Question

Lets get a good thread going on here.

Post mathematical questions here, and we all help each other out by solving them.

Answer 1

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

Answer 2

This isn't even a forum nignog

Answer 3

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

Answer 4

No, because 0.999 will never exceed by 1 point.

Answer 5

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

Answer 6

They eventually converge

Answer 7

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

Answer 8

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

Answer 9

@skullpatrol
.99999999... repeating is the same as 1.

Proof:

x=.9999999999
10x=9.999999999
10x-x=9.999999999-.99999999999
9x=9
x=1

Answer 10

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.

Answer 11

That's not a valid proof.

Answer 12

live footage of my waifu

Answer 13

http://en.wikipedia.org/wiki/0.999...

Answer 14

I got one for you guys.


1/0 = n

n x 0 = 1

Answer 15

Your 2nd equation violates the multiplicative property of 0.

How can 0 times any number be 1?

Answer 16

Because every number is 0, n must be infinite.

Answer 17

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

Answer 18

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

Answer 19

Mata, mata.

A x B = C
C/A = B
C/B = A

10 x 0 = 0
0/10 = 0
0/0 = 10

Answer 20

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.

Answer 21

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

Answer 22

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

Answer 23

What happened to the party?

Answer 24

How 'bout this:

Prove that \[\sum_{n=1}^{\infty} \frac{ 1 }{ k }\]diverges.

Answer 25

@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.

Answer 26

Remove the summation ("simplify"):

\[ \large \sum_{n=0}^{k}{2^k} \]

Think you're so tough? Simplify:

\[ \large \sum_{n=0}^{k}{3^k} \]

Answer 27

Er...

@muntoo the way you've written them the answers would just be \[k \times 2^{2}\]and\[k \times 3^{k}\]respectively.

Answer 28

Bump ~~

Answer 29

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

Answer 30

@muntoo

\[\sum_{n=0}^{k}2^{n}=1+2+4+8+16+32+...\frac{ 2^{k} }{ 4 }+\frac{ 2^{k} }{ 2 }+2^{k}\]\[=1+2(1+2+4+8+...+\frac{ 2^{k} }{ 8 }+\frac{ 2^{k} }{ 4 }+\frac{ 2^{k} }{ 2 })\]\[=2 \times 2^{k}(\frac{ 1 }{ 2 }+\frac{ 1 }{ 4 }+\frac{ 1 }{ 8 }+...+\frac{ 2 }{ 2^{k} }+\frac{ 1 }{ 2^{k} })+1\]\[=2 \times 2^{k} \times \sum_{n=1}^{k}\frac{ 1 }{ 2^{n} }+1=2 \times 2^{k} \times (1-2^{-k})+1\]\[=2 \times 2^{k}-2+1=2 \times 2^k-1\]

\[\sum_{n=0}^{k}3^{n}=1+3+9+27+81+...+\frac{ 3^{k} }{ 9 }+\frac{ 3^{k} }{ 3 }+3^{k}\]\[=3 \times 3^{k} \times (\frac{ 1 }{ 3 }+\frac{ 1 }{ 9 }+\frac{ 1 }{ 27 }+...+\frac{ 3 }{ 3^{k} }+\frac{ 1 }{ 3^{k} })+1\]\[=3 \times 3^{k} \times \sum_{n=1}^{k}\frac{ 1 }{ 3^{n}}+1\]\[=3 \times 3^{k} \times (\frac{ 1 }{ 2 }-\frac{ 1 }{ 2 \times 3^{k} })+1\]\[=\frac{ 3 }{ 2 }3^{k}-\frac{ 3 }{ 2 }=\frac{ 3 \times 3^{k} }{ 2 }-\frac{ 1 }{ 2 }\]