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That was as easy as

zetafunc wrote:

Βεν Γ. Κυθισ wrote:Do you mean rather than ?it should be (a^m)n^a.

DANG IT! I forgot to move the variables around! And I forgot to confirm it with a calculator or *my miiiiind*.

Malia123 wrote:

If 1500 tickets

I think the answer might be 1500 tickets because if 1500 tickets, then 1500 tickets!

I learnt that sine, cosine, and hyperbolic tangent all have attractive fixed points.

is the cardinality of the naturals and is the cardinality of the real numbers.

Ricky wrote:

And to make sure your understanding is right, you should realize why Cantor's diagonal proof for the reals does not work for the rationals (which can have infinitely many digits as well).

Another good question: Is there a set which is larger than the integers (can't be paired with them), but smaller than the reals?

There is no set with a cardinality greater than

and less than .In mathematical terms: .

zetafunc wrote:

Hi Βεν,

Nice contribution! Yes, the repeated iteration of the cosine function converges: actually, it converges to thefixed pointof the cosine function,

That makes so much sense! I also just tried

and it converged to its fixed point, 0. Tan has a fixed point but it is not attractive; sinh becomes a vertical line on 0 which means there is no attractive point, 0; cosh has no fixed point but it converges to ∞; 0 is tanh's attractive fixed point; cot has no attractive point due to it being tan with the x axis shifted; sec is related to tan because it shares half of its discontinuous points with tan suggesting it has no attractive point, and it does have no attractive point; csc is sec with the x axis shifted so it has no fixed point.turtleeey wrote:

I found this website by looking up help on how to solve math problems, of course,

I found this site in a similar way! I wanted to know what the trigonometric functions were, specifically *sine, cosine, *and *tangent*. And the Math is Fun page was one of the top results that wasn't a Wikipedia article that expects you to know everything else in the field.

**Βεν Γ. Κυθισ**- Replies: 1

My first name is Benjamin, and if you are able to read ancient Greek, you could find out my last name. I like to do stuff sometimes. I like to playz the gamez (who doesn't?). I like to program; I like to use Python, Lua, and Scratch to code my stuff. I like to think of stories that almost always connect somehow to this huge complex megastory that I haven't finished yet (and no, I haven't uploaded any of my stories yet). I like to play around with physics simulators like The Powder Toy and Algodoo. I think about math a lot (which explains my presence on the Math is Fun fourm). I also go outside sometimes.

What does this mean?

**Βεν Γ. Κυθισ**- Replies: 2

So if you have f_0 = cos(n) and then f_1 = cos(f_0) and so on to get f_x = cos(f_(x-1)), and then let n be any real number, f_x converges to approximately

If you know anything that might relate to this new irrational number please respond to this post!Edit:

**I found out what this is, so don't clog the post with what you think it means, its meaning has already been revealed.**

zetafunc wrote:

Hi Βεν,

Welcome to the forum! Thanks for your contribution. That looks like a nice list. Some comments:

I know what you mean, but this can be misleading: is multiplied by itself times. For example, , where gets multiplied by itself one time.a^n=a multiplied by itself n times

The brackets should go around the here, i.e. for even .If n is even then (-a^n)=a^n

On one side of the equation, the and are the wrong way round. It should read:a^n÷a^m=a^(m-n) (Makes sense right?)

If n>0 then (a^m)na=a^(m+n)

What did you mean here?

I fixed my errors and typos. My answer to

zetafunc wrote:

What did you mean here?

is that I messed up; it shouldn't be (a^m)na, it should be (a^m)n^a. I feel so stupid for not checking with a calculator.

I have a piece of paper that has good exponentiation equations to memorize.

This is what is written on it at the time of this post (edited ones have an underline).

a^n=a multiplied by itself

~~n times~~n-1 timesIf n is even then (-a)^n=a^n

(abc⋯)^n=a^n×b^n×c^n×⋯

(1÷a)^n=1÷a^n

(a+b)^2=a^2+2ab+b^2

a^m×a^n×⋯=a^(m+n+⋯)

a^n÷a^m=a^(n-m) (Makes sense right?)

((a^m)^n)^⋯=a^(mn⋯)

If n>0 then (a^m)a^n=a^(m+n)

2^n-2^(n-1)=(2^n)÷2

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