## Compassionate 3 years ago Lets get a good thread going on here. Post mathematical questions here, and we all help each other out by solving them.

1. Compassionate

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

2. bibby

This isn't even a forum nignog

3. skullpatrol

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

4. Compassionate

No, because 0.999 will never exceed by 1 point.

5. Compassionate

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

6. bibby

They eventually converge

7. Compassionate

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

8. bibby

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

9. Aylin

@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

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

That's not a valid proof.

12. bibby

live footage of my waifu

13. skullpatrol
14. Compassionate

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

15. skullpatrol

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

16. Compassionate

Because every number is 0, n must be infinite.

17. Aylin

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

18. skullpatrol

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

19. Compassionate

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

20. Aylin

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

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

22. 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

23. skullpatrol

What happened to the party?

24. Aylin

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

25. muntoo

@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

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

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

Bump ~~

29. muntoo

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

30. Aylin

@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 }$