Re: [math-fun] why can a line of slope 10^n determine n decimal digits of π?
Gary, it doesn't. I'm not sure what led you to think that the numbers are 31, 314, etc., but rather the actual numbers are only approximations to these integers. --Dan << Thanks, Joshua. But I still don't get it. Why does a line of slope -10 carve out exactly 31 arcs and not, say, 30?
Those who sleep faster get more rest.
Dan, I'm actually trying to figure out a conservation of linear momentum puzzle. Am not sure what it's called. It seems to trace the digits of pi exactly. Maybe you can see how this works. Say you have two masses M and m. You slide M into m, which in turn bounces off a wall and hits M. If M=m then after 3 bounces m will never hit M again. If M=100m then there's 31 bounces. If M=10000m there's 314 bounces. The puzzle is— why pi? Which is why I was drawing the circle and line. Kinetic energy and momentum are conserved as the velocities change. Kinetic energy traces an ellipse. Momentum follows a line. If you flatten the ellipse into a circle the initial momentum line* conveniently follows slope -1 when M=1 (ellipse already a circle in this case), slope -10 when M=100m, etc. * The line shifts because of the mass of the wall. Took me all day to realize this. On Wed, Nov 24, 2010 at 9:42 PM, Dan Asimov <dasimov@earthlink.net> wrote:
Gary, it doesn't. I'm not sure what led you to think that the numbers are 31, 314, etc., but rather the actual numbers are only approximations to these integers.
--Dan
<< Thanks, Joshua. But I still don't get it. Why does a line of slope -10 carve out exactly 31 arcs and not, say, 30?
Those who sleep faster get more rest.
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I haven't worked out how the limit is approached, but can show you why the limit gives pi. By rescaling time (or energy), the momenta P and p of the large and small masses lie on the ellipse P^2/M + p^2/m = 1 by conservation of energy. In the limit M >> m you can estimate the momentum transfer q from the narrow width of the ellipse as a function of P: P -> P - q q = 2 sqrt(m - (m/M)P^2) Also, in that limit P starts at +sqrt(M) and reaches -sqrt(M) before the collisions stop. Now consider a momentum range dP in this process, where dP is large compared to q but small on the scale sqrt(M) -- this is valid in the M >> m limit. The number of collisions in that range is dN = 2 dP/q The total number of collisions is therefore N = int dN = int_{-sqrt(M)}^{+sqrt(M)} dP/sqrt(m - (m/M)P^2) = sqrt(M/m) int_{-1}^{+1} dx/sqrt(1-x^2) = sqrt(M/m) pi Veit On Nov 25, 2010, at 2:09 AM, Gary Antonick wrote:
Dan,
I'm actually trying to figure out a conservation of linear momentum puzzle. Am not sure what it's called. It seems to trace the digits of pi exactly. Maybe you can see how this works.
Say you have two masses M and m. You slide M into m, which in turn bounces off a wall and hits M.
If M=m then after 3 bounces m will never hit M again.
If M=100m then there's 31 bounces.
If M=10000m there's 314 bounces. The puzzle is— why pi?
Which is why I was drawing the circle and line. Kinetic energy and momentum are conserved as the velocities change. Kinetic energy traces an ellipse. Momentum follows a line. If you flatten the ellipse into a circle the initial momentum line* conveniently follows slope -1 when M=1 (ellipse already a circle in this case), slope -10 when M=100m, etc.
* The line shifts because of the mass of the wall. Took me all day to realize this.
On Wed, Nov 24, 2010 at 9:42 PM, Dan Asimov <dasimov@earthlink.net> wrote:
Gary, it doesn't. I'm not sure what led you to think that the numbers are 31, 314, etc., but rather the actual numbers are only approximations to these integers.
--Dan
<< Thanks, Joshua. But I still don't get it. Why does a line of slope -10 carve out exactly 31 arcs and not, say, 30?
Those who sleep faster get more rest.
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participants (3)
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Dan Asimov -
Gary Antonick -
Veit Elser