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Compassionate Group Title

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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  1. Compassionate Group Title
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    It's one big math solving orgy in this thread.

    • one year ago
  2. bibby Group Title
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    This isn't even a forum nignog

    • one year ago
  3. skullpatrol Group Title
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    Does 0.999... = 1? Let the orgy begin.

    • one year ago
  4. Compassionate Group Title
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    No, because 0.999 will never exceed by 1 point.

    • one year ago
  5. Compassionate Group Title
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    Bibby, don't ruin the potential of this thread.

    • one year ago
  6. bibby Group Title
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    They eventually converge

    • one year ago
  7. Compassionate Group Title
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    If you try to derail my thread, I will steal your waifu

    • one year ago
  8. bibby Group Title
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    I don't even have a waifu. My question: is one-sided infatuation real love?

    • one year ago
  9. Aylin Group Title
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    @skullpatrol .99999999... repeating is the same as 1. Proof: x=.9999999999 10x=9.999999999 10x-x=9.999999999-.99999999999 9x=9 x=1

    • one year ago
  10. Compassionate Group Title
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    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
  11. Dido525 Group Title
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    That's not a valid proof.

    • one year ago
  12. bibby Group Title
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    live footage of my waifu

    • one year ago
  13. skullpatrol Group Title
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    http://en.wikipedia.org/wiki/0.999...

    • one year ago
  14. Compassionate Group Title
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    I got one for you guys. 1/0 = n n x 0 = 1

    • one year ago
  15. skullpatrol Group Title
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    Your 2nd equation violates the multiplicative property of 0. How can 0 times any number be 1?

    • one year ago
  16. Compassionate Group Title
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    Because every number is 0, n must be infinite.

    • one year ago
  17. Aylin Group Title
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    That's...not how vector fields work.

    • one year ago
  18. skullpatrol Group Title
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    Infinity is not a number. If it is give me its value?

    • one year ago
  19. Compassionate Group Title
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    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
  20. Aylin Group Title
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    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
  21. skullpatrol Group Title
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    I can replace your 10 with "any number" to get 0/0 = "any number."

    • one year ago
  22. skullpatrol Group Title
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    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
  23. skullpatrol Group Title
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    What happened to the party?

    • one year ago
  24. Aylin Group Title
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    How 'bout this: Prove that \[\sum_{n=1}^{\infty} \frac{ 1 }{ k }\]diverges.

    • one year ago
  25. muntoo Group Title
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    @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
  26. muntoo Group Title
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    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
  27. Aylin Group Title
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    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
  28. Compassionate Group Title
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    Bump ~~

    • one year ago
  29. muntoo Group Title
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    @Aylin Whoops: \[\Large \sum_{n=0}^{k}{2^n}\]\[\Large \sum_{n=0}^{k}{3^n}\]

    • one year ago
  30. Aylin Group Title
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    @muntoo \[\sum_{n=0}^{k}2^{n}=1+2+4+8+16+32+...\frac{ 2^{k} }{ 4 }+\frac{ 2^{k} }{ 2 }+2^{k}\]\[=1+2(1+2+4+8+...+\frac{ 2^{k} }{ 8 }+\frac{ 2^{k} }{ 4 }+\frac{ 2^{k} }{ 2 })\]\[=2 \times 2^{k}(\frac{ 1 }{ 2 }+\frac{ 1 }{ 4 }+\frac{ 1 }{ 8 }+...+\frac{ 2 }{ 2^{k} }+\frac{ 1 }{ 2^{k} })+1\]\[=2 \times 2^{k} \times \sum_{n=1}^{k}\frac{ 1 }{ 2^{n} }+1=2 \times 2^{k} \times (1-2^{-k})+1\]\[=2 \times 2^{k}-2+1=2 \times 2^k-1\] \[\sum_{n=0}^{k}3^{n}=1+3+9+27+81+...+\frac{ 3^{k} }{ 9 }+\frac{ 3^{k} }{ 3 }+3^{k}\]\[=3 \times 3^{k} \times (\frac{ 1 }{ 3 }+\frac{ 1 }{ 9 }+\frac{ 1 }{ 27 }+...+\frac{ 3 }{ 3^{k} }+\frac{ 1 }{ 3^{k} })+1\]\[=3 \times 3^{k} \times \sum_{n=1}^{k}\frac{ 1 }{ 3^{n}}+1\]\[=3 \times 3^{k} \times (\frac{ 1 }{ 2 }-\frac{ 1 }{ 2 \times 3^{k} })+1\]\[=\frac{ 3 }{ 2 }3^{k}-\frac{ 3 }{ 2 }=\frac{ 3 \times 3^{k} }{ 2 }-\frac{ 1 }{ 2 }\]

    • one year ago
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