Recent content by quadricode

I suppose I will give it a try. I need to take a little break. :/

I have that antiderivative in my notes. But, I don't know how to derive it. Moreover, that is not the antiderivative I want. I found an equivalent form which might lead to something interesting. ;)

I was evaluating the integral \frac{1}{2^n}\int^\infty_0 \mathrm{sech}^n\, x\, dx without any information (such as reduction formulae) beforehand. I needed an original idea for a project I am working on. If anyone wants to help me personally, I don't mind giving my notes (which contains the...

I have turned the whole thing...or at least most of the sum into binomial coefficients. And I looked for hours upon hours for some nice relations. But I could get to nothing. I even got an expression without any inverse (multiplicative) binomial coefficients (if I did my algebra correctly)...

You do not require Calculus in high school, but require at least three years of some type of math.

In the USA, Calculus can be taught in the junior and/or senior year of high school. The program is for students who are taking "an accelerated route" to mathematics. You can receive college credits by passing an exam called the Advanced Placement Calculus AB and/or BC exam, which is run by the...

The square brackets have no meaning (other than to group expressions); they are just used in the traditional fashion to avoid many parentheses. So, yes, [z]=(z)=z.

jostpuur: z! reads "z factorial." If z \in \mathbb{N}, then z!=z(z-1)(z-2)\cdots 2\cdot 1. For z=0, then 0!=1. This idea can also be extended to all real numbers (with a restriction). If z \in \mathbb{R}, then z!=\Gamma(z+1), where \Gamma(\cdot) is the Gamma function. Notice that z cannot be a...

I have tried well over 30 cumulative hours trying to evaluate this double series: S=\sum_{k=0}^{m}\sum_{j=0}^{k+m-1}(-1)^{k}{m \choose k}\frac{[2(k+m)]!}{(k+m)!^{2}}\frac{(k-j+m)!^{2}}{(k-j+m)[2(k-j+m)]!}\frac{1}{2^{k+j+m+1}}, for some integer m>0. I have simplified (or maybe complicated) the...