Vtyjkyuk

Actual problem is

"Is $theta=pi$ valid for polar coordinate representation of parabola"

Where it's equatioin is $theta(r)=arccos(fracar-1)$ a is some positive constant

My friend says if $ r to infty$ then it will yield $theta (r=infty)=pi$

but i don't think so(also, for i've seen several references of parabola omitting r goes infinity or $theta=(2n-1)pi$ or $2npi$
where n is integer)

Thus I want to see arccosine's behavior around -1 so that i'll show that

$limlimits_r to infty frac1r$ is not small enough to make $limlimits_r to inftyarccos(fracar -1)=pi$

I've thought about taylor expanding arccos(x) around x=-1

and i have found

arccos(-1+x)=$pi -frac1sqrt 1-z^2x -fracz(1-z^2)^3/2fracx^22$.......
where z=-1 x:positive and very small

it seems that every coefficients explode for z=-1 and i see that even every derivatives of coefficients explode when z=-1

so i think every coefficients are far greatly bigger term even multiplied with very small x which $lim x to 0$

Therefore I think inverse cosine is too sensitive to make $limlimits_r to inftyarccos(fracar-1)=pi$(or should i say sum doesn't converge however x is small in polynomial sense?)

I'm bit into math, and know little bit of analysis but i haven't studied analysis deeply

1. is $theta=pi$ valid for parabola (or $2npi$ but i'm sure you know what i mean) r=$fraca1+costheta$ <--primary question




2. want to know $limlimits_r to inftyarccos(fracar-1)=pi$ even if arccosine is super sensitive or whatever

The polar equation

$$r(costheta +1)=a$$ indeed represents parabolas, and close to $theta =pi$, you can write

$$a=r(cos(pi+delta) +1)=r(-cosdelta+1)approx fracrdelta^22.$$

When $r$ tends to infinity, $delta$ to zero and $theta$ to $pi$ (as fast as $r^-1/2$). There is no mystery here.

Q1. $theta=pi$ is of course not allowed, as $rcdot0=a$ has no solution.

Q2. By continuity of the arc cosine,

$$lim_rtoinftyarccosleft(frac ar-1right)=arccosleft(lim_rtoinftyfrac ar-1right)=pi.$$

Note that the Taylor expansion of the arc cosine around $-1$ does not exist.

Yes the limit will go to pi. It's not about sensitivity. In fact a function does not even have to be defined at the limit. The definition is for all epsilom > 0 there exists a finite r that gives an angle in pi +/- epsilom