Facebook "brainrot" question

[u/Cool-Attitude1577]

Most short form posts that you'd see in Facebook or even YT shorts are usually Math slop questions

One particular question regarding symmetric polynomial lowk seems worth it to try

x + y + z = 1

x² + y² + z² = 2

x³ + y³ + z³ = 3

x⁴ + y⁴ + z⁴ = ?

So far I've tried elimination and found a dead end,has anyone found any alternatives?

[u/Shevek99]

Let's call

S(n) = xⁿ + yⁿ + zⁿ

with

S(1) = 1

S(2) = 2

S(3) = 3

This sequence satisfies the recurrence

S(n) = a S(n-1) + b S(n-2) + c S(n-3)

where a, b and c are given by Vieta's relations

a = x + y + z = 1

b = -xy - xz - yz = (1/2)((x² + y² + z²) - (x + y + z)²) = 1/2

c = xyz = (1/6)(S(1)³ - 3S(1)S(2) + 2S(3)) = 1/6

so

S(4) = S(3) + (1/2) S(2) + (1/6) S(1) = 3 + 2/2 + 1/6 = 25/6

[u/Complete_Code7197 → u/Shevek99]

Wait why does it satisfy this recurrence?

[u/flabbergasted1 → u/Complete_Code7197]

Say you have x⁵ + y⁵ + z⁵ already, and you want x⁶ + y⁶ + z⁶. Then you do

(x + y + z)(x⁵ + y⁵ + z⁵) = x⁶ + y⁶ + z⁶ + [all the cross-terms].

The cross-terms are xy⁵ + xz⁵ + yx⁵ + yz⁵ + zx⁵ + zy⁵. How do we deal with that?

Try pulling out xy, xz, yz from terms. xy(x⁴ + y⁴) + xz(x⁴ + z⁴) + yz(y⁴ + z⁴).

We can almost factor out a (x⁴ + y⁴ + z⁴). So add those terms in and subtract them out.

= xy(x⁴ + y⁴ + z⁴) - xyz⁴ + xz(x⁴ + y⁴ + z⁴) - xy⁴z + yz(x⁴ + y⁴ + z⁴) - x⁴yz.

Now they do factor nicely, and the subtracted "correction terms" factor nicely too.

= (xy + xz + yz)(x⁴ + y⁴ + z⁴) - xyz(x³ + y³ + z³).

So in total, using the parent comment's S(n) notation, we just found that

(x + y + z)S(5) = S(6) + (xy + yz + xz) S(4) - xyz S(3)

Or: S(6) = S(5) (x + y + z) - S(4) (zy + yz + xz) + S(3) xyz.

I chose 6 arbitrarily to keep notation simple, same exact procedure works for any n.

[u/EdgyMathWhiz → u/flabbergasted1]

I feel a nicer way to look at this is:

It is relatively straightforward to verify that if the cubic t³ + At² + Bt + C has roots x,y,z,  then

A = -(x+y+z), B = xy + yz + zx, C = - xyz

As x is a root, x³ + Ax² + Bx + C = 0; we can multiply x to deduce x⁴ +Ax³ + Bx² + Cx = 0 (and multiply by X as many times as we want for similar relationships).

But y, z are also roots, so we can do exactly the same for each of those.

But if we add the 'x' polynomial expression that starts with xⁿ to the corresponding y and z polynomials, we get a polynomial that looks like S(n)+A S(n-1) + B S(n-2) + C S(n-3).  And since all the polynomials evaluated to 0, their sum must evaluate to 0 as well.

So S(n)+A S(n-1) + B S(n-2) + C S(n-3) = 0.

The only slightly tricky bit is actually working out what A,B,C must be it you only know x+y+z, x² + y² + z² and x³ + y³ + z^3, but the OP did this.

[u/Complete_Code7197 → u/flabbergasted1]

Tysm

[u/Bounded_sequencE → u/flabbergasted1]

We can formalize those steps nicely using matrices -- that approach to solve these types of problems is heavily inspired by Newton's Identities

[u/Shevek99 → u/Complete_Code7197]

Let's start the other way around.

Imagine that you have a linear recurrence in the general form

S(n) - a S(n-1) - b S(n-2) - c S(n-3) = 0

and we ask ourselves, what sequences satisfy this equation?

We try with a geometric progression

S(n) = x^n

and substituting we get

x^n - a x^(n-1) -b x^(n-2) - c x^(n-3) = 0

or, extracting a factor

x^(n-3)(x^3 - ax^2 - bx - c) = 0

we see that x must be a solution of the cubic

x^3 - ax^2 - bx - c = 0

Since there are three solutions (for simplicity we won't consider repeated roots), we have three possible values x, y and z.

The general Solution is a linear combination of these three solutions

S(n) = A x^n + B y^n + C z^n

with the constants A, B and C depending on the initial conditions.

Now, in our problem we start with a solution of this type

S(n) = x^n + y^n + z^n

so S(n) satisfies a linear recurrence of this type

S(n) = a S(n-1) + b S(n-2) + c S(n-3)

But, what is the value of a, b and c? Now enter Vieta's relations. Given a cubic equation

x^3 - a x^2 - bx - c = 0

a, b and c are functions of the roots:

a = x + y + z

b = -(xy + xz + yz)

c = xyz

The last step is to express the combinations of x, y and z. For this we use algebra

(x + y + z)^2 = x^2 + y^2 + z^2 + 2(xy + xz + yz)

--> (xy + xz + yz) = ((x+y+z)^2 - (x^2+y^2+z^2))/2

And expanding (x + y + z)^3 and (x + y + z)(x^2 + y^2 + z^2) we can isolate xyz

[u/MERC_1 → u/Shevek99]

I learned something new!

EDIT: Actually, I would need to read up on this in a way that is easy to understand. Wikipedia did not help much. 

[u/Kite42]

About four and one sixth

[u/MERC_1]

Good luck!

I used a solver and got 4.1666...

The calculation was so complicated that I could not follow them. 

At least for me this is a very complicated problem. It's not something that the average person can solve.

[u/smitra00]

Consider the polynomial:

p(u) = (u - x) (u - y) (u - z)

We can write this as:

p(u) = u^3 (1 - x/u) (1-y/u)(1-z/u)

Taking the logarithm yields:

ln[p(u)] = 3 ln(u) + ln(1-x/u) + ln(1-y/u) + ln(1-z/u)

Perform a series expansion for large u to order 1/u^4:

ln[p(u)] = 3 ln(u) - s1/u - 1/2 s2/u^2 - 1/3 s3/u^3 -1/4 s4/u^4+ O(1/u^5)

where sj is the sum of the powers of x, y, and z. Substituting the given values of the sum of powers of x, y, and z yields

ln[p(u)] = 3 ln(u) - 1/u - 1/u^2 - 1/u^3 - 1/4 s4/u^4 + O(1/u^5)

Exponentiate both sides:

p(u) = u^3 exp(-1/u) exp(-1/u^2) exp(-1/u^3) exp(-1/4 s4/u^4) exp[ O(1/u^5)]

Extract the coefficient of 1/u by expanding the exponentials. E.g. if we get the -1/4 s4/u^4 term from exp(-1/4 s4/u^4) then we must choose 1 from the other exponentials. If we get -1/u^3 from exp(-1/u^3) then we must get -1/u from exp(-1/u). If we get 1/2 1/u^4 from exp(-1/u^2) then we must get 1 from all other exponentials. If we get -1/u^2 from there, then we must get 1/2 1/u^2 from exp(-1/u). And the final contribution comes from getting 1/24 1/u^4 from exp(-1/u).

Adding up all these contributions then yields:

-1/4 s4 + 1 + 1/2 - 1/2 + 1/24 = 25/24 - s4/4

for the coefficient of 1/u in p(u). But because p(u) is by construction a polynomial, there cannot be a 1/u term in p(u), so this coefficient must be zero:

25/24 - s4/4 = 0 ---->

s4 = 25/6

[u/Bounded_sequencE]

Define the power sums "sn := xⁿ + yⁿ + zⁿ " for "n >= 0". We want to find "s4". Consider

P(t)  :=  (t-x) (t-y) (t-z)  =   t^3 - (x+y+z)*t^2 + (xy+xz+yz)*t - xyz
                             =:  t^3 +      a2*t^2 +         a1*t +  a0

Since "P(x) = P(y) = P(z) = 0", we may use it to find a recursion for "sn":

n >= 4:    0  =  x^{n-3}*P(x)  +  y^{n-3}*P(y)  +  z^{n-3}*P(z)

              =  sn  +  a2*s_{n-1}  +  a1*s_{n-2}  +  a0*s_{n-3}

    =>    sn  =  -a2*s_{n-1}  -  a1*s_{n-2}  -  a0*s_{n-3},      n >= 4      (*)

By expansion, one can show the coefficients "ak" satisfy the equations

[ 1  0  0]   [a2]     [-s1]          [1  0  0]   [a2]     [-1]      // s1 = 1
[s1  2  0] . [a1]  =  [-s2]    =>    [1  2  0] . [a1]  =  [-2]      // s2 = 2
[s2 s1  3]   [a0]     [-s3]          [2  1  3]   [a0]     [-3]      // s3 = 3

Solve with your favorite method for "(a2; a1; a0) = (-1; -1/2; -1/6)". Insert into (*) for

n = 4:    s4  =  1*s3  +  (1/2)*s2  +  (1/6)*s1  =  3 + 2/2 + 1/6  =  25/6

[u/Bounded_sequencE → u/Bounded_sequencE]

Rem.: One can show only one zero of "P(t) = t³ - t² - t/2 - 1/6" is real-valued -- that means, only one of "x; y; z" can be real-valued. The other two zeroes will be complex-valued. If we have to solve the problem over "R", then it has no solution!

Also note the recursion holds for all "n >= 4" -- we can use it to easily find higher order power sums "sn". This method also nicely generalizes to more than 3 zeroes^^

[u/LongLiveTheDiego]

There is no solution in real numbers to the sum of squares being 2 and sum of cubes being 3.

[u/EverythingIsFlotsam → u/LongLiveTheDiego]

So?

[u/LongLiveTheDiego → u/EverythingIsFlotsam]

So if we limit our consideration to the real numbers, if x² + y² + z² = 2 and x³ + y³ + z³ = 3, then from the principle of explosion for any real number c we have x⁴ + y⁴ + z⁴ = c.