Girls vs boys maths challenge

Ambitious
Learn to draw? That would take me forever

I keep thinking "That can't be too hard!" - after all, it's just lines on paper. I'm getting a little bit better without any practice at all, too

I have no idea what just happened . . .

If only… the trouble is that it's hard enough that I can't easily get started on it

Haha. I guess then, unlike me, you're not a malco.
Graph sketches? Not for me.
My art teacher in year 8 told me to drop art

Don't worry, graph sketches don't have to be accurate, at any level. If the basic features are all there (cuts the axes/has turning points/asymptotes (if you don't know what any of those words mean, they aren't relevant to you yet - these rules will keep working right through to doing a maths degree) the right number of times in the right order, does vaguely like what it's supposed to do as it goes to +/- infinity, , and that's about it), you're fine. It doesn't have to be pretty. If you have to make the gap between "0" and "1" be twice the size of the gap between "1" and "2" on some axis, that's fine. Just start off with the basic bits - draw the x-axis, the points where the graph cuts that axis, ditto for the y-axis, what it does at the ends, where it goes flat, anything else interesting that happens, then just join them all up and you're done.

I keep thinking "That can't be too hard!" - after all, it's just lines on paper. I'm getting a little bit better without any practice at all, too

Drawing technique is not hard, it can be developed quite rapidly. It is being able to see and portray whatever you are drawing or painting that takes years. Similarly, a lot of mathematical technique (using this term loosely) can be learned quite quickly. However, it doesn't necessarily make you 'see'. Take many A-level mathematics students who can routinely apply methods but struggle where thinking is involved, the situation is similar. So just like for mathematics, the curriculum for art also fails to equip students with the ability to 'see', that ability takes practice and is more rewarding.

Yeah, I think I know what you mean by "see" - I've slowly been gaining this ability through my attempts to become a better introspector and rationalist. I can, with some effort of will, "just see the lines" that are in front of me - as a corollary to my learning to see what is really there in a metaphorical sense (as in, see what the evidence presents you, not what you want to get out of the evidence, in rationality training).

I'm so confused . . .

Solution

Using the integral definition of the Gamma function,

$\displaystyle \begin{aligned} \sum_{n=1}^{\infty} \frac{(-1)^n \Gamma \left(2n-\frac{4}{5} \right)}{(2n)!} = \int_0^{\infty} \left( \sum_{n=1}^{\infty} \frac{(-1)^n t^{2n}}{(2n)!} \right) t^{-\frac{9}{5}} e^{-t} \ dt = \int_0^{\infty} t^{-\frac{9}{5}} e^{-t} ( \cos t - 1) \ dt \end{aligned}[br]\displaystyle \begin{aligned} = \frac{1}{2} \int_0^{\infty} t^{-\frac{9}{5}} \left(e^{-t(1-i)}+e^{-t(1+i)}-2e^{-t} \right) \ dt = \frac{1}{2} \Gamma \left( - \frac{4}{5} \right) \left( \frac{1}{(1-i)^{-\frac{4}{5}}}+\frac{1}{(1+i)^{-\frac{4}{5}}}-2 \right) \end{aligned}[br]\displaystyle = \frac{1}{4} \Gamma \left(- \frac{4}{5} \right) (2^{\frac{2}{5}} (1+\sqrt{5})-4)$

Edit: Just realised the justification of this is 'sketchy' as we are inventing an integral representation for $\Gamma(z)$ for Re(z)<0 but, as we define it as such, the consequences are also valid by the Bohr-Möllerup theorem and it's application in extending the definition of the Gamma function to these aloes through series representation (i.e. as the answer exists, an invalid representation of the Gamma function actually gives an answer when we use the evaluative formula with the assumption that it is valid and the representations are consistent, we must deduce that this answer is correct).

$\displaystyle \begin{aligned} \Rightarrow 2^{\frac{3}{5}} \left(1+ 4\sum_{n=0}^{\infty} \frac{(-1)^{n+1} \Gamma \left(2n-\frac{4}{5} \right)}{5 (2n)! \Gamma \left( \frac{1}{5} \right)} \right) =2^{\frac{3}{5}} \left(1-\frac{\Gamma \left( - \frac{4}{5} \right)}{5 \Gamma \left( \frac{1}{5} \right)} (2^{\frac{2}{5}} (1+\sqrt{5})-4) \right) = \frac{1+ \sqrt{5}}{2} \ \square \end{aligned}$

Yes, the thread's been derailed somewhat. . .

Problem:
Find: (without using IBP)
$\displaystyle\int e^{ax}\cos(bx) \ dx$
and:
$\displaystyle\int e^{ax}\sin(bx) \ dx$

dafaq is that?

Erm, what level of maths have you come across?

Yes, the thread's been derailed somewhat. . .

Problem:
Find: (without using IBP)
$I = \displaystyle\int e^{ax}\cos(bx) \ dx$
and:
$J = \displaystyle\int e^{ax}\sin(bx) \ dx$

just finished gcse maths!

dafaq is that?

$I+iJ = \displaystyle \int e^{ax} (\cos(bx)+i \sin(bx)) dx = \int e^{ax + ibx} dx = \dfrac{1}{a+ib} e^{(a+ib) x} + C+iD = e^{a x} \dfrac{a - i b}{a^2 + b^2} e^{ib x}+C+iD$.
So $I = \text{Re}(I+iJ) = e^{a x} \cos(bx) \dfrac{a}{a^2+b^2} + e^{ax} \dfrac{b}{a^2+b^2} \sin(bx) + C$; $J = \text{Im}(I+iJ) = e^{a x} \sin(bx) \dfrac{a}{a^2+b^2} - e^{a x} \cos(b x) \dfrac{b}{a^2+b^2} + D$.

Yup, looks good (of course), although it's less scary in factorised form.

Factorisation is overrated :P
(I lie, of course; factorisation is one of the single most useful tools there is.)