# Calculation beyond computional limits

1. Jun 2, 2010

### short circut

So i am ttying to calculate a velocity in this problem i dreamt up. Only problem i cant get a computer to solve it. So i was hoping someone here could help

i am trying to find the value of x such that

$$\frac{\sqrt{1-x^2}}{x}=1.0967*10^{-86}$$

This arose in me trying to calculate somewhat relativistically how fast a raindrop would have to travel to for into a black hole. My result is very crude.

edit:

I think i have it now. And i cant find delete

Last edited: Jun 2, 2010
2. Jun 2, 2010

### rasmhop

Let a = 1.0967*10^{-86}

Square the equation to get it to get
$$\frac{1-x^2}{x^2} =a^2$$
Now solve for x^2 and take the positive square root (x must be positive since a is positive).

x is going to be VERY close to 1 (in fact I expect most software will report x=1 if you ask for a numerical answer).

3. Jun 2, 2010

### short circut

Yeah everything is reporting 1. Thats what i was trying to get around. I dont think i can get around it. Because that gives me a ratio of sqrt(1/(1.(86 zeros)1)) Which i cant even imagine. So basically you have to get that v/c ratio of the speed of light to turn a raindrop into a blackhole semirelativistically. I am not well versed in relativistic fluid dynamics so i imagine this is quite a ways off. Plus not to mention i only have 3 sig figs. So i could never actually measure this anyways.

4. Jun 2, 2010

### litup

Don't you have a computing center in your school? Oh, maybe you are in HS, sorry then. Most colleges have computers that with something like Mathematica, can give quadruple and more precision. At 16 digits per precision, I guess you would need sextuple precision or so, it certainly can be done with the right software.

5. Jun 2, 2010

### rasmhop

Well if you really want a numeric approximation I just asked Maple to compute 250 digits of this. I'm not sure exactly how Maple does its floating-point computation, but I suspect the long string of 0's at the end is a sign that it doesn't handle such high-precision numbers by default (but up to something like 150 digits it seems correct).

0.9999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999398624555000000000000000000000000000000000000000000000000000000000000000000000

I have no idea how this could be of any use however.

6. Jun 3, 2010

### uart

No need need for any computational power here. You can solve it using a binomial expansion to a VERY close approximation as :

$$x \simeq 1 - \frac{a^2}{2}$$

where $a= 1.0967 \times 10^{-86}$

Last edited: Jun 3, 2010