Compassionate
  • Compassionate
Lets get a good thread going on here. Post mathematical questions here, and we all help each other out by solving them.
Mathematics
katieb
  • katieb
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Compassionate
  • Compassionate
It's one big math solving orgy in this thread.
bibby
  • bibby
This isn't even a forum nignog
skullpatrol
  • skullpatrol
Does 0.999... = 1? Let the orgy begin.

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Compassionate
  • Compassionate
No, because 0.999 will never exceed by 1 point.
Compassionate
  • Compassionate
Bibby, don't ruin the potential of this thread.
bibby
  • bibby
They eventually converge
Compassionate
  • Compassionate
If you try to derail my thread, I will steal your waifu
bibby
  • bibby
I don't even have a waifu. My question: is one-sided infatuation real love?
anonymous
  • anonymous
@skullpatrol .99999999... repeating is the same as 1. Proof: x=.9999999999 10x=9.999999999 10x-x=9.999999999-.99999999999 9x=9 x=1
Compassionate
  • Compassionate
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.
anonymous
  • anonymous
That's not a valid proof.
skullpatrol
  • skullpatrol
http://en.wikipedia.org/wiki/0.999...
Compassionate
  • Compassionate
I got one for you guys. 1/0 = n n x 0 = 1
skullpatrol
  • skullpatrol
Your 2nd equation violates the multiplicative property of 0. How can 0 times any number be 1?
Compassionate
  • Compassionate
Because every number is 0, n must be infinite.
anonymous
  • anonymous
That's...not how vector fields work.
skullpatrol
  • skullpatrol
Infinity is not a number. If it is give me its value?
Compassionate
  • Compassionate
Mata, mata. A x B = C C/A = B C/B = A 10 x 0 = 0 0/10 = 0 0/0 = 10
anonymous
  • anonymous
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.
skullpatrol
  • skullpatrol
I can replace your 10 with "any number" to get 0/0 = "any number."
skullpatrol
  • skullpatrol
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
skullpatrol
  • skullpatrol
What happened to the party?
anonymous
  • anonymous
How 'bout this: Prove that \[\sum_{n=1}^{\infty} \frac{ 1 }{ k }\]diverges.
anonymous
  • anonymous
@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
  • anonymous
Remove the summation ("simplify"): \[ \large \sum_{n=0}^{k}{2^k} \] Think you're so tough? Simplify: \[ \large \sum_{n=0}^{k}{3^k} \]
anonymous
  • anonymous
Er... @muntoo the way you've written them the answers would just be \[k \times 2^{2}\]and\[k \times 3^{k}\]respectively.
Compassionate
  • Compassionate
Bump ~~
anonymous
  • anonymous
@Aylin Whoops: \[\Large \sum_{n=0}^{k}{2^n}\]\[\Large \sum_{n=0}^{k}{3^n}\]
anonymous
  • anonymous
@muntoo \[\sum_{n=0}^{k}2^{n}=1+2+4+8+16+32+...\frac{ 2^{k} }{ 4 }+\frac{ 2^{k} }{ 2 }+2^{k}\]\[=1+2(1+2+4+8+...+\frac{ 2^{k} }{ 8 }+\frac{ 2^{k} }{ 4 }+\frac{ 2^{k} }{ 2 })\]\[=2 \times 2^{k}(\frac{ 1 }{ 2 }+\frac{ 1 }{ 4 }+\frac{ 1 }{ 8 }+...+\frac{ 2 }{ 2^{k} }+\frac{ 1 }{ 2^{k} })+1\]\[=2 \times 2^{k} \times \sum_{n=1}^{k}\frac{ 1 }{ 2^{n} }+1=2 \times 2^{k} \times (1-2^{-k})+1\]\[=2 \times 2^{k}-2+1=2 \times 2^k-1\] \[\sum_{n=0}^{k}3^{n}=1+3+9+27+81+...+\frac{ 3^{k} }{ 9 }+\frac{ 3^{k} }{ 3 }+3^{k}\]\[=3 \times 3^{k} \times (\frac{ 1 }{ 3 }+\frac{ 1 }{ 9 }+\frac{ 1 }{ 27 }+...+\frac{ 3 }{ 3^{k} }+\frac{ 1 }{ 3^{k} })+1\]\[=3 \times 3^{k} \times \sum_{n=1}^{k}\frac{ 1 }{ 3^{n}}+1\]\[=3 \times 3^{k} \times (\frac{ 1 }{ 2 }-\frac{ 1 }{ 2 \times 3^{k} })+1\]\[=\frac{ 3 }{ 2 }3^{k}-\frac{ 3 }{ 2 }=\frac{ 3 \times 3^{k} }{ 2 }-\frac{ 1 }{ 2 }\]

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