The Math Problems Thread :D

[miyano_shiho]

Hello!  ;D

I thought creating this thread/topic would be helpful. As suggested by Shubho from another topic about a math problem, this thread would be the place to put any future math problems.

If you have a problem you can't solve, a topic you want to discuss, an interesting fact you want to share, or even a puzzle you want to post just for fun, you can freely post it here, as long as it's about math. Maybe some people, including me, may want to answer it.

Spoiler:

BTW, I know until calculus but nothing more advanced than that. (not a math major  :-\)

I hope other people who also like math find their place here. I will also occasionally post math trivia (or something similar) here, since I love math very much.

[Jecka]

I think I'll use this when I start college

[miyano_shiho]

Then I hope this thread will still be alive by then.
(And I hope it's just not only me who posts here. *first time creating a thread* )

[Jecka]

At least you've created one unlike me

I'll support you though

[miyano_shiho]

Thanks Jecka! 

[miyano_shiho]

Well, to start with, here are some random things I posted in the "Count to Forever" thread.

Trivia Set # 1 

0! = ?

Spoiler: Answer and Proof

Answer is 1. Since (n-1)! = n!/n, 0! = (1-1)! = 1!/1 = 1

0.99999... = 1

Spoiler:

Note it is equal not approximately equal. There are many proofs, including the infinite sum. (0.9 + 0.09 + 0.009 + ... = 1) Although personally, it still looks weird for me. 

Spoiler: Solution

Spoiler:

Answer is 2. Try to solve this one yourself. (It's the same as the previous one. )

More to come soon... 

[shubhoshinichi]

Well, miyano.. Count me in. And i guess you have already counted me in. Well as you know i'm in love with physics. But, that definitely doesn't really mean i don't like maths. I really love maths and i think the creator of this thread knows that and i will always be there say in problems, suggestion, solving and creating as well. Thanks miyano.

[miyano_shiho]

Hahaha By default, you should be part of this thread (and your name was mentioned in the first post anyway...). Thanks too

[Kleene Onigiri]

0.99999... = 1

Spoiler:

Note it is equal not approximately equal. There are many proofs, including the infinite sum. (0.9 + 0.09 + 0.009 + ... = 1) Although personally, it still looks weird for me. 

Best explanation for me was that:

1/3= 0.333...

1/3 + 1/3 + 1/3 = 1
0.333.... +0.333... +0.333... = 0.999... = 1

[Giogio]

A math thread

Here's a nice one (worked on the follow-ups of this with my best friend for quite some time, eventually causing me to decide to study math ^^):

((x)) := x^x^x^x^x^x^x^x^x^x....
with the smallest brackets being on the very right side, like this
((x)) := x^(x^(x^(x^(x^(x^(x^(x^(x^(x^....(x^x)
Also, lets note the n^th step with an index, like ((x))₃ = x^(x^x)

We called them iterated powers.

So, obviously
((1)) =1
((2)) -> +oo
((0)) jumps between 0 and 1  ((0))₁ = 1, ((0))₂=0

It's clear for every x>1 ((x)) diverges versus +infinity, for 0 it jumps, and in between ((x)) converges, right?
Wrong.
Try 1.1, or 0.1.
So, for which x will ((x)) turn out to diverge, conerge or oszillate? Try to find out, it's fun
I'll post the solution and some more stuff next week when I have some time. If you're interested, ofc.

[meitantei_mayu]

I don't remember any of these types of mathematical problems anymore. :X

*_* You all are mathematical geniuses!

Count me in if it's an accounting problem though. 

[Giogio]

I'm not used to the english vocabulary here - by domain, do you mean "the set of input values for which the function is defined"?

In that case, yes, if you were not yet introduced to complex numbers, |R would be the answer.
|R\{1, -1} would be |R without 1 and -1;
(|R/{1, -1} does not make a lot of sense to me, there are such notations in group theory, but with {1,-1} those don't fit)

Hm, what were the other choices?

[Kor]

This thread made me realize how little I actually remember from high school math (three years ago)

[miyano_shiho]

...This thread actually had posts! 

@Kleene: That proof seems more convincing. 
@Meitantei_mayu: Thanks!
@Conan324: I agree with Giogio. |R would be my answer too.
@Kor: 

@Giogio:

It's clear for every x>1 ((x)) diverges versus +infinity, for 0 it jumps, and in between ((x)) converges, right?
Wrong.
Try 1.1, or 0.1.
So, for which x will ((x)) turn out to diverge, conerge or oszillate? Try to find out, it's fun
I'll post the solution and some more stuff next week when I have some time. If you're interested, ofc.

Indeed, not all values for 0<x<1 converge. As already mentioned, for 0, indexes with odd numbers have an answer equal to 0, even numbers, 1. I haven't really worked out the problem yet, and this is actually new to me (higher calculus, perhaps).
Please do post the solution... I'm really interested about it. 
Thanks for offering to put some more math stuff in this thread. 

Spoiler:

I wish I were a math major too ... Don't ask me why I'm not, it breaks my heart. 

[Akonyl]

((0)) jumps between 0 and 1  ;D ((0))₁ = 1, ((0))₂=0

I'd just like to point out here that the assumption that I assume you're making here of 0^0=1 (so that it oscillates between 0 and 1) isn't always correct. There's a good post about it here, but it boils down to "0^0=1 because we really wish it did because it makes things pretty", not "0^0=1 because x^0=1". The assumption that 0^0=1 is only made because it would be inconvenient to make equations account for both possibilities.

not to mention, you define ((x))₃ as x^x^x, which means that ((x))₂ would be x^x, which means that ((x))₁ is just x, in which case, how is ((0))₁ == 1? I could see maybe if you meant ((0))₀ though, or if you improperly defined ((x))₃ as x^x^x.

also, I'm not sure what you mean by "Try 1.1, or 0.1", because those seem to fit the assumption of "1.1 goes to +inf, and 0.1 converges" that you said they break, so  :-X


edit: And I knew your operator already existed, but I was having trouble remembering where I saw it, but here: http://en.wikipedia.org/wiki/Tetration, 2^2^2^2 is just denoted as ⁴2.