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

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Compassionate
 3 years ago
Best ResponseYou've already chosen the best response.0It's one big math solving orgy in this thread.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0This isn't even a forum nignog

skullpatrol
 3 years ago
Best ResponseYou've already chosen the best response.0Does 0.999... = 1? Let the orgy begin.

Compassionate
 3 years ago
Best ResponseYou've already chosen the best response.0No, because 0.999 will never exceed by 1 point.

Compassionate
 3 years ago
Best ResponseYou've already chosen the best response.0Bibby, don't ruin the potential of this thread.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0They eventually converge

Compassionate
 3 years ago
Best ResponseYou've already chosen the best response.0If you try to derail my thread, I will steal your waifu

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I don't even have a waifu. My question: is onesided infatuation real love?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0@skullpatrol .99999999... repeating is the same as 1. Proof: x=.9999999999 10x=9.999999999 10xx=9.999999999.99999999999 9x=9 x=1

Compassionate
 3 years ago
Best ResponseYou've already chosen the best response.0Bibby, 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.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0That's not a valid proof.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0live footage of my waifu

Compassionate
 3 years ago
Best ResponseYou've already chosen the best response.0I got one for you guys. 1/0 = n n x 0 = 1

skullpatrol
 3 years ago
Best ResponseYou've already chosen the best response.0Your 2nd equation violates the multiplicative property of 0. How can 0 times any number be 1?

Compassionate
 3 years ago
Best ResponseYou've already chosen the best response.0Because every number is 0, n must be infinite.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0That's...not how vector fields work.

skullpatrol
 3 years ago
Best ResponseYou've already chosen the best response.0Infinity is not a number. If it is give me its value?

Compassionate
 3 years ago
Best ResponseYou've already chosen the best response.0Mata, mata. A x B = C C/A = B C/B = A 10 x 0 = 0 0/10 = 0 0/0 = 10

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Incorrect. 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.

skullpatrol
 3 years ago
Best ResponseYou've already chosen the best response.0I can replace your 10 with "any number" to get 0/0 = "any number."

skullpatrol
 3 years ago
Best ResponseYou've already chosen the best response.010 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

skullpatrol
 3 years ago
Best ResponseYou've already chosen the best response.0What happened to the party?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0How 'bout this: Prove that \[\sum_{n=1}^{\infty} \frac{ 1 }{ k }\]diverges.

anonymous
 3 years ago
Best 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.

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Er... @muntoo the way you've written them the answers would just be \[k \times 2^{2}\]and\[k \times 3^{k}\]respectively.

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

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