I'm no math major but...

[Cruiser]

...yesterday at 12 o'clock the SNOTEL snow depth reading for bert was 61.9". As of 12 o'clock today it was 76.6". The decimel points give me problems, but I'm pretty sure that there is a lot of snow falling up there. Who is going up in the morning?

[lemon boy]

Nobody, the pass is closed. and pssst...don't look at the snow depth, it's next to useless. What is useful though is the water content.

[ptavv]

If the snotel is anything like we have here it's more susecptible to blown snow than it is to fallen snow.

Inches of H2O is the number that matters.

[Steven S. Dallas]

Inches of H2O is the number that matters.

Length or width?

Bow-chicka-wow-wow, you're getting boned by a huge snow dick.

[Odin]

you're getting boned by a huge snow dick.

Where's the subtlety?

NOTE:

Huge Snow WANG!

Second Note:

I am an EXPERT Math Major.

[Buster Highmen]

Where's the subtlety?

I am an EXPERT Math Major.

OK, so what's the real connective k-theory of Brown Gitler Spectra?

[Odin]

With or without redundant tangental illuminent spectra?

[Cruiser]

Nobody, the pass is closed.

If they keep it closed into the morning tomorrow, there's allways butler gulch. Anyone?

[Odin]

how bout the mod 2 cohomology of the Brown-gitler spectra

Let a=HF2(hf2) denote the mod 2 Steenrod algebra. Let B(n) denote the nth brown-gitler spectrum over HF2. For each integer n > 0, B(n) is a Z/2-complete specturm. Set B(0) =s0 completed at 2, and b(wn) ~ B(2n + 1). Then for N >2, there is an isomorphism of the left a-modules

[Buster Highmen]

With or without redundant tangental illuminent spectra?

Alpha blended with a mousse of equivariant normal bundles.

[Buster Highmen]

how bout the mod 2 cohomology of the Brown-gitler spectra

Let a=HF2(hf2) denote the mod 2 Steenrod algebra. Let B(n) denote the nth brown-gitler spectrum over HF2. For each integer n > 0, B(n) is a Z/2-complete specturm. Set B(0) =s0 completed at 2, and b(wn) ~ B(2n + 1). Then for N >2, there is an isomorphism of the left a-modules

You've found my PhD thesis.

Let [n/2] denote the integer less than or equal to n/2 for an integer n.

Consider the sequence generated by {s(i+1)=s(i)+s([i/2]):s(0)=0, s(1)=1}.

Does lim(i->infinty) s(i+1)/s(i) converge?

[ptavv]

What does "Let [n/2] denote the integer less than or equal to n/2 for an integer n." have to do with:

Consider the sequence generated by {s(i+1)=s(i)+s([i/2]):s(0)=0, s(1)=1}.

Does lim(i->infinty) s(i+1)/s(i) converge?

???

[Buster Highmen]

What does "Let [n/2] denote the integer less than or equal to n/2 for an integer n." have to do with:

Consider the sequence generated by {s(i+1)=s(i)+s([i/2]):s(0)=0, s(1)=1}.

Does lim(i->infinty) s(i+1)/s(i) converge?

???

A famous sequence is the Fibonnacci sequence.
It looks like: 1, 1, 2, 3, 5, 8, 13, 21,... where the next element of the sequence is generated by adding the previous 2.

In that notation, I'd write {f(i+1) = f(i)+f(i-1), f(0)=1, f(1)=1}.
Note that lim(i->inf) f(i+1)/f(i) = (1+sqrt(5))/2, tao the golden proportion.

So the sequence I'm talking about is generated by adding the previous element with one from just below or equal to halfway down the previous elements.

[Odin]

infinty) s(i+1)/s(i) converge?

Lets avoid trying to stee rhis coversation into a philosophical question.

I kant take it

[ptavv]

A famous sequence is the Fibonnacci sequence.
It looks like: 1, 1, 2, 3, 5, 8, 13, 21,... where the next element of the sequence is generated by adding the previous 2.

In that notation, I'd write {f(i+1) = f(i)+f(i-1), f(0)=1, f(1)=1}.
Note that lim(i->inf) f(i+1)/f(i) = (1+sqrt(5))/2, tao the golden proportion.

So the sequence I'm talking about is generated by adding the previous element with one from just below or equal to halfway down the previous elements.

Aside from the fact that the limit as i goes to infinity of the fibonnacci sequence is phi, not tao, my point was that "Let [n/2] denote the integer less than or equal to n/2 for an integer n." has nothing to do with the sequence you outlined.

[bio-smear]

threeve
...

[Steven S. Dallas]

You guys are ruining a perfectly good thread about giant snow dicks.

[Buster Highmen]

Aside from the fact that the limit as i goes to infinity of the fibonnacci sequence is phi, not tao, my point was that "Let [n/2] denote the integer less than or equal to n/2 for an integer n." has nothing to do with the sequence you outlined.

The limit of the quotient of the successive elements in the Fibonnacci sequence has been "tao" for some time.

And it has everything to do with it. Unless you can index a sequence by rational numbers.

[lemon boy]

I like beer.

[ptavv]

to get this back on topic

[iskibc]

I betcha my snow dick is bigger than yours.

[ptavv]

And it has everything to do with it. Unless you can index a sequence by rational numbers.

I just didn't see the point in it since we're an internet forum and not a mathematical journal.

Also: Evidently tau and phi both refer to the same golden ratio. Phi is simply more common. The more you knew. doo dee doo.

[xkyzero]

[jrbd]

What does "Let [n/2] denote the integer less than or equal to n/2 for an integer n." have to do with:

Consider the sequence generated by {s(i+1)=s(i)+s([i/2]):s(0)=0, s(1)=1}.

Does lim(i->infinty) s(i+1)/s(i) converge?

???

Look for the square brackets in s definition: it uses [i/2] which is the floor of i/2, as defined earlier for n. But the answer is 42.

[Odin]

I just didn't see the point in it since we're an internet forum and not a mathematical journal.

Also: Evidently tau and phi both refer to the same golden ratio. Phi is simply more common. The more you knew. doo dee doo.

if....

I = a math journal then icemang must also = a math gournal which means that everyone is a math journal.