Sharp EL-W506T vs. Sharp EL-W516T

[Mjim]

(11-19-2019 04:19 PM)Pjwum Wrote:  Mjim, your result is correct. I messed up km and m.

(Of course 1/r^2 converges and so does the indefinite integral over 1/r^2 from rE to inf. Express the energy to bring a mass probe from Earth to infinity in space as kinetic energy at rE and you get the so called escape velocity which is independant from the mass probe (11.2 km/s for our planet)).

1/r^2 drops really fast and is reduced to 1/10th at about 1/20th the distance to the moon or 0.05 light seconds from Earth. Poor EL-W506 takes its default Simpson n=100 and averages the first trapezoid over a distance of 1/100 ly = 315360 light seconds, which results in a huge error right from the start.

As far as I have seen the Sharp doesn't offer default settings for its integration algorithm, as Bob asked. You have to - optionally - make an educated guess for n every time you start an integration instead of specifying your required eps. Of course this is beyond any high school level.

Patrick

To be honest I've been stuck teaching myself calculus II (stuck on the center of mass section for several weeks now), so the Simpson integration method, error function, convergence...well they're relatively fresh in my head. As they're new concepts, I still find them pretty interesting and challenging, even if they are part of an introductory level to Calculus.

The question about working out the energy of an object traveling in space was very helpful in a way as it was a fairly practical application of many new concepts I had learned.

As you pointed out having 1/100 intervals over 1 light year (each would be 3.6525 light days), is worthless in this case, since just about all the energy required for the entire trip happens within the first interval giving rise to the huge inaccuracies. I know the Casio fx-570ms/991ms will at least alter the number of divisions based on the integral and it's interval, but I'm guessing it doesn't break up the integral into multiple pieces since it refused to solve this problem ('math error').

I tried breaking up the integral and managed to get a better result on the Sharp using the default sample rate of n=100:

(C = G*mE*1000 = 3.98589*10^17, r = distance to center of earth (and what we are integrating with respect to), c = speed of light)

integral(C/r^2, 6.371*10^6, c/10) +
integral(C/r^2, c/10, c) +
integral(C/r^2, c, 10c) +
integral(C/r^2, 10c, 100c) +
integral(C/r^2, 100c, 1000c) +
integral(C/r^2, 1000c, 3600c) + (3600c = 1 light hour)
integral(C/r^2, 3600c, 3600c*24) + (3600c*24 = 1 light day)
integral(C/r^2, 3600c*24, 3600c*24*10) +
integral(C/r^2, 3600c*24*10, 3600c*24*100)+
integral(C/r^2, 3600c*24*100, 3600c*24*365.25) (3600c*24*365.25 = 1 light year)
= 6.256305944*10^10

So 2000 divisions (10 integrals of n=100 (which is 200 divisions) ) will give a better result than 32,768 divisions if you concentrate the samples where most of the change is happening (at the start).

Really interesting stuff, which shows just how important good algorithm design is for calculators.

EDIT:
Oh, wow, just reread your first point, plugging this into the formula E=1/2*m*v^2 (m=1000kg, E = Energy required to travel 1ly from the earth) does give the escape velocity (11,186m/s is what I got). Great to see how these fit together (I don't have a general physics background, so seeing how these equations fit together is educational; thanks!).