Ahnfeltia
almost human
★★★★
- Joined
- Oct 6, 2022
- Posts
- 3,825
truethis thread is for nerdcels
truethis thread is for nerdcels
Is this complex variables?Integrate ( x*ln(x) − x + 1 ) / ( sqrt(x)*(x − 1)^2 ) from x = 0 to infinity.
Is this complex variables?
Maybe it's possible, but I dunno.what's the derivative of ln(x) / (x − 1)
Is it pi^2/2Maybe it's possible, but I dunno.
Yep. Nicely done.Is it pi^2/2
there's thisDoes anyone have good way of calculating week of day based on dates like 1st Jan 2000. Looked into Zellers algorithm but it's too hard. Easiest thing I came across is this https://artofmemory.com/blog/how-to-calculate-the-day-of-the-week/
I forgot to add that f has to be continuous like a dumbass. My bad.Let f be a function from the reals to the reals and let f^n denote the n-fold composition of f with itself. We say that x is a fixed point of order n if f^n(x) = x but f^k(x) ≠ x for all positive integers k < n. Show that if f has a fixed point of order 3 it must also have a fixed point of order 2.
Looking into itthere's this
Doomsday rule - Wikipedia
en.wikipedia.org
Let f be a function from the reals to the reals and let f^n denote the n-fold composition of f with itself. We say that x is a fixed point of order n if f^n(x) = x but f^k(x) ≠ x for all positive integers k < n. Show that if f has a fixed point of order 3 it must also have a fixed point of order 2.
I forgot to add that f has to be continuous like a dumbass. My bad.
C' contains the points (c, b) and (b, a), and is a function from y to x. Let D be the curve on f from (m, m) to (c, a). Since b < m < c, D and C' intersect by the intermediate value theorem. The intersection is a fixed point of order 2 since it can be mapped back onto the mirror image of the curve C.
I'll assume the image of f is an interval.For all functions f: R^2 -> R, prove that there exist vectors v1 and v2 with |f(v2) - f(v1)| < |v2 - v1|.
Bonus:
Replace the inequality above with |f(v2) - f(v1)| < |v2 - v1|^p. For what values of p is the statement still true?
- I'm not convinced the IVT applies here. After all, the IVT applies to a function. What's your function in question?
- How do you know that the intersection point ain't (m, m)?
- Naively there should be 6 cases. How did you reduce down to 2?
Prove that no Hölder continuous functions with exponent > 1 exist.
yes sorryI think you mean no non-constant functions.
false e.g. |x| is Lipschitz but not differentiable at 0For this proof use the property that all Holder continuous functions with a positive exponent are differentiable.
Integrate (1 - cos(x)) / (x * sqrt(x)) from x = 0 to infinity.
Too crude for my taste. For those who want more of a challengeGiven a ball B of radius 1 in R^d, prove that there exists a set of 6^d balls of radius 1/2 with centers on the surface of B which cover the entire surface of B.
Idk, can this be solved using modular arith and dirichlet's?. Or is the solution in the realm of higher mathFor a positive integer n, given n integers (not necessarily distinct), prove that there is at least 1 non empty subset sum that is divisible by n.
Idk, can this be solved using modular arith and dirichlet's?. Or is the solution in the realm of higher math
Anyways, comfort problem for the masses:
Let (S) be the sphere
(x-a)^2+(y-b)^2+(z-c)^2=1
With a,b,c satisfying
a^2+(b-5)^2+(c-200)^2=16
For all a,b,c (S) is tangent to two stationary spheres with radii R1, R2 (R2>R1)
Calculate P=sqrt(R2^2-R1^2)
I should've just said (S), not the sphere (S).There are 2 ways: a simple way, and a more complicated way (that I came up with).
Both are elementary.
This doesnt make sense.
Anyways, comfort problem for the masses:
Let (S) be the sphere
(x-a)^2+(y-b)^2+(z-c)^2=1
With a,b,c satisfying
a^2+(b-5)^2+(c-200)^2=16
For all a,b,c (S) is tangent to two stationary spheres with radii R1, R2 (R2>R1)
Calculate P=sqrt(R2^2-R1^2)
honestly no idea if this is rightFor a positive integer n, given n integers (not necessarily distinct), prove that there is at least 1 non empty subset sum that is divisible by n.
719746294729748173917391739273When Chad and Stacy have a baby it will inherit 50% of Chad DNA.
Chad's grandkid will be only 25% his DNA, and so on...
At some point down the genetic road, his DNA will be like a single chromosome that can mutate after being hit by cosmic radiation, making Chad's offspring indistinguishable from a theorethical descendant of you and I.
How many generations it takes for Chad's offspring to accidentally become my genetic son?
100% correct, very well donehonestly no idea if this is right
n = 1 case is satisfied because any number can be divided by 1
cases where n > 1 would need pigeonhole principle
consider a set of n integers {a1, a2 .. ,an}
and the amount of sums it can produce:
a1 - 1 st
a1 + a2 - 2 nd
a1 + a2 + a3 .. + an = n th
the total amount of possible sums here is n. and the total amount of distinct non-zero remainders would be n-1 because that's the maximum possible number of remainders you can get out of a number n. the pigeonhole principle states that at least two of these sums would have the same remainder when divided by n. This is good because if you subtract the two sums you are left with a remainder of zero meaning that the new subset would be divisible by n.
therefore we can prove that there exists a subset within {a1, a2 .. ,an} that it has a sum divisible by n (which is equivalent to the initial statement)
nice, what are the other ways of doing it?100% correct, very well done
The only other way I know of is significantly more cumbersome and involves first proving the statement for primes and subsequently showing that it holds for a product if it holds for the factors of said product. I learned it from @mindlessselfindulgenice, what are the other ways of doing it?
Integrate sqrt(x) * e^(-k * (x + (1 / x))) from 0 to infinity given k > 0
sqrt(π/k) exp(-2k) (1+1/(2k))
one way to obtain this is to use example 2 at <https://sos440.blogspot.com/2017/01/glassers-master-theorem.html> and differentiate under the integral sign w.r.t. s twice
another way is to substitute u = x + 1/x - 2 for 0 < x < 1 and x > 1 separately and simplify until you end up with exp(-2k) times the Laplace transform of (u+1) / sqrt(u)
my answer for the sqrt(x) integral is the same as that of Wolfram Alpha and both of the methods I outlined can easily be adapted to solve for the sqrt(1/x) integral, again concurring with Wolfram AlphaSo I wrote the problem wrong, it was a variant of Putnam 1985 B5, and the integrand should have sqrt(1 / x), not sqrt(x)
This makes it significantly easier to solve
For the new problem, my answer is wrong according to wolfram alpha:
integrate sqrt(x) * e^(-k * (x + 1/x)) from 0 to infinity - Wolfram|Alpha
Wolfram|Alpha brings expert-level knowledge and capabilities to the broadest possible range of people—spanning all professions and education levels.www.wolframalpha.com
prove that for every y > 1 there exist distinct x > 1 and z > 1 such that x^y^z = z^y^x
correctx * ln(y) + ln(ln(z)) = z * ln(y) + ln(ln(x))
ln(y) = (ln(ln(z)) - ln(ln(x))) / (z - x)
The derivative of ln(ln(x)) takes all positive real values.
There exists a tangent line of ln(ln(x)) with slope ln(y). Shifting this line down and finding the points of intersection with ln(ln(x)) gives values of x and z.
this is just the integral of x * floor(1/x) as x runs from 0 to 1, which can be written as the sum of n times the integral of x from x = 1/(n+1) to 1/n as n goes from 1 to infinity = the sum of (2n+1) / (2n(n+1)^2) as n goes from 1 to infinity = ½ * ( sum ( 1/n - 1/(n+1) ) + sum 1/(n+1)^2 ) = ½ * sum 1/n^2 = pi^2/12x is uniformly distributed on (0, 1)
Find E[x * floor(1 / x)]