@Behaviour Well, one can get the pizza hat without outside input. Despite all the encouragement I gave to get mine, I think I got four answers from people outside this room in the first half hour (plus yours, plus Huy's).
I meant I'm not about to ask questions that can be answered within 30 minutes, let alone by 5 different people. My questions are supposed to be hard. :)
When you integrate the equality w.r.t $t$, that's what you get I think: $$\displaystyle \dfrac{1}{1-y^2} \dfrac{\text{d}y}{\text{d}t} = 1 \implies \int \dfrac{1}{1-y^2} \text{ d}y = \int \text{d}t$$
@KhallilBenyattou: In case you are interested in solving it, it is a lot easier if you know a lot of ansatz-functions (still not trivial then, imo). I can give you the proper ansatz, if you want
@KhallilBenyattou: I'm making my way to bed. Do you want the ansatz, or do you want to try till tomorrow? Or should I mail it to you or something and you can look at it whenever you want?
@KhallilBenyattou: I'm making my way to bed. Do you want the ansatz, or do you want to try till tomorrow? Or should I mail it to you or something and you can look at it whenever you want?
Could someone hep me to understand this proof? http://math.stackexchange.com/questions/1076161/the-ring-is-a-principal-ideal-domain-especially-an-integral-domain
it is the partition generating function * $\binom{\text{number of partitions of }n}{\text{into an even number of parts}}-\binom{\text{number of partitions of }n}{\text{into an odd number of parts}}$
@DonLarynx I am trying to find the truncation to degree 10
@DonLarynx Therefore, I need to expand out that power series representation into a power series
@DonLarynx still there?
i've noticed that this chat tends to slow down after 6
nobody types things anymore, hahah
@DonLarynx here it is in better form. i realized that some of the exponents didn't turn into LaTeX. $\frac{1}{(1+x)(1−x)(1+x^2)(1−x^2)(1+x^3)(1−x^3)....}$
If $k$ is an odd divisor of $2b$, call $2k$ the *friend* of $k$. Split the divisors of $2b$ into pairs of friends. For example, if $b=45$, we have the following pairs of friends.
We have split the divisors of $2b$ into pairs of friends. Each pair has one odd number and one even number, so $2b$ has exactly as many odd divisors as even divisors.
But my question is: How does this ensure we didn't leave out any even divisors?
Suppose $f$ is an even divisor we left out. Then we have $f = 2e$ where e is some odd number. But this is a friend, and so this is included in our original list.
Then suppose $e$ is even. Then $f = 2*2*g$ for some integer. but $b$ can't be divisible by 4. RAA
@robjohn sensei can you take a look at the second part of this question please if it is possible to derive the coefficients of the polynomial via combinatorial identities (avoid reference to Cheby polynomials) :-) I tried for a while but couldn't get anything useful !
I'm stuck! How do I expand this out to get the truncation of degree 10? $(1−x+x^2−x^3...)(1+x+x^2+x^3+....)(1−x^2+x^4−x^6....)(1+x^2+x^4+x^6....)(1−x^3+x^6−x^9....)(1+x^3+x^6+x^9+....)....$
@JohnMerlino because cross-multiplication is simply multiplying the reciprocal by its reciprocal on each side. For example, $\frac{5}{7} = \frac{40}{56} \Longrightarrow 56*\frac{5}{7} = 40$ and so on.
@anon (3x + 1)/4*8x/(9x^2 – 1). You are allowed to cancel the 4 and 8 even though they are different fractions. I understand to cancel a fraction within itself, but being allowed to cancel two different fractions with each, it's hard to see why it's allowed
@JohnMerlino because (a/b)(c/d) equals (ac)/(bd); all the top things are up top and all the bottom things are down below, it doesn't matter in which fraction a term originated, only whether it's up top or down below
So far, I've unlocked one of the secret hats. I'm wondering what other secret hats are out there and how to earn them?
