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for you mathematicians

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the check is for

1279$
0
No votes
55$
0
No votes
5053$
0
No votes
536$
0
No votes
143$
1
14%
super pissed
3
43%
fk math
3
43%
 
Total votes : 7

for you mathematicians

Postby wyosurf » Fri Sep 19, 2008 9:06 am

Image
i have the answer, so who can tell me how pissed randall is at verizon?
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Postby Hermgruf » Fri Sep 19, 2008 11:03 am

deleting all my posts.
Last edited by Hermgruf on Fri Feb 03, 2012 3:57 pm, edited 1 time in total.
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Postby wyosurf » Fri Sep 19, 2008 11:25 am

story is, don't tick off a mathematician (thats all i know)
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Postby stinkbug » Fri Sep 19, 2008 11:40 am

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.
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Postby fossilgrom » Fri Sep 19, 2008 12:50 pm

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.
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Postby stinkbug » Fri Sep 19, 2008 2:51 pm

fossilgrom-

You should conclude your post with:

WHAT NOW, B!TCHES?
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Postby pra_ggresion » Sat Sep 20, 2008 1:56 pm

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 .
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Postby stinkbug » Sat Sep 20, 2008 4:41 pm

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.
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Postby pra_ggresion » Tue Sep 23, 2008 11:40 am

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.
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Postby holddown » Thu Sep 25, 2008 7:46 am

Pra, you still have a sign error. Check Euler's identity.
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Postby navier-stokes » Fri Sep 26, 2008 3:10 am

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


I agree.....
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Postby pra_ggresion » Tue Sep 30, 2008 9:36 am

dagnab I quit. cos(pi)=-1
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