/sci/ - Science & Math

this one's easy

I'm not going to say it, but I guess you could check them

>>4292533

3984

What on earth does the first d/dx notation even mean? Oh infinitesimal to a power, u so crazy.

>>4292533
Totally. I figured it out in less than a minute.

I'm not going to post the answer, though, because I want other people to do it.

Was this A1 or B1?

I've got it, but I wouldn't say it was unusually easy for a 1st problem.

I would post solution, but I'm guessing that's bad etiquette. How long do you normally wait? (first day here.)

>>4292684
Fucking redditors. Please just leave.
>mfw that fucking reaction image has been used thousands of times and still no one has noticed the spot between the arms that isn't filled in with the right color

i don't frequent /sci/ too much, but i've been coming back more and more lately. what level of comprehension of mathematics is required for these problems? high school math faggot here. teach me your ways, /sci/.

>>4292703
>>mfw that fucking reaction image has been used thousands of times and still no one has noticed the spot between the arms that isn't filled in with the right color
damn. never noticed that.

>>4292700
>Was this A1 or B1?
B4, and one of the harder questions on the test

>I'm guessing that's bad etiquette.
The usual etiquette is to rip the question for being too easy, implying heavily that you can solve it, post no actual solution, and then wait until one of the handful of actual smart people here solve it so you can claim that that's the answer you had in mind all along.

Ohhhhh shiiiiiiiiiit! Postan in a sticky!

(I am in precalculus and what is this)

>>4292703

I'm on the 12th derivative of that fraction. This quotient and product rule business is a mess but lets see how it goes. I'll keep you guys posted

>>4292824
lol

>>4292726
I see. OK, I'm writing up a soln right now...

Let q(x) = x^3 - x.
Then
p(x) = a(x)q(x) + r(x)
for some r(x) of deg ≤ 2.

A partial fraction decomposition yields

p(x) / q(x) = a(x) + A/x + B/(x-1) + C/(x+1)

[here we used that r(x) cannot have roots in common with q(x), lest q(x) and p(x) share roots.]

Let n = 1992.

D^n (p/q) = 0 + n!(-1)^n [ A/x^(n+1) + B/(x-1)^(n+1) + C/(x+1)^(n+1) ].

Let
a1(x) = [(x+1)(x-1)]^(n+1),
a2(x) = [x(x-1)]^(n+1),
a3(x) = [x(x+1)]^(n+1).

Then
D^n (p/q) = n!(-1)^n [ A a1(x) + B a2(x) + C a3(x) ] / g(x),
where g(x) = (x^3 - x)^(n+1).

Since the g.c.d. of these three polynomials is 1, there exist constants A, B and C such that

A a1(x) + B a2(x) + C a3(x) = 1.

Using these particular A,B and C, we may take

p(x) = A(x-1)(x+1) + Bx(x+1) + Cx(x-1),

and we have

D^n (p/q) = n!(-1)^n /g(x).

So the min degree possible is zero.
[I wrote this fast and didn't double check it -- expect errors!]

>>4292876
>Since the g.c.d. of these three polynomials is 1, there exist constants A, B and C such that...

Nope. Polynomials, yes, but not constants.

>>4292892
Dammit, you're right. I can't believe I did that.

I was just looking over it and something didn't seem right...
OK, I'll try to salvage.

>>4292824
UPDATE PLEASE

>>4292824

>>4292908
OK, I've corrected my above solution. [But again, I did it quickly and didn't double check.] Here goes:
Answer: 3892.
Let q(x) = x^3 - x.
Then
p(x) = a(x)q(x) + r(x)
for some r(x) of deg ≤ 2.

A partial fraction decomposition yields

p(x) / q(x) = a(x) + A/x + B/(x-1) + C/(x+1)

for some constants A, B and C, none of which is zero. [Here we used that r(x) cannot have roots in common with q(x), lest q(x) and p(x) share roots.]

Let n = 1992.

D^n (p/q) = 0 + n!(-1)^n [ A/x^(n+1) + B/(x-1)^(n+1) + C/(x+1)^(n+1) ].

let a1(x) = [(x+1)(x-1)]^(n+1), a2(x) = [x(x+1)]^(n+1), a3(x) = [x(x-1)]^(n+1).

Then
D^n (p/q) = n!(-1)^n [ A a1(x) + B a2(x) + C a3(x) ] / g(x),
where g(x) = (x^3 - x)^(n+1).

Note that the numerator and denominator share no factors if none if A, B and C are all nonzero.

Using the binomial theorem, we get:
A a1(x) + B a2(x) + C a3(x)
= (A+B+C) x^(2n+2)
+ (n+1)(B-C) x^(2n+1)
+ [(n+1)A + Comb(n+1,2)(B+C)] x^(2n)
+ Comb(n+1,3)(B - C) x^(2n-1)
+ [Comb(n+1,2)A + Comb(n+1,4)(B+C)] x^(2n-2)
+ ...

The system obtained by setting the first three coefficients to zero,
A + B + C = 0
B - C = 0
A + (n/2)B + (n/2)C = 0,

has a unique solution (determinant ≠ 0). For this solution, A, B and C are all nonzero, and
[Comb(n+1,2)A + Comb(n+1,4)(B+C)] ≠ 0.

Thus the min possible degree of f(x) is 2n-2 = 2*1992 - 2 = 3982. This is achieved if we take

p(x) = A(x-1)(x+1) + Bx(x+1) + Cx(x-1).

>>4293120
>Note that the numerator and denominator share no factors if none if A, B and C are all nonzero.
Ignore the "if none".
And replace "factors" with "nonconstant factors".

>>4293120
>has a unique solution (determinant ≠ 0)
yes
>For this solution, A, B and C are all nonzero
no. The unique solution is A=B=C=0

>>4293120
Assuming my solution is correct, I'd say B4 sounds about right. I'm guessing
>>4292533
>>4292681
>>4292684
were jokes. Hilarious.

>>4293133
Fuck! another embarrassing error!

>>4293133
Then it becomes immediately clear that the min degree is 2 greater than what I said: 3984.

Just set the rhs of the last equation of that system,
A + (n/2)B + (n/2)C = 0,
to be nonzero.
*Then* the unique solution to the system has all of A, B and C nonzero.

>>4292550
Doh. I was far from the first....

>>4293144
winrar winrar chicken dinrar

>>4293148
>implying that guy didn't just look it up

>>4293150
>implying that guy didn't just look it up
Hmm. I guess I shouldn't be surprised...but I just don't get *why* someone would do that on an anonymous forum.

>>4293158
Though I guess it's for the same reason people use cheat engines to win high scores on flash games. Whatever the fuck that reason is.