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Lets get a good thread going on here.
Post mathematical questions here, and we all help each other out by solving them.
 one year ago
 one year ago
Lets get a good thread going on here. Post mathematical questions here, and we all help each other out by solving them.
 one year ago
 one year ago

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CompassionateBest ResponseYou've already chosen the best response.0
It's one big math solving orgy in this thread.
 one year ago

bibbyBest ResponseYou've already chosen the best response.0
This isn't even a forum nignog
 one year ago

skullpatrolBest ResponseYou've already chosen the best response.0
Does 0.999... = 1? Let the orgy begin.
 one year ago

CompassionateBest ResponseYou've already chosen the best response.0
No, because 0.999 will never exceed by 1 point.
 one year ago

CompassionateBest ResponseYou've already chosen the best response.0
Bibby, don't ruin the potential of this thread.
 one year ago

bibbyBest ResponseYou've already chosen the best response.0
They eventually converge
 one year ago

CompassionateBest ResponseYou've already chosen the best response.0
If you try to derail my thread, I will steal your waifu
 one year ago

bibbyBest ResponseYou've already chosen the best response.0
I don't even have a waifu. My question: is onesided infatuation real love?
 one year ago

AylinBest ResponseYou've already chosen the best response.5
@skullpatrol .99999999... repeating is the same as 1. Proof: x=.9999999999 10x=9.999999999 10xx=9.999999999.99999999999 9x=9 x=1
 one year ago

CompassionateBest ResponseYou've already chosen the best response.0
Bibby, of course it's real love. Don't you get butterflies in your stomach when your waifu comes to cheer you up? Also, pic related, Mai Waifu is drinking tea. Ayin, nope, your equation is nonsensical.
 one year ago

Dido525Best ResponseYou've already chosen the best response.0
That's not a valid proof.
 one year ago

bibbyBest ResponseYou've already chosen the best response.0
live footage of my waifu
 one year ago

skullpatrolBest ResponseYou've already chosen the best response.0
http://en.wikipedia.org/wiki/0.999...
 one year ago

CompassionateBest ResponseYou've already chosen the best response.0
I got one for you guys. 1/0 = n n x 0 = 1
 one year ago

skullpatrolBest ResponseYou've already chosen the best response.0
Your 2nd equation violates the multiplicative property of 0. How can 0 times any number be 1?
 one year ago

CompassionateBest ResponseYou've already chosen the best response.0
Because every number is 0, n must be infinite.
 one year ago

AylinBest ResponseYou've already chosen the best response.5
That's...not how vector fields work.
 one year ago

skullpatrolBest ResponseYou've already chosen the best response.0
Infinity is not a number. If it is give me its value?
 one year ago

CompassionateBest ResponseYou've already chosen the best response.0
Mata, mata. A x B = C C/A = B C/B = A 10 x 0 = 0 0/10 = 0 0/0 = 10
 one year ago

AylinBest ResponseYou've already chosen the best response.5
Incorrect. If you start with A x B = C, you can only say C/A=B if you exclude the possibility that A might be 0. Likewise with C/B=A.
 one year ago

skullpatrolBest ResponseYou've already chosen the best response.0
I can replace your 10 with "any number" to get 0/0 = "any number."
 one year ago

skullpatrolBest ResponseYou've already chosen the best response.0
10 x 0 = 0 or a*0 = 0 0/10 = 0 or 0/a = 0 0/0 = 10 or 0/0 = a where a = any number
 one year ago

skullpatrolBest ResponseYou've already chosen the best response.0
What happened to the party?
 one year ago

AylinBest ResponseYou've already chosen the best response.5
How 'bout this: Prove that \[\sum_{n=1}^{\infty} \frac{ 1 }{ k }\]diverges.
 one year ago

muntooBest ResponseYou've already chosen the best response.0
@Aylin assuming n=k, http://en.wikipedia.org/wiki/Harmonic_series_(mathematics)#Divergence If n is not equal to k, I give up. The problem is too difficult for me. Maybe some Wiles will come along and write a 300 page proof.
 one year ago

muntooBest ResponseYou've already chosen the best response.0
Remove the summation ("simplify"): \[ \large \sum_{n=0}^{k}{2^k} \] Think you're so tough? Simplify: \[ \large \sum_{n=0}^{k}{3^k} \]
 one year ago

AylinBest ResponseYou've already chosen the best response.5
Er... @muntoo the way you've written them the answers would just be \[k \times 2^{2}\]and\[k \times 3^{k}\]respectively.
 one year ago

muntooBest ResponseYou've already chosen the best response.0
@Aylin Whoops: \[\Large \sum_{n=0}^{k}{2^n}\]\[\Large \sum_{n=0}^{k}{3^n}\]
 one year ago

AylinBest ResponseYou've already chosen the best response.5
@muntoo \[\sum_{n=0}^{k}2^{n}=1+2+4+8+16+32+...\frac{ 2^{k} }{ 4 }+\frac{ 2^{k} }{ 2 }+2^{k}\]\[=1+2(1+2+4+8+...+\frac{ 2^{k} }{ 8 }+\frac{ 2^{k} }{ 4 }+\frac{ 2^{k} }{ 2 })\]\[=2 \times 2^{k}(\frac{ 1 }{ 2 }+\frac{ 1 }{ 4 }+\frac{ 1 }{ 8 }+...+\frac{ 2 }{ 2^{k} }+\frac{ 1 }{ 2^{k} })+1\]\[=2 \times 2^{k} \times \sum_{n=1}^{k}\frac{ 1 }{ 2^{n} }+1=2 \times 2^{k} \times (12^{k})+1\]\[=2 \times 2^{k}2+1=2 \times 2^k1\] \[\sum_{n=0}^{k}3^{n}=1+3+9+27+81+...+\frac{ 3^{k} }{ 9 }+\frac{ 3^{k} }{ 3 }+3^{k}\]\[=3 \times 3^{k} \times (\frac{ 1 }{ 3 }+\frac{ 1 }{ 9 }+\frac{ 1 }{ 27 }+...+\frac{ 3 }{ 3^{k} }+\frac{ 1 }{ 3^{k} })+1\]\[=3 \times 3^{k} \times \sum_{n=1}^{k}\frac{ 1 }{ 3^{n}}+1\]\[=3 \times 3^{k} \times (\frac{ 1 }{ 2 }\frac{ 1 }{ 2 \times 3^{k} })+1\]\[=\frac{ 3 }{ 2 }3^{k}\frac{ 3 }{ 2 }=\frac{ 3 \times 3^{k} }{ 2 }\frac{ 1 }{ 2 }\]
 one year ago
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