Consider the curves:

$$\displaystyle f: \Big(\frac{x-c}{a}\Big)^{2k+1} + \Big(\frac{y-d}{b}\Big)^{2k+1}=1$$ $$\displaystyle g: y = -\frac{b}{a}x +d+ \frac{bc}{a}$$ Where $a$,$b$,$c$,$d$,$k$ $\in$ $N.$
Let the area enclosed by the curves $f$ and $g$ be $\psi(k).$
Also, $$\displaystyle \lim_{n \to \infty} \psi(k) = \xi(a,b).$$ Find:$$\phi(2)=\displaystyle \sum_{\substack{a,b \geq 1 \\ a\geq b}} \xi \Big(\frac{1}{a^2},\frac{1}{b^2}\Big)$$ Anyone who wishes to generalize this can compute- $$\displaystyle \phi(m)= \sum_{\substack{a,b \geq 1 \\ a\geq b}} \xi \Big(\frac{1}{a^m},\frac{1}{b^m}\Big)$$ Where $m>1$.

After seeing the parameters involved I have no idea how to proceed answering this question. Can somebody please suggest an approach for solving this question?


closed as off-topic by TheSimpliFire, RRL, max_zorn, Cesareo, mrtaurho May 7 at 13:43

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  • $\begingroup$ Both $\;f,g\;$ definitions are hard to understand for me: what exactly does $\;f=something=1\;$ mean?! It means $\;f\;$ is the constant function equal to $\;1\;$ ...? I doubt it. And you call them "curves", not "surfaces" or something like that. I really don't understand. $\endgroup$ – DonAntonio May 6 at 8:09
  • $\begingroup$ I also don't think so $\endgroup$ – Ramone Estevez May 6 at 8:25
  • 2
    $\begingroup$ @DonAntonio I read that as "$f$ is given by the equation $something = 1$". I know it's a bit of abuse of the $:=$ symbol, but at least they do not have an actual equality symbol there. $\endgroup$ – Arthur May 6 at 8:26
  • $\begingroup$ @Arthur That sounds reasonable. $\endgroup$ – DonAntonio May 6 at 8:42
  • $\begingroup$ Can you guys give any suggestions ? $\endgroup$ – Ramone Estevez May 6 at 11:33

Maybe the following is a correct start. Note that the curve $g$ is defined by $y-d=-\frac{b}{a}(x-c)$. In that sense, $c$ and $d$ are just vertical and horizontal shift parameters, respectively, and they will not change the value of any enclosed area. So we can just take $c=d=0$. Also, since $2k+1$ is odd, substituting $y=-\frac{b}{a}$ into the formula for $f$ yields the equation $0=1$, so there is no other possibility than that $f$ and $g$ don't intersect so that $g$ is an asymptote of $f$. Then we already obtain $\xi(a,b)=b\cdot\lim_{k\to\infty}\int_{-\infty}^\infty\left(\sqrt[2k+1]{1-\left(\frac{x}{a}\right)^{2k+1}}+\frac{x}{a}\right)dx$. I suspect that this integral limit is equal to $2a$ - but don't immediately know how to prove this. In the end, we would have $\xi(a,b)=2ab$ so that $\phi(m)=2\cdot\sum_{a=1}^\infty\frac{1}{a^m}\sum_{b=a}^{\infty}\frac{1}{b^m}$. I don't have a quick idea about how to solve this - maybe something with zeta functions?


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