## 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

1. Compassionate Group Title

It's one big math solving orgy in this thread.

2. bibby Group Title

This isn't even a forum nignog

3. skullpatrol Group Title

Does 0.999... = 1? Let the orgy begin.

4. Compassionate Group Title

No, because 0.999 will never exceed by 1 point.

5. Compassionate Group Title

Bibby, don't ruin the potential of this thread.

6. bibby Group Title

They eventually converge

7. Compassionate Group Title

If you try to derail my thread, I will steal your waifu

8. bibby Group Title

I don't even have a waifu. My question: is one-sided infatuation real love?

9. Aylin Group Title

@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 Group Title

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 Group Title

That's not a valid proof.

12. bibby Group Title

live footage of my waifu

13. skullpatrol Group Title
14. Compassionate Group Title

I got one for you guys. 1/0 = n n x 0 = 1

15. skullpatrol Group Title

Your 2nd equation violates the multiplicative property of 0. How can 0 times any number be 1?

16. Compassionate Group Title

Because every number is 0, n must be infinite.

17. Aylin Group Title

That's...not how vector fields work.

18. skullpatrol Group Title

Infinity is not a number. If it is give me its value?

19. Compassionate Group Title

Mata, mata. A x B = C C/A = B C/B = A 10 x 0 = 0 0/10 = 0 0/0 = 10

20. Aylin Group Title

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 Group Title

I can replace your 10 with "any number" to get 0/0 = "any number."

22. skullpatrol Group Title

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 Group Title

What happened to the party?

24. Aylin Group Title

How 'bout this: Prove that $\sum_{n=1}^{\infty} \frac{ 1 }{ k }$diverges.

25. muntoo Group Title

@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 Group Title

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 Group Title

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 Group Title

Bump ~~

29. muntoo Group Title

@Aylin Whoops: $\Large \sum_{n=0}^{k}{2^n}$$\Large \sum_{n=0}^{k}{3^n}$

30. Aylin Group Title

@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 }$