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i have the answer, so who can tell me how pissed randall is at verizon?

deleting all my posts.

story is, don't tick off a mathematician (thats all i know)

Classic!

I couldn't do the math...but I had to look it up to see wtf.

Dude was pissed.

We'll need another $ amount in the poll.

Well I'll take a stab.

I think the answer is 0.002.

The last term appears to be asymptotic and would approach the value of one.

If that is the case, then the equation is really 0.002 + Eulers Identity (which=0).

Thus 0.002.

fossilgrom-

You should conclude your post with:

WHAT NOW, B!TCHES?

sigma(1/2^n,1,infinity)<or = to integral(1/n^2,1,infinity)dn, let b=infinity
--> <or = to limit,b-->infinity of integral(1/n^2,1,b)dn
--> <or = to limit,b-->infinity of 1/n),1,b] by fundamental theorem of calculus
--> < or = to limit,b-->infinity of (-1/1)-(-1/b), 1 divided by infinity=0
Therefore sigma(1/2^n,1,infinity)<or= -1, assume equals b/c test for divergence sucks and it's a power series anyways.

e^(i*pi)=cos(pi)+i*sin(pi)=1

0.002+sigma(1/2^n,1,infinity)+e^(i*pi)=0.002-1+1
=$0.002

therefore, Randall got jipped .

pra_ggresion wrote:sigma(1/2^n,1,infinity)<or = to integral(1/n^2,1,infinity)dn, let b=infinity
--> <or = to limit,b-->infinity of integral(1/n^2,1,b)dn
--> <or = to limit,b-->infinity of 1/n),1,b] by fundamental theorem of calculus
--> < or = to limit,b-->infinity of (-1/1)-(-1/b), 1 divided by infinity=0
Therefore sigma(1/2^n,1,infinity)<or= -1, assume equals b/c test for divergence sucks and it's a power series anyways.

e^(i*pi)=cos(pi)+i*sin(pi)=1

0.002+sigma(1/2^n,1,infinity)+e^(i*pi)=0.002-1+1
=$0.002

therefore, Randall got jipped .

", B!TCHES!" C'mon, pra, close the deal.

pra_ggresion wrote:sigma(1/2^n,1,infinity)<or = to integral(1/n^2,1,infinity)dn, let b=infinity
--> <or = to limit,b-->infinity of integral(1/n^2,1,b)dn
--> <or = to limit,b-->infinity of 1/n),1,b] by fundamental theorem of calculus
--> < or = to limit,b-->infinity of (-1/1)-(-1/b), 1 divided by infinity=0
Therefore sigma(1/2^n,1,infinity)<or= -1, assume equals b/c test for divergence sucks and it's a power series anyways.

e^(i*pi)=cos(pi)+i*sin(pi)=1

0.002+sigma(1/2^n,1,infinity)+e^(i*pi)=0.002-1+1
=$0.002

therefore, Randall got jipped .

A hour later I realized I made a mistake. I was thinking to myself, "self how could a summation of only positive values yeild a negative value?"
I accidentally switched the lower and upper limit of integral while implementing the Fundamenal Theorem of Calc. So the Summation should be 0-(-1)=1 making
0.002+sigma(1/2^n,1,infinity)+e^(i*pi)=0.002+1+1=2.002
Randall still got jipped IMHO.

Pra, you still have a sign error. Check Euler's identity.

holddown wrote:Pra, you still have a sign error. Check Euler's identity.

I agree.....

dagnab I quit. cos(pi)=-1