n = 1, then m = 2, creating the triple 3, 4, 5.

If n = 1, then the set {m - 1, m, m + 1} must contain 1 power of 2, 1 power of 3, and 1 power of 5. m has to be the power of 2. Since m^2 + 1 is represented as 4^k + 1 for some k, it can not be divisible by 2 or 3, so it must be a power of 5. However, since m^2 - 1 = (m - 1)(m + 1) is divisible by 5, m^2 + 1 is 2 mod 5, which is a contradiction.

If m - n = 1, then the set {m, n, m + n} must contain 1 power of 2, 1 power of 3, and 1 power of 5. This set can be rewritten as {m - 1, m, 2m - 1}. The value m^2 + n^2 can be rewritten as 2m^2 - 2m + 1. Since 2m^2 - 2m + 1 is 1 mod m - 1, 1 mod m, and m mod 2m - 1, it is not divisible by any of 2, 3, 5, so it must be divisible by different prime number(s).

Very nice. Very different than what I had in mind, but probably cleaner.

 

[mindlessselfindulge]

Overkill, but yes, that works.

Very nice. Very different than what I had in mind, but probably cleaner.

What's your solution?

 

[Ahnfeltia]

What's your solution?

Essentially just using modular arithmetic. Since it's more onerous than your solution, I won't bother with the details.

 

[mindlessselfindulge]

Prove that (abc)^2 + a^2 + b^2 + c^2 + 2 >= 2(ab + bc + ca) for real numbers a, b, c

 

[mindlessselfindulge]

Prove that (abc)^2 + a^2 + b^2 + c^2 + 2 >= 2(ab + bc + ca) for real numbers a, b, c

Restrict a, b, c to positive reals since changing all of their signs to positive only increases the RHS and if any of them are 0 the problem is trivial.

AM-GM on (abc)^2 + 1 + 1 gives the new inequality:

a^2 + b^2 + c^2 + 3 (abc)^(2/3) >= 2(ab + bc + ca)

Since this is homogeneous with degree 2, let abc = 1. Then:

3 >= 2/a + 2/b + 2/c - a^2 - b^2 - c^2

Let a = e^x, b = e^y, c = e^z with x+y+z=0

The function f(x) = 2e^(-x) - e^(2x) has an inflection point at -ln(2) / 3

Spamming Jensens to maximize f(x) + f(y) + f(z) gives the maximum at x=y=z=0, where the sum is 3.

 

[Ahnfeltia]

The function f(x) = 2e^(-x) - e^(2x) has an inflection point at -ln(2) / 3

Spamming Jensens to maximize f(x) + f(y) + f(z) gives the maximum at x=y=z=0, where the sum is 3.

Doesn't Jensen only tell you that

3 >= 2/a + 2/b + 2/c - a^2 - b^2 - c^2

holds so long as x & y & z are all greater than the inflection point?

 

[mindlessselfindulge]

Doesn't Jensen only tell you that

holds so long as x & y & z are all greater than the inflection point?

Should have wrote out steps

x <= y <= z

If x <= y <= inflection point, z is kept constant and y is shifted towards the inflection point and x is shifted negative

If inflection point <= y <= z, then x is constant and y, z are shifted to (y + z) / 2

Let m = (y + z) / 2 and x = -2m.

Then maximize 2f(m) + f(-2m) at m = 0

 

[Ahnfeltia]

If inflection point <= y <= z, then x is constant and y, z are shifted to (y + z) / 2

Let m = (y + z) / 2 and x = -2m.

Then maximize 2f(m) + f(-2m) at m = 0

because z ⩾ y ⩾ x implies that m ⩾ −2m, which in turn implies that m ⩾ 0 yes

If x <= y <= inflection point, z is kept constant and y is shifted towards the inflection point and x is shifted negative

I don't think it's trivial that this increases f(x) + f(y) but yes.

Spoiler: solution

It suffices to show that f(x − p) + f(y + p) is increasing in p while y + p < inflection point. Differentiating w.r.t. p yields the expression g(x − p) − g(y + p) where g(w) = 2*exp(2w) + 2*exp(−w). It now suffices to show that g(w) is decreasing while w < inflection point, which can again be obtained via differentiation.

 

[Ahnfeltia]

Integrate ( x*ln(x) − x + 1 ) / ( sqrt(x)*(x − 1)^2 ) from x = 0 to infinity.

 

[Ahnfeltia]

Integrate ( x*ln(x) − x + 1 ) / ( sqrt(x)*(x − 1)^2 ) from x = 0 to infinity.

@CountBleck @Grim_Reaper

 

[Snahhan]

A physics problem:

Given a = 7m/s3 ( variable acceleration) and 2s, find v final from rest.

 

[Ahnfeltia]

A physics problem:

Given a = 7m/s3 ( variable acceleration) and 2s, find v final from rest.

Assuming a constant jolt (or jerk) of 7 m/s^3. If the initial acceleration and velocity are both 0, then the acceleration over time is a(t) = \int_0^t 7 ds = 7t, and the velocity over time is v(t) = \int_0^t a(s) ds = \int_0^t 7s ds = 7t^2/2. Plugging in t = 2 gives a final velocity of 14 m/s.

 

[Ventingblackpiller]

2+2=

Post all math related problems and solutions here. Don't cheat.

First problem: If the coefficients of x³ and x^4 in the expansion of (1+ ax+ bx² ) (1−2x) ^18 in powers of x are both zero, then (a, b) is equal to?

 

[Bianor]

2+2=

 

[Ventingblackpiller]

View attachment 1069241

Cope it's 5

 

[Ahnfeltia]

A physics problem:

Given a = 7m/s3 ( variable acceleration) and 2s, find v final from rest.

welcome to the forum btw

 

[Ahnfeltia]

Cope it's 5

 

[Snahhan]

welcome to the forum btw

Hi!

I must admit that I am a complete fool when it comes to math. I do not even understand the questions of these math problems , let alone the solutions.

But I have read about physics in online forums and blogs and found some interesting things there which could be also interesting to you.

Here is another physics problem:

Given a = 7m/s2 ( non variable acceleration) and 2s, find v final from rest.

 

[Ahnfeltia]

Given a = 7m/s2 ( non variable acceleration) and 2s, find v final from rest.

v(2) = \int_0^2 7 ds = 14

 

[Snahhan]

v(2) = \int_0^2 7 ds = 14

Now another physics problem:

How can a body accelerated to cube give the same answer as the body which is accelerated to square. Do not you agree that a body which accelerates at rising rate should be moving faster after 2 seconds?

 

[Ahnfeltia]

Do not you agree that a body which accelerates at rising rate should be moving faster after 2 seconds?

not necessarily because its initial acceleration is slower

 

[Raider919]

fuck you your thread sucks

 

[Spooky_Heejin]

Q(p_m+1) Q(p_i) =/= Q(p_j)

 

[CountBleck]

Integrate ( x*ln(x) − x + 1 ) / ( sqrt(x)*(x − 1)^2 ) from x = 0 to infinity.

honestly i have no idea.

 

[Awake]

this thread is for nerdcels

 

[Ahnfeltia]

honestly i have no idea.

Spoiler: hint

what's the derivative of ln(x) / (x − 1)