/sci/ - Science & Math

[Anonymous No.11201209]

Well, lets see how many smart fellas we have on this board.

[Anonymous No.11201223]

P=NP

[Anonymous No.11201236]

tfw can't solve this

[Anonymous No.11201237]

n for n =4,5,6,7...

[Anonymous No.11201238]

>>11201209
x,y,z={5,4,1,-4,-5}

[Anonymous No.11201278]

n=igger

[Anonymous No.11201340]

>>11201209
n is for niggers

[Anonymous No.11201344]

what are x y and z? why do they say what n is but not x y and z?

retarded OP

[Anonymous No.11201357]

>>11201344
The point is that you don't need to know this. For example

[eqn]x^4 + y^4 + z^4 = \frac{(x+y+z)^4 - 6 (x+y+z)^2 (x^2+y^2+z^2) + 3 (x^2+y^2+z^2)^2 + 8 (x+y+z)(x^3+y^3+z^3)}{6} = \frac{1 - 12 + 12 + 24}{6} = \frac{25}{6}
[/eqn]

[Anonymous No.11201372]

>>11201357

you are retarded. x y and z could be complex numbers
that's the whole point of mathematical structures lol to make sure you know what oyu are dealing with
and you niggers are just writing shit

[Anonymous No.11201689]

x^5+y^5+z^5=6

[Anonymous No.11201717]

>>11201209
[math]x^n+y^n+z^n=n[/math]

Quite easy to be honest

[Anonymous No.11201722]

There should be a general formula that, when given fixed values for the sum of the first n powers of n variables
x_1 + x_2 + ... + x_n,
x_1^2 + x_2^2 + ... + x_n^2,
...
x_1^n + x_2^n + ... + x_n^n,
tells you what the sum of the higher powers are.

It probably involves some ugly-looking matrix.

[Anonymous No.11201742]

>>11201344
>>11201372
holy shit you're retarded

[Anonymous No.11201936]

>>11201742
>>11201357
Nice post, anon~

[Anonymous No.11202454]

N = (x + y + z) n

[Anonymous No.11202540]

>>11201237
>>11201223
>>11201278
>>11201340
>>11201372
>>11201717
>>11202454
Haha sadboi brainlet cope

[Anonymous No.11202545]

4
You're welcome for the 4

[Anonymous No.11202642]

>>11202540
Reta r

[Anonymous No.11202688]

6

[Anonymous No.11202737]

>>11201209
ill top you stupid shit
[eqn]\displaystyle \sum_{n=1}^{N} n^3 = \Bigg( \displaystyle \sum_{n=1}^{N} n \Bigg)^2[eqn]

now prove the above statement.

[Anonymous No.11202743]

>>11202737
>[eqn]\displaystyle \sum_{n=1}^{N} n^3 = \Bigg( \displaystyle \sum_{n=1}^{N} n \Bigg)^2[/eqn]

forgot slash

[Anonymous No.11202762]

>>11202540
>>11202540
You mad cuz it's true.

[Anonymous No.11203000]

>>11202737
If you're gonna post a literal freshman algebra problem at least don't fuck up the [math]\LaTeX[/math].

[Anonymous No.11203005]

>>11201209
You can just compute the trace of X^n where X is the 3x3 diagonal with x,y,z on the diagonal.
Computing the characteristic polynomial of X gives X^3 - X^2 - X/2 - 1/6.
Solving X^3 - X^2 - X/2 - 1/6 = 0 will give x,y,z.

If you don't want to solve the cubic, use the recurrence A(n+3) = A(n+2) + A(n+1)/2 + A(n)/6.

This gives A(4) = A(3) + A(2)/2 + A(1)/6 = 3 + 1 + 1/6 = 25/6
(agreeing with >>11201357)

A(5) = A(4) + A(3)/2 + A(2)/6 = (4 + 1/6) + (3/2) + (2/6) = 6
(agreeing with >>11201689)

[Anonymous No.11203078]

>>11202762
>i'm mad because it's true you guys are all doing the brainlet cope
confirmed brainlet

[Anonymous No.11203172]

>>11201278
>>11201340
>haha nigger lol XD
kys

[Anonymous No.11204320]

bmup

[Anonymous No.11204326]

Approaches infinity, can't be complex as it clearly states that n starts at one and sticks to whole numbers with no complex part to be found.

[Anonymous No.11204348]

>>11204326
? is two of them are complex conjugates, it works

[Anonymous No.11205143]

>>11204348
Yes. they are roots of a cubic with real coefficients so they would come in pairs of conjugates.
Wolfram gives
x ~ 1.4308
y ~ -0.21542 + 0.26471i
z ~ -0.21542 - 0.26471i

[Anonymous No.11205321]

xyz^n

[Anonymous No.11206212]

>>11205143
For large n, A(n) ~ 1.4308495662^n since the other roots have absolute value less than 1.