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Compassionate

  • 2 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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  1. Compassionate
    • 2 years ago
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    It's one big math solving orgy in this thread.

  2. bibby
    • 2 years ago
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    This isn't even a forum nignog

  3. skullpatrol
    • 2 years ago
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    Does 0.999... = 1? Let the orgy begin.

  4. Compassionate
    • 2 years ago
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    No, because 0.999 will never exceed by 1 point.

  5. Compassionate
    • 2 years ago
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    Bibby, don't ruin the potential of this thread.

  6. bibby
    • 2 years ago
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    They eventually converge

  7. Compassionate
    • 2 years ago
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    If you try to derail my thread, I will steal your waifu

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

  9. Aylin
    • 2 years ago
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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

  10. Compassionate
    • 2 years ago
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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.

  11. Dido525
    • 2 years ago
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    That's not a valid proof.

  12. bibby
    • 2 years ago
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    live footage of my waifu

  13. skullpatrol
    • 2 years ago
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    http://en.wikipedia.org/wiki/0.999...

  14. Compassionate
    • 2 years ago
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    I got one for you guys. 1/0 = n n x 0 = 1

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

  16. Compassionate
    • 2 years ago
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    Because every number is 0, n must be infinite.

  17. Aylin
    • 2 years ago
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    That's...not how vector fields work.

  18. skullpatrol
    • 2 years ago
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    Infinity is not a number. If it is give me its value?

  19. Compassionate
    • 2 years ago
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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

  20. Aylin
    • 2 years ago
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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.

  21. skullpatrol
    • 2 years ago
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    I can replace your 10 with "any number" to get 0/0 = "any number."

  22. skullpatrol
    • 2 years ago
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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

  23. skullpatrol
    • 2 years ago
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    What happened to the party?

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

  25. muntoo
    • 2 years ago
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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.

  26. muntoo
    • 2 years ago
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    Remove the summation ("simplify"): \[ \large \sum_{n=0}^{k}{2^k} \] Think you're so tough? Simplify: \[ \large \sum_{n=0}^{k}{3^k} \]

  27. Aylin
    • 2 years ago
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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.

  28. Compassionate
    • 2 years ago
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    Bump ~~

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

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

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