I kinda figured. You did already spoil the answer however.
3.5 ?
Here is another one: X/(Cos(x)(Sin(x) + Cos(x)) From 0 to pi/4
Correct
There's either a bracket too many or there's one missing
Is there a clever way too solve this without a boatload of algebra?
Of course. And you know it. I only tried because it was possible while rotting on bed and browsing other shit jfl
Newton's identity? It's the only way I know of solving that
Lol. Just assume a b and c as 1 1 and -2 and plug those values in the equation
That doesn't prove that the expression always equals 7/2 for any a + b + c = 0
Yes. Its not a proof
That sounds like a boatload of algebra. I couldn't come up with any other way either.
Proving that was kinda part of the question if I'm not mistaken
Depends on the setting.
There is a way that you don't need to divide the derivative by the original polynomial to find the sum of the powers.
Here is the original problem:
View: https://www.youtube.com/watch?v=gga590HOEPA&t=412s
Not so much algebra involved if you follow the steps of this guy.
It doesn't explicitly ask for a proof, but it was asked in an olympiad, so an explanation might be needed
this is largely what I had in mind
Applying the king's property yields the integral from 0 to pi/4 of (pi/4 - x)/(cos(x)(sin(x) + cos(x))). Taking their average yields that the integral equals pi/8 times the integral from 0 to pi/4 of sec(x)^2/(tan(x)+1). Substituting t = tan(x) yields that the integral equals pi/8 times the integral from t = 0 to 1 of 1/(t+1) -- i.e., the integral equals pi*ln(2)/8.
Correct
The total number of arrangements is 10! / 2^5 = 113400. The number of those arrangements that feature at least one row boasting two cars of the same team is 5*5*8! / 2^5 = 31500. Ergo, the number of arrangements where no row features two cars of the same team is 113400 - 31500 = 81900.
Wrong.
Why did you divide by 2^5? The total cases are 10!, given that all cars are different
I assumed cars in the same team were identical. In case they aren't removing the divisions by 2^5 should yield the correct answer, which should then be 2620800.
Wrong, but you are getting closer
My approach was indeed too simplistic. First of all, it should have been 5 rows times 10 pairs times 8! possibilities for the rest instead of 5 pairs (it matters which is on the left and which is on the right). Secondly, that's still wrong because I've overcounted every configuration in which multiple rows contain a pair.
I can't one-two-three come up with a nice way of counting the total number. Brute forcing the problem yields 2088960 as the answer.
i was doing x -->(pi/2-x)
...
Sure, but brute forcing the problem ain't worth any credit. Edit: just now seeing your edit. No, I had the computer just calculate all the possibilities (this is called brute forcing) because I couldn't figure out how to make inclusion-exclusion work.Correct
Did you use the inclusion-exclusion principle?
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@trying to ascend What is 2+2?
it's over
it's over
L
@TouhouMathcel
Find the generating function for the sequence
a(n)-(n+1)a(n-1) =0
Where a(0)=1
a(n)-(n+1)a(n-1) =0
Where a(0)=1
Let G(x) be the ordinary (or did you mean exponential?) generating function of a(n) -- i.e., G(x) = sum a(n)x^n (from n = 0 to oo). Then
G(x) = 1 + sum a(n+1)x^(n+1) = 1 + sum (n+2)a(n)x^(n+1) = 1 + d/dx sum a(n)x^(n+2) = 1 + d/dx [ x^2G(x) ] = 1 + 2xG(x) + x^2G'(x)
so we have to solve the first-order linear ordinary differential equation y = 1 + 2xy + x^2y' with y(0) = 1. While doing so is standard (integrating factor) the solution is ugly, involving an exponential integral. The exponential generating function seems as tho it'll be even uglier, however. Am I missing something?
Upon further inspection, the initial condition y(0) = 1 is superfluous. It's actually very easy to solve the recursion directly -- i.e., a(n) = (n+1)!
As such, it's not at all surprising that the ordinary generating function is ugly, given that it's radius of convergence at 0 has to be 0.
I was wrong about the exponential generating function. It'll be G(x) = sum (n+1)x^n = d/dx sum x^(n+1) = d/dx [ x/(1-x) ] = 1/(1-x)^2.
Based
This was not the original problem i was working on i think i made a mistake while defining recursive relation the problem was if you begin a 4*4 matrix with all entries 1 and then multiply it by a 4*4 matrix with all entries 2 and continue doing it till n matrices the base case a1 is entry of first matrix and a2 is entry of (1)4*4 x (2)4*4 and then next is a3 is (1)4*4x(2)4*4x(3)4*4 and so a4...like this where ($)4*4 represents all matrix entries are $
I think the nth product will be 2^nxn! But i think there is no generating function for an=2^nxn!
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You're close. Your a(n) = 4^(n-1)n! as can be easily checked via eigendecomposition. As for there not being a generating function, there's indeed no sensible ordinary generating function, but once again there is an exponential generating function that's nice. Indeed, the exponential generating function works out to
G(x) = sum a(n)x^n/n! = sum 4^(n-1)x^n = ¼ sum (4x)^n
which is just a good old geometric series with radius of convergence ¼.
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Lol like an inverse relationship, if you are good at math chances are you are a professional or are studying in STEM and you are basically a betabuxxx bitch which makes you a volcel for not trying hard enough in your academic/professional area and rotting here instead
Just be a betabuxx theory
L
when multiplied by 0 the answer is 0
L
multiply with 0 you get 0
L
0 when you multiply with 0
L
0 because a(n)-(n+1)a(n-1) =0 * 0
3+2
Even university girls that choose to remain single and independent still have sex with Chads behind closed doors. Meanwhile, nerds at university have no social life and are rotting in loneliness
L
=2+3
you can be bad at math and still be good at STEM if you're a some surgeon or doctor
betabuxxing is based, i wanna betabuxx a fat piss haired woman.
That's medicine, not STEM
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