Modern-day Easter is derived from two ancient traditions: one Judeo-Christian and the other Pagan. Both Christians and Pagans have celebrated death and resurrection themes following the Spring Equinox for millennia. Most religious historians believe that many elements of the Christian observance of Easter were derived from earlier Pagan celebrations.
The equinox occurs each year on March 20, 21 or 22. Both Neopagans and Christians continue to celebrate religious rituals linked to the equinox in the present day. Wiccans and other Neopagans usually hold their celebrations on the day or eve of the equinox. Western Christians wait until the Sunday on or after the next full moon. The Eastern Orthodox churches follow a different calculation; their celebration is often many weeks after the date selected by the Western churches.
Therefore to you all on this Easter Sunday (except you wiccans; because that wicker chair you sold me collapsed) a very happy time too you.
Mark:
I wrote a simulation and get something a bit greater than .521.
Rap:
I wrote a simple routine. After > 50,000 iterations my routine gets somewhat greater than 0.521.
Dagnabit! The hardest question in the world; and you both come up with the correct answer. And Rap got a sympathy ((hug))! Happy Easter dudes
The answer is [2^1/2 + 2 + 5*ln(1+2^1/2)] / 15 =~ 0.521405433
And this is why
First, let I(x,y) stand for "the integral from x to y of ...", and sqr stand for "the square root of...".
To find the answer we must solve I(0,1) I(0,1) I(0,1) I(0,1) sqr((x-a)^2+(y-b)^2) db da dy dx.
Even the first integration of this is very messy, requiring a big table of integrals, and I think at best a small book of integrals would only take you through the second one. So we need to be more creative to make the problem more manageable. First we introduce a change of variables:
u=x-a
s=x
v=y-b
t=y
Or:
x=s
y=t
a=s-u
b=t-v
(1): Now we have I(0,1) I(t,t-1) I(0,1) I(s,s-1) sqr(u^2 + v^2) * |J| du ds dv dt
Where |J| = the Jacobian =
| db/du db/ds db/dv du/dt | | 0 0 -1 1 |
| | | |
| da/du da/ds da/dv da/dt | | -1 1 0 0 |
| | = | | = -1
| dy/du dy/ds dy/dv dy/dt | | 0 0 0 1 |
| | | |
| dx/du dx/ds dx/dv dx/dt | | 0 1 0 0 |
Now let's solve the following inner two integrals of equation (1):
(2): I(0,1) I(s,s-1) sqr(u^2 + v^2) du ds.
The area of integration is bounded by the quadrilateral with vertices (0,0), (1,1), (0,1), and (-1,0), with s being the horizontal axis and u being the vertical.
Changing the order of integral we get:
I(-1,0) I(0,u+1) sqr(u^2 + v^2) ds du + I(0,1) I(u,1) sqr(u^2 + v^2) ds du =
I(1,0) sqr(u^2 + v^2) * (1+u) du + I(0,1) sqr(u^2 + v^2) * (1-u) du =
-I(1,0) sqr(p^2 + v^2) * (1-p) dp + I(0,1) sqr(u^2 + v^2) * (1-u) du (letting p=-u in the first integral) =
I(0,1) sqr(p^2 + v^2) * (1-p) dp + I(0,1) sqr(u^2 + v^2) * (1-u) du (Notice the two terms are equal)=
2* I(0,1) sqr(u^2 + v^2) * (1-u) du.
Now put this into equation (1), and moving the constant terms to the left:
-2* I(0,1) I(t,t-1) I(0,1) sqr(u^2 + v^2) * (1-u) du dv dt =
-2* I(0,1) I(0,1) I(t,t-1) sqr(u^2 + v^2) * (1-u) dv dt du (changing the order of integration)=
-2* I(0,1) (1-u) I(0,1) I(t,t-1) sqr(u^2 + v^2) dv dt du =
Note that the inner two integral are of the same form as equation (2), thus:
4* I(0,1) I(0,1) sqr(u^2 + v^2) * (1-u) * (1-v) dv du =
4* I(0,1) I(0,1) sqr(u^2 + v^2) * (1-u) * (1-v) du dv (changing the order of integration)
Next we make another substitution:
u=r*sin(w)
v=r*cos(w)
This time the Jacobian shall equal:
| du/dr du/dw | | sin(w) r*cos(w) |
| | = | | = -r
| dv/dr dv/dw | | cos(w) -r*sin(w) |
Thus the integral as a function of w and r is:
-8 * I(0,pi/4) I(0,1/cos(w)) -r * sqr(r*sin2(w) + r*cos2(w)) * (1-r*cos(w)) * (1-r*sin(w)) dr dw.
Note that we halved the area of integration, going from 0 to pi/4 as opposed to pi/2, and multiplying by 8 instead of the 4. This is permissible because of the symetry of the two regions.
= 8 * I(0,pi/4) I(0,1/cos(w)) r^2 * (1-r*cos(w)) * (1-r*sin(w)) dr dw.
= 8 * I(0,pi/4) I(0,1/cos(w)) r^2 - r^3*(cos(w)+sin(w)) + r^4*(cos(w)+sin(w)) dr dw.
= 8 * I(0,pi/4) r^3/3 - r^4*(cos(w)+sin(w))/4 + r^5*(cos(w)+sin(w))/5 ( from 0 to 1/cos(w) ) dw.
= 8 * I(0,pi/4) (3*cos^3(w))-^1 - (cos(w)+sin(w))/(4*cos^4) + (cos(w)+sin(w))/(5*cos5(w)) dw.
= 8 * I(0,pi/4) (3*cos^3(w))-^1 - (4*cos^3(w))-^1 - sin(w)*(4*cos^4(w))-^1 + sin(w)*(5*cos4(w))-^1 dw.
= 8 * I(0,pi/4) (12*cos^3(w))-^1 - sin(w)*(20*cos^4(w))-^1 dw.
(3) = 2/15 * I(0,pi/4) 5/cos^3(w) - 3*sin(w)/cos^4(w) dw.
Next we must dust off our table of integrals to help us with the integral of cos-^3(w).
I dx/cos^n(x) = (1/(n-1)) * sin(x)/cos^n-1(x) + ((n-2)/(n-1)) * I dx/cos^n-2(x).
In our case n=3 so, I dx/cos^3(x) = (1/2) * sin(x)/cos^2(x) + (1/2) * I sec(x) dx.
Now lets look up the integral of sec(x):
I sec(x) dx = (1/2) * ln((1+sin(x))/(1-sin(x))).
Now we are ready to integrate cos-^3(w):
I(0,pi/4) cos-^3(w) dw = sin(w)/(2*cos^2(w)) (from 0 to pi/4) + (1/2)*I(0,pi/4) sec(x) dx.
= (1/2)*(2-^1/2/2-^1 - 0) + (1/4)*ln((2+sqr(2))/(2-sqr(2))).
= (1/2)*sqr(2) + (1/4)*ln(((2+sqr(2))/2+sqr(2))/((2-sqr(2))*(2+sqr(2))).
= 1/sqr(2) + (1/4)*ln((4+4*sqr(2)+2)/(4-2)).
= 1/sqr(2) + (1/4)*ln(3+2*sqr(2)).
= 1/sqr(2) + (1/2)*ln(3+2*sqr(2))^1/2.
= 1/sqr(2) + (1/2)*ln(1+sqr(2)).
Now lets integrate sin(w)/cos^4(w).
let m=cos(w), thus dm=-sin(w) dw.
I(0,pi/4) sin(w)/cos^4(w).
= I(1,1/sqr(2)) -1*m^-4 dm.
= 1/(3*m^3) from 0 to 1/sqr(2).
= (1/3) * (2*sqr(2) - 1).
Not it is time to plus these integrals into equation (3):
(2/15) * [ 5*(1/sqr(2) + (1/2)*ln(1+sqr(2))) + 3*((1/3) * (2*sqr(2) - 1)) ]
= (2/15) * [ 5*sqr(2)/2 + (5/2)*ln(1+sqr(2)) - 4*sqr(2)/2 + 1 ]
= (2/15) * [ sqr(2)/2 + (5/2)*ln(1+sqr(2)) + 1 ]
= (sqr(2) + 2 + 5*ln(1+sqr(2))) / 15.
Easy freeking peasy!
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