How big is an egg, anyway?restart;IntroductionLUklbXJvd0c2Iy9JK21vZHVsZW5hbWVHNiJJLFR5cGVzZXR0aW5nR0koX3N5c2xpYkdGJzYjLUkjbWlHRiQ2JVEhRicvJSdpdGFsaWNHUSV0cnVlRicvJSxtYXRodmFyaWFudEdRJ2l0YWxpY0YnEllipses have fascinated mathematicians and scientists for ages, in part because of celestial mechanics (closed orbits) and other uses, such as cross sections of the Earth. They are also a good deal more mysterious than their closest cousin, the circle.Let 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be the equation of an ellipse, with 0 < b < 1 (so it's oriented horizontally). The case b=1 is a circle. It's common to define the eccentricity of the ellipse by e = sqrt(1-b^2);so that e=0 is a circle and e=1 is an ellipse so squashed that it has become a line segment. (Today's LUklbXJvd0c2Iy9JK21vZHVsZW5hbWVHNiJJLFR5cGVzZXR0aW5nR0koX3N5c2xpYkdGJzYjLUkjbWlHRiQ2JVEiZUYnLyUnaXRhbGljR1EldHJ1ZUYnLyUsbWF0aHZhcmlhbnRHUSdpdGFsaWNGJw== has nothing to do with the base of the natural exponential function.)The area of this ellipse is simply Pi*b. However, the perimeter (length of the curve) is considerably more complicated. There is an exact expression for it, in terms of a power series in e. The coefficient of the LUklbXJvd0c2Iy9JK21vZHVsZW5hbWVHNiJJLFR5cGVzZXR0aW5nR0koX3N5c2xpYkdGJzYjLUkjbWlHRiQ2JVEibkYnLyUnaXRhbGljR1EldHJ1ZUYnLyUsbWF0aHZhcmlhbnRHUSdpdGFsaWNGJw==th term isc:= n->-1/(2*n-1) * ((2*n)! / (2^n*n!)^2)^2 ;and the perimeter series itself isPerim:= 2*Pi * Sum( c(n)*e^(2*n), n=0..infinity );(Again, e in all of these formulas is eccentricity, a variable, and has nothing to do with the number exp(1).) We used a capitalized version of the Sum command, because with the lowercase command Maple tries to actually find the value of the sum. Numerically, here is how the series starts (after division by 2 Pi to keep things simpler).sum( c(n)*e^(2*n), n=0..6 );Now we use value to turn the Sum into a sum. The power series does converge, but not to any familiar function.Perim:= value(Perim);Because this "elliptic function" is not so easily understood, simpler approximate formulas were sought. One of the simplest isP[1]:= Pi*sqrt( 2*(1+b^2) - (1-b)^2/2 );If we convert this to a function of e, we findP[1]:= eval( P[1], b=sqrt(1-e^2) );We now take its Taylor series centered at e=0 to gettaylor( P[1]/(2*Pi), e=0, 12 );By comparison to , we can see that LUklbXJvd0c2Iy9JK21vZHVsZW5hbWVHNiJJLFR5cGVzZXR0aW5nR0koX3N5c2xpYkdGJzYjLUklbXN1YkdGJDYlLUkjbWlHRiQ2JVEiUEYnLyUnaXRhbGljR1EldHJ1ZUYnLyUsbWF0aHZhcmlhbnRHUSdpdGFsaWNGJy1GIzYjLUkjbW5HRiQ2JFEiMUYnL0Y2USdub3JtYWxGJy8lL3N1YnNjcmlwdHNoaWZ0R1EiMEYn gets the first four nonzero Taylor terms correct. The difference looks likeerr1:= Perim-P[1]: taylor( err1/(2*Pi), e=0, 12);Recall that LUklbXJvd0c2Iy9JK21vZHVsZW5hbWVHNiJJLFR5cGVzZXR0aW5nR0koX3N5c2xpYkdGJzYlLUkjbWlHRiQ2JVEiZUYnLyUnaXRhbGljR1EldHJ1ZUYnLyUsbWF0aHZhcmlhbnRHUSdpdGFsaWNGJy1JI21vR0YkNjBRIjxGJy9GM1Enbm9ybWFsRicvJSZmZW5jZUdRJmZhbHNlRicvJSpzZXBhcmF0b3JHRj0vJSlzdHJldGNoeUdGPS8lKnN5bW1ldHJpY0dGPS8lKGxhcmdlb3BHRj0vJS5tb3ZhYmxlbGltaXRzR0Y9LyUnYWNjZW50R0Y9LyUlZm9ybUdRJmluZml4RicvJSdsc3BhY2VHUS90aGlja21hdGhzcGFjZUYnLyUncnNwYWNlR0ZPLyUobWluc2l6ZUdRIjFGJy8lKG1heHNpemVHUSlpbmZpbml0eUYnLUkjbW5HRiQ2JEZURjk=. For values of e close to zero, 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is very small, and the other terms are smaller still. So this formula is excellent for ellipses that are near-circular. But the error does grow as we approach e=1; here we take a look at the relative error in the approximation as a function of LUklbXJvd0c2Iy9JK21vZHVsZW5hbWVHNiJJLFR5cGVzZXR0aW5nR0koX3N5c2xpYkdGJzYjLUkjbWlHRiQ2JVEiZUYnLyUnaXRhbGljR1EldHJ1ZUYnLyUsbWF0aHZhcmlhbnRHUSdpdGFsaWNGJw==.plot( err1/Perim, e=0.01..1 );The error stays less than 1% until we get a fairly eccentric ellipse:fsolve( err1/Perim = 0.01, e );In fact, we know that as e approaches 1, the perimeter of the ellipse approaches twice the length of the interval [-1,1].eval( Perim, e=1);Our approximation is not very good there.eval( P[1], e=1 );evalf(%);For celestial mechanics at least, even error of just 1% can be unacceptable. We can do better if we switch to another variable. Defineh = (1-b)^2 / (1+b)^2;Then a better approximation, known as Lindner's formula, isP[2]:= Pi*(1+b)*(1 + h/8)^2;In terms of eccentricity, this is P[2]:= eval( P[2], h = (1-b)^2 / (1+b)^2 );P[2]:= simplify( eval( P[2], b=sqrt(1-e^2) ) );This formula gets a very impressive first 6 terms right at e=0 (the powers 0, 2, 4, 6, 8, and 10).taylor( (Perim-P[2])/(2*Pi), e=0, 16 );Even at an eccentricity of 1/2, this error is negligible.eval( 1/1048576*e^12, e=0.5 ); In fact, the error is a good bit smaller than the previous estimate, even at larger eccentricities.err2:= Perim-P[2]: plot( [err1/Perim,err2/Perim], e=0.9..1 , color=[red,blue] );Exercises1.The self-taught genius Ramanujan in 1914 proposedP[3]:= Pi*(1+b)*(1 + 3*h/(10+sqrt(4-3*h)));Convert this as above into an expression in e. Find out how many Taylor terms are correct at e=0. Plot the relative error for e in (0,1).LUklbXJvd0c2Iy9JK21vZHVsZW5hbWVHNiJJLFR5cGVzZXR0aW5nR0koX3N5c2xpYkdGJzYjLUkjbWlHRiQ2JVEhRicvJSdpdGFsaWNHUSV0cnVlRicvJSxtYXRodmFyaWFudEdRJ2l0YWxpY0Yn2. Earth's orbit is an ellipse with e=0.016718. Using the first term of the Taylor series for the error in Ramanujan's formula, estimate the error in calculating the length of Earth's orbit this way. (By comparison, quantum physics says that space ceases to be continuous at the scale of Planck's constant, which is about 10^(-35) meters.)LUklbXJvd0c2Iy9JK21vZHVsZW5hbWVHNiJJLFR5cGVzZXR0aW5nR0koX3N5c2xpYkdGJzYjLUkjbWlHRiQ2JVEhRicvJSdpdGFsaWNHUSV0cnVlRicvJSxtYXRodmFyaWFudEdRJ2l0YWxpY0YnLUklbXJvd0c2Iy9JK21vZHVsZW5hbWVHNiJJLFR5cGVzZXR0aW5nR0koX3N5c2xpYkdGJzYjLUkjbWlHRiQ2JVEhRicvJSdpdGFsaWNHUSV0cnVlRicvJSxtYXRodmFyaWFudEdRJ2l0YWxpY0YnLUklbXJvd0c2Iy9JK21vZHVsZW5hbWVHNiJJLFR5cGVzZXR0aW5nR0koX3N5c2xpYkdGJzYjLUkjbWlHRiQ2JVEhRicvJSdpdGFsaWNHUSV0cnVlRicvJSxtYXRodmFyaWFudEdRJ2l0YWxpY0Yn