e=MC squared.
Einstein stole my formula.
e=MC squared.
Einstein stole my formula.
Jim Ryun = 3:24
It is impossible to separate a cube into two cubes, or a fourth power into two fourth powers, or in general, any power higher than the second, into two like powers. I have discovered a truly marvelous proof of this, which this forum is too narrow to contain
Sally Vixens wrote:
We can't even see Question 5.6 in preview - and you call someone else's response "pathetic?"
Sally, even if you saw it, you wouldn't understand its statement, hence its acceptable omission. Regardless, here's the statement,
Let U be an open cover of X, acyclic with respect to sheaf F, i.e.,
for each simplex s, if r > 0, then H^r(U_s, F|U_s) = 0.
Then H^*(U, F) is isomorphic to H^*(X, F),
where H^*(U, F) is the Cech cohomology.
I was messing around on a napkin and figured out that no three positive integers a, b, and c can satisfy the equation a^n + b^n = c^n for any integer value of n greater than two.
Probably no big deal, but it's kind of cool.
How many girls to approach in a night to get a one night stand,
How many minutes to talk to my girl each day to make her not bitch about me not paying her attention. Pretty complex math involved here
frank reynolds wrote:
Sally Vixens wrote:We can't even see Question 5.6 in preview - and you call someone else's response "pathetic?"
Sally, even if you saw it, you wouldn't understand its statement, hence its acceptable omission. Regardless, here's the statement,
Let U be an open cover of X, acyclic with respect to sheaf F, i.e.,
for each simplex s, if r > 0, then H^r(U_s, F|U_s) = 0.
Then H^*(U, F) is isomorphic to H^*(X, F),
where H^*(U, F) is the Cech cohomology.
Well, genuis Frank ... how about showing the solution here? Or, have you even bought the book yet? Please don't look in the back for the answer. Please solve on your own if able to.
Sally Vixens wrote:
frank reynolds wrote:Sally, even if you saw it, you wouldn't understand its statement, hence its acceptable omission. Regardless, here's the statement,
Let U be an open cover of X, acyclic with respect to sheaf F, i.e.,
for each simplex s, if r > 0, then H^r(U_s, F|U_s) = 0.
Then H^*(U, F) is isomorphic to H^*(X, F),
where H^*(U, F) is the Cech cohomology.
Well, genuis Frank ... how about showing the solution here? Or, have you even bought the book yet? Please don't look in the back for the answer. Please solve on your own if able to.
Sally, please go play with your puppy dogs and ponytails now and leave real thought to the men.
By the way, you're welcome for introducing you to the world of modern mathematics, a la Alexander Grothendieck. Have you even heard of a sheaf or cohomology, or a something as basic as an open cover before? No! Go away and add some integers or something.
5 cents + 5 cents
Still haven't figured that one out yet. But, my teacher says its okay to graduate me anyway because 80% of the class knows the answer to this question.
Solve the Schrödinger equation, Hˆψ = Eψ, as a second order
differential equation
frank reynolds wrote:
We can't even see Question 5.6 in preview - and you call someone else's response "pathetic?"
Well, genuis Frank ... how about showing the solution here? Or, have you even bought the book yet? Please don't look in the back for the answer. Please solve on your own if able to.[/quote]
Sally, please go play with your puppy dogs and ponytails now and leave real thought to the men.
By the way, you're welcome for introducing you to the world of modern mathematics, a la Alexander Grothendieck. Have you even heard of a sheaf or cohomology, or a something as basic as an open cover before? No! Go away and add some integers or something.[/quote]
How clever of you Frank Reynolds to tell me to go play with my puppy dogs and ad integers. What a really swell guy you appear to be - a real winner.
You still have not solved that problem. Have you still not bought that book? You calling that one response pathetic then telling me to play with puppy dogs. What a POS you are. Actually - my knowledge of calculus perhaps better than yours. I entered business but excelled in mathematics in spite of that. Please show solution. Oh wait - you are a poser - you never really do stuff - you stuff pretend to be smart. Idiot!
Sally Vixens wrote:
frank reynolds wrote:Sally, even if you saw it, you wouldn't understand its statement, hence its acceptable omission. Regardless, here's the statement,
Let U be an open cover of X, acyclic with respect to sheaf F, i.e.,
for each simplex s, if r > 0, then H^r(U_s, F|U_s) = 0.
Then H^*(U, F) is isomorphic to H^*(X, F),
where H^*(U, F) is the Cech cohomology.
Well, genuis Frank ... how about showing the solution here? Or, have you even bought the book yet? Please don't look in the back for the answer. Please solve on your own if able to.
Complexity is totally subjective. Just because someone else doesn't know algebraic geometry doesn't make them stupid and claiming they are only shows your ignorance. In fact, I suspect you're just some snotty undergraduate who thinks they know everything. How about you go try to prove the main theorem of Complex Multiplication. Or the main theorem of Iwasawa Theory. Or go calculate bounds on the arithmetic rank of an elliptic curve with non-trivial Tate-Shaferavich group. You do know how to do 2-descent don't you?
I once thought to solve all the
Jackson's ElectroMagnetics exercises because I thought all the Chinese students had an answer-book. Got halfway through chapter 2.
Per our illustrious president, BO, it is 2-1.
a2 + b2 = c2
PF wrote:
It is impossible to separate a cube into two cubes, or a fourth power into two fourth powers, or in general, any power higher than the second, into two like powers. I have discovered a truly marvelous proof of this, which this forum is too narrow to contain
Hey, you too? Small world.
j-invariant wrote:
Sally Vixens wrote:Well, genuis Frank ... how about showing the solution here? Or, have you even bought the book yet? Please don't look in the back for the answer. Please solve on your own if able to.
Complexity is totally subjective. Just because someone else doesn't know algebraic geometry doesn't make them stupid and claiming they are only shows your ignorance. In fact, I suspect you're just some snotty undergraduate who thinks they know everything. How about you go try to prove the main theorem of Complex Multiplication. Or the main theorem of Iwasawa Theory. Or go calculate bounds on the arithmetic rank of an elliptic curve with non-trivial Tate-Shaferavich group. You do know how to do 2-descent don't you?
Hi, j-invariant. Please read non-maths more carefully. I neither implicity or explicitly stated Ms. Sally Vixens was stupid. In fact, I agree wholeheartedly with you; having experience with or exposure to research mathematics is no sign of intelligence, just a sign of experience and desire (and loneliness). Go make a friend. In the world of living, breathing people, not academia wannabes.
And Sally, I'll leave you alone now. Please go pursue your business ventures with your renewed self-esteem. You're welcome. Which calculus do you refer to exactly? Tensor calculus, index calculus, Kirby calculus? Oh, perhaps you mean business calculus. Hold your head high, little girl. You are worthy.
Honestly...
probably something like this:
-2[-3(x − 2y) + 4y]
I have numerical dyslexia so the above could look like:
(2x + 3)(x2 − x − 5).
for all I know.
If I spent a very long time and maybe an entire sheet of paper I could maybe solve it.
might as well jump wrote:
Theorem: If {s1, s2, s3, ...} converges to S and {t1, t2, t3, ...} converges to T, then {(s1+t1), (s2+t2), (s3+t3), ...} converges to S + T.
Proof:. Choose ∂ > 0. Since {sn} -> S, there exists an integer N1 such that |sn - S| N1. Since {tn} -> T, there exists an integer N2 such that |tn - T| N2. Hence, if N = max{N1, N2}, then
|(sn+tn) - (S + T)| = |(sn - S) + (tn - T)| ≤ |sn - S| + |tn - T| `` ∂
whenever n ≥ N, which shows {(sn+tn)} -%% S+T.
Though I can follow this.
Whatever/
Well I started a theorem, but it's incomplete.