Here's the question you clicked on:
Compassionate
Lets get a good thread going on here. Post mathematical questions here, and we all help each other out by solving them.
It's one big math solving orgy in this thread.
This isn't even a forum nignog
Does 0.999... = 1? Let the orgy begin.
No, because 0.999 will never exceed by 1 point.
Bibby, don't ruin the potential of this thread.
If you try to derail my thread, I will steal your waifu
I don't even have a waifu. My question: is one-sided infatuation real love?
@skullpatrol .99999999... repeating is the same as 1. Proof: x=.9999999999 10x=9.999999999 10x-x=9.999999999-.99999999999 9x=9 x=1
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.
That's not a valid proof.
I got one for you guys. 1/0 = n n x 0 = 1
Your 2nd equation violates the multiplicative property of 0. How can 0 times any number be 1?
Because every number is 0, n must be infinite.
That's...not how vector fields work.
Infinity is not a number. If it is give me its value?
Mata, mata. A x B = C C/A = B C/B = A 10 x 0 = 0 0/10 = 0 0/0 = 10
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.
I can replace your 10 with "any number" to get 0/0 = "any number."
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
What happened to the party?
How 'bout this: Prove that \[\sum_{n=1}^{\infty} \frac{ 1 }{ k }\]diverges.
@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.
Remove the summation ("simplify"): \[ \large \sum_{n=0}^{k}{2^k} \] Think you're so tough? Simplify: \[ \large \sum_{n=0}^{k}{3^k} \]
Er... @muntoo the way you've written them the answers would just be \[k \times 2^{2}\]and\[k \times 3^{k}\]respectively.
@Aylin Whoops: \[\Large \sum_{n=0}^{k}{2^n}\]\[\Large \sum_{n=0}^{k}{3^n}\]
@muntoo \[\sum_{n=0}^{k}2^{n}=1+2+4+8+16+32+...\frac{ 2^{k} }{ 4 }+\frac{ 2^{k} }{ 2 }+2^{k}\]\[=1+2(1+2+4+8+...+\frac{ 2^{k} }{ 8 }+\frac{ 2^{k} }{ 4 }+\frac{ 2^{k} }{ 2 })\]\[=2 \times 2^{k}(\frac{ 1 }{ 2 }+\frac{ 1 }{ 4 }+\frac{ 1 }{ 8 }+...+\frac{ 2 }{ 2^{k} }+\frac{ 1 }{ 2^{k} })+1\]\[=2 \times 2^{k} \times \sum_{n=1}^{k}\frac{ 1 }{ 2^{n} }+1=2 \times 2^{k} \times (1-2^{-k})+1\]\[=2 \times 2^{k}-2+1=2 \times 2^k-1\] \[\sum_{n=0}^{k}3^{n}=1+3+9+27+81+...+\frac{ 3^{k} }{ 9 }+\frac{ 3^{k} }{ 3 }+3^{k}\]\[=3 \times 3^{k} \times (\frac{ 1 }{ 3 }+\frac{ 1 }{ 9 }+\frac{ 1 }{ 27 }+...+\frac{ 3 }{ 3^{k} }+\frac{ 1 }{ 3^{k} })+1\]\[=3 \times 3^{k} \times \sum_{n=1}^{k}\frac{ 1 }{ 3^{n}}+1\]\[=3 \times 3^{k} \times (\frac{ 1 }{ 2 }-\frac{ 1 }{ 2 \times 3^{k} })+1\]\[=\frac{ 3 }{ 2 }3^{k}-\frac{ 3 }{ 2 }=\frac{ 3 \times 3^{k} }{ 2 }-\frac{ 1 }{ 2 }\]