[math-fun] why can a line of slope 10^n determine n decimal digits of π?
Just noticed this yesterday and can't figure out why it works. Draw a circle with radius 1 around 0,0. Draw a line with slope -10 that hits the circle at 1,0. This line will also hit the circle again and make a little arc. The length of the arc will be about 1/31 the perimeter of the circle. Now try again with a line of slope -100. The line will create an arc 1/314 of the circle. Starting to look a bit like pi. Am wondering why this is. In case I made a mistake here's what I'm trying to describe.. if the circle were a clock the first line would cut an arc from 15 minutes (3:00) to almost the 13 minute mark.
tan theta = theta for small theta. --Joshua On Tue, Nov 23, 2010 at 9:55 PM, Gary Antonick <gantonick@post.harvard.edu> wrote:
Just noticed this yesterday and can't figure out why it works.
Draw a circle with radius 1 around 0,0. Draw a line with slope -10 that hits the circle at 1,0. This line will also hit the circle again and make a little arc. The length of the arc will be about 1/31 the perimeter of the circle.
Now try again with a line of slope -100. The line will create an arc 1/314 of the circle.
Starting to look a bit like pi. Am wondering why this is.
In case I made a mistake here's what I'm trying to describe.. if the circle were a clock the first line would cut an arc from 15 minutes (3:00) to almost the 13 minute mark.
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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? I'll try to think about some more. On Wed, Nov 24, 2010 at 2:01 AM, Joshua Zucker <joshua.zucker@gmail.com> wrote:
tan theta = theta for small theta.
--Joshua
On Tue, Nov 23, 2010 at 9:55 PM, Gary Antonick <gantonick@post.harvard.edu> wrote:
Just noticed this yesterday and can't figure out why it works.
Draw a circle with radius 1 around 0,0. Draw a line with slope -10 that hits the circle at 1,0. This line will also hit the circle again and make a little arc. The length of the arc will be about 1/31 the perimeter of the circle.
Now try again with a line of slope -100. The line will create an arc 1/314 of the circle.
Starting to look a bit like pi. Am wondering why this is.
In case I made a mistake here's what I'm trying to describe.. if the circle were a clock the first line would cut an arc from 15 minutes (3:00) to almost the 13 minute mark.
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The angle in radians = arctan of the slope. Maybe it's easier if you draw the line to the center of the circle and see that the slope is about 1/10 or 1/100 or so on, so the arctan is about 1/10 or 1/100 radians, and of course there are pi radians in a semicircle. --Joshua On Wed, Nov 24, 2010 at 9:08 PM, Gary Antonick <gantonick@post.harvard.edu> wrote:
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?
I'll try to think about some more.
On Wed, Nov 24, 2010 at 2:01 AM, Joshua Zucker <joshua.zucker@gmail.com> wrote:
tan theta = theta for small theta.
--Joshua
On Tue, Nov 23, 2010 at 9:55 PM, Gary Antonick <gantonick@post.harvard.edu> wrote:
Just noticed this yesterday and can't figure out why it works.
Draw a circle with radius 1 around 0,0. Draw a line with slope -10 that hits the circle at 1,0. This line will also hit the circle again and make a little arc. The length of the arc will be about 1/31 the perimeter of the circle.
Now try again with a line of slope -100. The line will create an arc 1/314 of the circle.
Starting to look a bit like pi. Am wondering why this is.
In case I made a mistake here's what I'm trying to describe.. if the circle were a clock the first line would cut an arc from 15 minutes (3:00) to almost the 13 minute mark.
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Joshua, Dan, Veit— thanks. All very helpful. Tiny thetas are now a lot more clear. But I still don't understand what's going on with *big* thetas. What happens when M/m=1, for example, or M/m=10. A line of slope of -1 should cut out about 1/3 of the circle. A line of slope of -10 should cut out about 1/31. But what do these lines *actually* cut? That's what I'm not quite clear about. A slope of -1 appears to cut the circle into four equal slices. But pi doesn't begin with a 4. How about slope -10? Cuts the circle into about 31 slices. But how many exactly? Fewer than 30.9? More than 31.1? I actually did make some progress on this since yesterday. One thing I realized is that the sum of these differences will never be more than the first slice. Which means the digits of pi are jotted out exactly. I also realized what this sum means: that the two balls M and m will always be traveling at different velocities after the final collision. The circle will always be divided into some 10^n pi equal wedges plus one extra (smaller) wedge. In pondering I came up with the following scenario which I think is kind of a fun intuition test. As M/m approaches infinity the size of the extra small wedge approaches zero. Does this mean at the limit the two balls will have the same final velocity? If so, what is this final velocity? (And how is this scenario reversible?) On Thu, Nov 25, 2010 at 11:58 AM, Joshua Zucker <joshua.zucker@gmail.com> wrote:
The angle in radians = arctan of the slope.
Maybe it's easier if you draw the line to the center of the circle and see that the slope is about 1/10 or 1/100 or so on, so the arctan is about 1/10 or 1/100 radians, and of course there are pi radians in a semicircle.
--Joshua
On Wed, Nov 24, 2010 at 9:08 PM, Gary Antonick <gantonick@post.harvard.edu> wrote:
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?
I'll try to think about some more.
On Wed, Nov 24, 2010 at 2:01 AM, Joshua Zucker <joshua.zucker@gmail.com> wrote:
tan theta = theta for small theta.
--Joshua
On Tue, Nov 23, 2010 at 9:55 PM, Gary Antonick <gantonick@post.harvard.edu> wrote:
Just noticed this yesterday and can't figure out why it works.
Draw a circle with radius 1 around 0,0. Draw a line with slope -10 that hits the circle at 1,0. This line will also hit the circle again and make a little arc. The length of the arc will be about 1/31 the perimeter of the circle.
Now try again with a line of slope -100. The line will create an arc 1/314 of the circle.
Starting to look a bit like pi. Am wondering why this is.
In case I made a mistake here's what I'm trying to describe.. if the circle were a clock the first line would cut an arc from 15 minutes (3:00) to almost the 13 minute mark.
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On Friday 26 November 2010 22:17:54 Gary Antonick wrote:
As M/m approaches infinity the size of the extra small wedge approaches zero. Does this mean at the limit the two balls will have the same final velocity? If so, what is this final velocity? (And how is this scenario reversible?)
In the limit, the final velocity of both balls is zero. The bigger M/m is, the greater the fraction of the energy and momentum that ends up belonging to the bigger ball; as its mass tends to infinity, the corresponding velocity goes to zero. Of course a final velocity *equal* to zero makes for an irreversible scenario, but that's only the limit. In any actual case with finite M, the balls end up with unequal, nonzero final velocities. -- g
Gareth, At the same time I think a reasonable argument can be made for the final velocities to be -V at the limit. I think this is what the circle and tangent line are suggesting. Which means, if I've got it right, that the wall (the third ball) would have to be very, very heavy. - Gary On Fri, Nov 26, 2010 at 4:05 PM, Gareth McCaughan <gareth.mccaughan@pobox.com> wrote:
On Friday 26 November 2010 22:17:54 Gary Antonick wrote:
As M/m approaches infinity the size of the extra small wedge approaches zero. Does this mean at the limit the two balls will have the same final velocity? If so, what is this final velocity? (And how is this scenario reversible?)
In the limit, the final velocity of both balls is zero. The bigger M/m is, the greater the fraction of the energy and momentum that ends up belonging to the bigger ball; as its mass tends to infinity, the corresponding velocity goes to zero.
Of course a final velocity *equal* to zero makes for an irreversible scenario, but that's only the limit. In any actual case with finite M, the balls end up with unequal, nonzero final velocities.
-- g
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Steve Landsburg posted about this very problem today at http://www.thebigquestions.com/2014/06/18/follow-the-bouncing-ball/ He references the paper http://ics.org.ru/doc?pdf=440&dir=e by Gregory Galperin which does indeed prove that the number of collisions does give the first N digits of pi. -Thomas C On Fri, Nov 26, 2010 at 11:48 PM, Gary Antonick <gantonick@post.harvard.edu> wrote:
Gareth,
At the same time I think a reasonable argument can be made for the final velocities to be -V at the limit. I think this is what the circle and tangent line are suggesting.
Which means, if I've got it right, that the wall (the third ball) would have to be very, very heavy.
- Gary
On Fri, Nov 26, 2010 at 4:05 PM, Gareth McCaughan <gareth.mccaughan@pobox.com> wrote:
On Friday 26 November 2010 22:17:54 Gary Antonick wrote:
As M/m approaches infinity the size of the extra small wedge approaches zero. Does this mean at the limit the two balls will have the same final velocity? If so, what is this final velocity? (And how is this scenario reversible?)
In the limit, the final velocity of both balls is zero. The bigger M/m is, the greater the fraction of the energy and momentum that ends up belonging to the bigger ball; as its mass tends to infinity, the corresponding velocity goes to zero.
Of course a final velocity *equal* to zero makes for an irreversible scenario, but that's only the limit. In any actual case with finite M, the balls end up with unequal, nonzero final velocities.
-- g
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participants (4)
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Gareth McCaughan -
Gary Antonick -
Joshua Zucker -
Thomas Colthurst