SuicideFuel Math thread problem (official)

[Diddy29]

Alberta?

Ontario

 

[trying to ascend]

Be P(x) = x^6 + bx^5 + cx^4 + dx^3 + ex^2 + fx + g a polinomial with integer coefficients and that P (sqrt(2) + cubic root(3)) = 0.

The polynomial R(x) is the remainder of the division of P(x) by x³ - 3x - 1.

Find the sum of the coefficients of R(x)

 

[DespressedCurryCel1]

Ontario

ye look at a grade 11 math text book for math 20-1 quadratic functions and equations you’ll find questions like ones I gave you

 

[Diddy29]

ye look at grade 11 math in Alberta quadratic functions and equations

Yeah I learned the same stuff as you in Grade 11. I only remembered now the Grade 9 questions you posted, you also have to do them in Grade 11-12, using the calculus way.

 

[Ahnfeltia]

Be P(x) = x^6 + bx^5 + cx^4 + dx^3 + ex^2 + fx + g a polinomial with integer coefficients and that P (sqrt(2) + cubic root(3)) = 0.

The polynomial R(x) is the remainder of the division of P(x) by x³ - 3x - 1.

Find the sum of the coefficients of R(x)

Using that {1, sqrt(2), cbrt(3), cbrt(9), sqrt(2)*cbrt(3), sqrt(2)*cbrt(9)} is linearly independent over the rational numbers*, P(sqrt(2) + cbrt(3)) = 0 yields the following six equations in six unknowns:

• 17 + 60b + 4c + 3d + 2e + g = 0
• 120 + 4b + 12c + 2d + f = 0
• 90 + 20b + 3c + 6d + f = 0
• 60 + 3b + 12c + e = 0
• 24 + 15b + 8c + 2e = 0
• 18 + 20b + 3d = 0

This system of equations of course begs to be solved using linear algebra, which yields the following (unique) solution: b = 0, c = -6, d = -6, e = 12, f = -36, g = 1 -- i.e., p(x) = x^6 - 6x^4 - 6x^3 + 12x^2 - 36x + 1. Doing polynomial long division yields that R(x) = 3x^2 - 54x - 4, whose coefficients sum to -55.

* I highly recommend you to try and think about how one could prove that {1, sqrt(2), cbrt(3), cbrt(9), sqrt(2)*cbrt(3), sqrt(2)*cbrt(9)} is linearly independent over the rational numbers. It's pretty tricky, but really nice. Do let me know if you managed to figure it out.

 

[trying to ascend]

Using that {1, sqrt(2), cbrt(3), cbrt(9), sqrt(2)*cbrt(3), sqrt(2)*cbrt(9)} is linearly independent over the rational numbers*, P(sqrt(2) + cbrt(3)) = 0 yields the following six equations in six unknowns:

• 17 + 60b + 4c + 3d + 2e + g = 0
• 120 + 4b + 12c + 2d + f = 0
• 90 + 20b + 3c + 6d + f = 0
• 60 + 3b + 12c + e = 0
• 24 + 15b + 8c + 2e = 0
• 18 + 20b + 3d = 0

This system of equations of course begs to be solved using linear algebra, which yields the following (unique) solution: b = 0, c = -6, d = -6, e = 12, f = -36, g = 1 -- i.e., p(x) = x^6 - 6x^4 - 6x^3 + 12x^2 - 36x + 1. Doing polynomial long division yields that R(x) = 3x^2 - 54x - 4, whose coefficients sum to -55.

* I highly recommend you to try and think about how one could prove that {1, sqrt(2), cbrt(3), cbrt(9), sqrt(2)*cbrt(3), sqrt(2)*cbrt(9)} is linearly independent over the rational numbers. It's pretty tricky, but really nice. Do let me know if you managed to figure it out.

Correct

I can't prove that, I'm not on that level yet

 

[Ahnfeltia]

I can't prove that, I'm not on that level yet

Provided you know what linear independence means, you should theoretically be able to do it if you understood what I did. It's not that much more difficult.

 

[LeDepravedCel]

Math is for niggers

 

[Ahnfeltia]

Math is for niggers

 

[DespressedCurryCel1]

SOS HELP ME NIGGERS PLS PLS

 

[DespressedCurryCel1]

Some nigger help now pls!!!! Thanks in advance,


despressed(not depressed) currycel1

 

[DespressedCurryCel1]

HELP!!

 

[DespressedCurryCel1]

SOMEONE PLS SEND HELP

 

[Diddy29]

Some nigger help now pls!!!! Thanks in advance,


despressed(not depressed) currycel1

View attachment 713132

6x9

 

[DespressedCurryCel1]

6x9

Solution????

 

[Diddy29]

Solution????

Ye length is 9" and width is 6"

 

[DespressedCurryCel1]

Ye length is 9" and width is 6"

Pls screen shot how you got that

 

[Diddy29]

Pls screen shot how you got that

(12-4x)(12-2x)=54
144-72x+8x^2=54
90-72x+8x^2=0
Solve for x using quadratic formula and you get 1.5 and 7.5
7.5 doesn't make sense so x is 1.5
then sub 1.5 in the first equation and you get your measurements.

 

[trying to ascend]

Find X, given that 1/X = sin(pi/14)sin(3pi/14)sin(5pi/14)

 

[Diddy29]

Find X, given that 1/X = sin(pi/14)sin(3pi/14)sin(5pi/14)

8

 

[trying to ascend]

8

Correct

What was your solution?

 

[Diddy29]

Correct

What was your solution?

I used trig identities and subsitutions

 

[Med Amine]

1*1=0

 

[trying to ascend]

1*1=0

Fake news

 

[LonelyRobot]

This thread is so pozzed. None of these are math problems, they are lame exercises.
https://www.maa.org/external_archive/devlin/LockhartsLament.pdf

 

[Ahnfeltia]

This thread is so pozzed. None of these are math problems, they are lame exercises.
https://www.maa.org/external_archive/devlin/LockhartsLament.pdf

I only read bits and pieces of the article you linked. Regarding what you said, "problem" is often used synonymously with "exercise" in mathematics and physics. Besides, asking fellowcels to solve the Riemann Hypothesis or the Collatz conjecture is out of the question, so I'm not exactly sure what you wanted from this thread.

 

[LonelyRobot]

the article

Its not an article, its an essay. Articles are written and read by low IQ jews LARPing as high IQ intellectuals.

asking fellowcels to solve the Riemann Hypothesis or the Collatz conjecture is out of the question, so I'm not exactly sure what you wanted from this thread

It can be a simple question, like, which is true for the following:
View: https://imgur.com/a/rLZVleL
the circles in the left square have a greater, less great, or equal area to the circle in the right square.

"problem" is often used synonymously with "exercise" in mathematics and physics

I see your point

 

[Ahnfeltia]

Its not an article, its an essay. Articles are written and read by low IQ jews LARPing as high IQ intellectuals.

Potayto potahto as far as I'm concerned. All I care about is the content.

It can be a simple question, like, which is true for the following:
View: https://imgur.com/a/rLZVleL
the circles in the left square have a greater, less great, or equal area to the circle in the right square.

I fail to see what distinguishes the problem (or whatever you wanna call it) you gave from the ones previously given in this thread. Not to mention, your problem is probably the easiest in the entire thread. The areas are obviously equal.

 

[Sneir]

Potayto potahto as far as I'm concerned. All I care about is the content.

I fail to see what distinguishes the problem (or whatever you wanna call it) you gave from the ones previously given in this thread. Not to mention, your problem is probably the easiest in the entire thread. The areas are obviously equal.

solved (in video game)

 

[Old Ironsides]

I don't need this thread...
All I need to do is fucking ask @NoLooksNoLife



it's over

 

[Med Amine]

SuicideFuel​

 

[trying to ascend]

Find X: 1 + 2x + 3x² + 4x³ ... = 1/2 + 3/4 + 5/8 + 7/16 ...

 

[Ahnfeltia]

Find X: 1 + 2x + 3x² + 4x³ ... = 1/2 + 3/4 + 5/8 + 7/16 ...

The right-hand side is presumably the sum of (2n-1)/2^n from n=1 to infinity, which is 4S-1, where S is the sum of n/2^(n+1) from n=1 to infinity, which, upon differentiating the identity y/(1-y) = sum of 1/y^n from y=1 to infinity w.r.t. y and subsequently plugging in y=2, is readily seen to be 1. As such, the right-hand side equals 4*1-1 = 3.

As for the left-hand side, differentiating the identity x/(1-x) = sum x^n for n=1 to infinity w.r.t. x readily yields that the left-hand side equals 1/(1-x)^2. Note that x needs to be between +1 & -1 for the left-hand side to be convergent. The original equation now becomes 1/(1-x)^2 = 3, which has solutions x = 1±1/√3. Since 1+1/√3 > 1, x = 1-1/√3.

 

[trying to ascend]

The right-hand side is presumably the sum of (2n-1)/2^n from n=1 to infinity, which is 4S-1, where S is the sum of n/2^(n+1) from n=1 to infinity, which, upon differentiating the identity y/(1-y) = sum of 1/y^n from y=1 to infinity w.r.t. y and subsequently plugging in y=2, is readily seen to be 1. As such, the right-hand side equals 4*1-1 = 3.

As for the left-hand side, differentiating the identity x/(1-x) = sum x^n for n=1 to infinity w.r.t. x readily yields that the left-hand side equals 1/(1-x)^2. Note that x needs to be between +1 & -1 for the left-hand side to be convergent. The original equation now becomes 1/(1-x)^2 = 3, which has solutions x = 1±1/√3. Since 1+1/√3 > 1, x = 1-1/√3.

Correct

 

[Michael15651]

i used to slay problems when i was preparing for university

Now you slay solutions

 

[Med Amine]

can't relate to this thread a lot tbh since i was in a medicine college so no math

 

[trying to ascend]

can't relate to this thread a lot tbh since i was in a medicine college so no math

How much dinars do you make?

 

[Med Amine]

How much do you make?

i am not a doctor, i dropped out after 2 years of college because i was too deformed to fit in with other students.
my brother is still studying in STEM uni tho and will be going to work in US when he finish.
now i have a normal wageslave job and i make than half what i was going to make if i continues studying. i work in a big food company , probably will get a promotion by the end of 2023

 

[TouhouMathcel]

Does math logic and theory of computations problems count as math problems? Such as lambda calculus and combined logic?

 

[Ahnfeltia]

Does math logic and theory of computations problems count as math problems? Such as lambda calculus and combined logic?

I'd say so, but I doubt even most mathcels here are familiar with those topics.

 

[TouhouMathcel]

A Turing machine is defined by the following scheme. Formulate an algorithm according to which this machine operates

 

[TouhouMathcel]

I'd say so, but I doubt even most mathcels here are familiar with those topics.

Okay, thank you. Maybe someone will be interested in getting acquainted with those topics

 

[TouhouMathcel]

Calculate the linear algorithmic complexity of the following Turing machine

 

[trying to ascend]

Consider three sets E1 = {1, 2, 3}, F1 = {1, 3, 4} and G1 = {2, 3, 4, 5}. Two elements are chosen at random, without replacement, from the set E1 and let S1 denote the set of these chosen elements. Let E2 = E1 - S1 and F2 = F1 ∪ S1. Now, two elements are chosen at random, without replacement, from the set F2 and let S2 denote the set of these chosen elements. Let G2 = G1 ∪ S2. Finally, two elements are chosen at random, without replacement, from the set G2 and let S3 denote the set of these chosen elements. Let E3 = E2 ∪ S3. Given that E1 = E3, let p be the conditional probability of the event S1 = {1, 2}. Then the value of p is

 

[Ahnfeltia]

A Turing machine is defined by the following scheme. Formulate an algorithm according to which this machine operates

View attachment 738404

While I have a passing familiarity with finite state automata and Turing machines, I'm at a bit of a loss regarding some of the conventions. Does the machine operate on finite binary strings? Is "B" a blank symbol? What happens when a state reads a symbol for which there's no arrow? Does it reject in that case?

If the answer to all of those questions is affirmative, then I believe the machine accepts only those strings of the form "some number of 0s followed by an equal number of 1s". Am I correct?

 

[Ahnfeltia]

Consider three sets E1 = {1, 2, 3}, F1 = {1, 3, 4} and G1 = {2, 3, 4, 5}. Two elements are chosen at random, without replacement, from the set E1 and let S1 denote the set of these chosen elements. Let E2 = E1 - S1 and F2 = F1 ∪ S1. Now, two elements are chosen at random, without replacement, from the set F2 and let S2 denote the set of these chosen elements. Let G2 = G1 ∪ S2. Finally, two elements are chosen at random, without replacement, from the set G2 and let S3 denote the set of these chosen elements. Let E3 = E2 ∪ S3. Given that E1 = E3, let p be the conditional probability of the event S1 = {1, 2}. Then the value of p is

5/26 I believe.

 

[trying to ascend]

5/26 I believe.

Wrong, you are close though

 

[Ahnfeltia]

Wrong, you are close though

I indeed made a stupid mistake. It should be 6/29, right?

 

[trying to ascend]

I indeed made a stupid mistake. It should be 6/29, right?

No

 

[Ahnfeltia]

No

Damn it, I miscounted the size of G1. I thought it went up to 6 for some reason. 1/5. Third time's gotta be the charm.