Re: [math-fun] Golden Ratio discovered in uterus
Here's a piecewise linear spiral approximation to a nautilus (shown me by Jack Holloway): http://www.amazon.com/The-Irrationals-Story-Numbers-Count/dp/0691143420/ref=... --rwg Also, quarter-circles (in "four point ellipses") do interesting things<http://gosper.org/pumpstill.gif> . ----- Hi Jim, Yes, there is more to be said, but I had to use that curve (consisting of quarter circles) because the point I wanted to make is specifically about that curve. The curve is only a ruler-and-compass approximation to an exponential spiral and the differences may be important in some contexts. But it is a much larger error to claim that curve is a model for the nautilus, and that big error is continuously being copied and re-published. E.g., they both appear on the cover of Livio's popular book The Golden Ratio, and inside, his text about them confusingly does not clarify the difference. As to the curve not being a "true spiral" I don't know of a standard definition of "true spiral" that includes the many types of spiral that people often study, but doesn't include that assemblage of quarter circles. As far as I am concerned, it is a "true spiral" that happens not to be C2 continuous. Georgehttp://georgehart.com/ On 8/15/2012 11:02 PM, James Propp wrote:>>From 0:20 to 0:40 in his video, George shows a famous picture of the> "golden spiral". But as many of you probably know (no doubt including> George), this picture is bogus. Specifically, it conflates two different> curves: a curve made up of quarter-circular arcs (which isn't a true> spiral) and a curve made of true spiral arcs (which isn't tangent to the> rectangles).>> George doesn't discuss this in his video, since minor issues like this> would distract from his main point. But still, I wish the video hadn't> used the inaccurate picture.>> Yes, I'm being a bit picky. At times like this I am reminded of a motto I> saw on a tee-shirt once: "I'm not pompous. I'm pedantic. There's a> difference." :-)>> Jim>> On Tue, Aug 14, 2012 at 4:22 PM, George Hart <george@georgehart.com <http://gosper.org/webmail/src/compose.php?send_to=george%40georgehart.com>> wrote:>>> Hi Alex,>>>> Unfortunately, as a consequence of this being hosted by the Guardian, I>> expect all the folks who uncritically see the golden ratio everywhere will>> now be blathering about golden uteri. I've started making a series of math>> videos and here's one rebuttal to the golden ratio cult:>>>> http://www.youtube.com/watch?**v=_gxC8OjoQkQ<http://www.youtube.com/watch?v=_gxC8OjoQkQ>>>>> As to the image on that page, I hope you don't mind if I interpret your>> use of the middle finger as an editorial comment. I feel it's misleading>> to show precise metal calipers and print 1.618, as if finger measurements>> were precise enough to support four significant digits.>>>> George>> http://georgehart.com/>>>>>>>>>> On 8/14/2012 10:23 AM, Alex Bellos wrote:>>>>> I've just launched a maths blog for the Guardian>>>>>> Here's my first post: the research is serious, but any conclusions need>>> to be taken with a medium to large dosage of salt.>>>>>> http://www.guardian.co.uk/**science/alexs-adventures-in-**>>> numberland/2012/aug/14/golden-**ratio-uterus<http://www.guardian.co.uk/science/alexs-adventures-in-numberland/2012/aug/14/golden-ratio-uterus>>>>>>> alex>>> ______________________________**_________________
On 8/16/2012 4:52 PM, Bill Gosper wrote:
Here's a piecewise linear spiral approximation to a nautilus (shown me by Jack Holloway): http://www.amazon.com/The-Irrationals-Story-Numbers-Count/dp/0691143420/ref=...
Hi Bill, That's a "Spiral of Theodorus", which closely approximates an Archimedean spiral, not an exponential spiral. See: http://en.wikipedia.org/wiki/Spiral_of_Theodorus If the image on that book cover included another revolution, it would become obvious how it doesn't match the nautilus it is superimposed on. A nautilus grows by a multiplicative factor of roughly three each revolution, while the square root spiral grows by roughly an additive increment of Pi each revolution. It is very misleading for them to show just one revolution, so it looks like a reasonable model, but I don't think this error is commonly made. Do you know other places where the Spiral of Theodorus is used as a model of the nautilus? George http://georgehart.com/
http://en.wikipedia.org/wiki/Spiral_of_Theodorus gives just two terms of the expansion Sum[ArcTan[1/Sqrt[n]], {n, 1, k}] == 1411/(12672*k^(9/2)) - 1147/(8064*k^(7/2)) + 167/(840*k^(5/2)) - 41/(120*k^(3/2)) + 7/(6*Sqrt[k]) + H + 2*Sqrt[k] with H, Hlawka's Schneckenkonstante ( A105459 - The On-Line Encyclopedia of Integer Sequences(tm) (OEIS *...* <http://oeis.org/A105459/internal> ), given as thirteen digits of -2.1577829966594462209291427868295777235041... I'm surprised that this constant term (which is not the leading term) is the only difficult one to compute. With encouragement from Corey, Neil, and Mozart, H==-Sqrt[-B + k] + (-1 + B - k)*ArcCot[Sqrt[-B + k]] + Sum[ArcTan[1/Sqrt[n]], {n, 1, k}] where k is any positive integer and B is Bernoulli's umbra. I.e., pick, say, k=1, expand around B=0, then Normal[%]/. B^(p_: 1) -> BernoulliB[p]]. Catch: This process only works to about B^d, where d ~ 2πk, and gives about 2πk/(ln 10) digits. Depending on the relative costs of arctan vs Bernoulli terms, it's cheaper to shoot for higher k and lower d. This would have involved a messy numerical integration but for the minor miracle that ArcTan[x^rational] integrates in closed form. There's a hint in OEIS that H comes out in half-integer zetas, but no hits from ries nor http://isc.carma.newcastle.edu.au/ . On Thu, Aug 16, 2012 at 7:41 PM, George Hart <george@georgehart.com> wrote:
On 8/16/2012 4:52 PM, Bill Gosper wrote:
Here's a piecewise linear spiral approximation to a nautilus (shown me by Jack Holloway): http://www.amazon.com/The-**Irrationals-Story-Numbers-** Count/dp/0691143420/ref=pd_**sim_b_25#reader_0691143420<http://www.amazon.com/The-Irrationals-Story-Numbers-Count/dp/0691143420/ref=pd_sim_b_25#reader_0691143420>
Hi Bill,
That's a "Spiral of Theodorus", which closely approximates an Archimedean spiral, not an exponential spiral. See:
http://en.wikipedia.org/wiki/**Spiral_of_Theodorus<http://en.wikipedia.org/wiki/Spiral_of_Theodorus>
If the image on that book cover included another revolution, it would become obvious how it doesn't match the nautilus it is superimposed on. A nautilus grows by a multiplicative factor of roughly three each revolution, while the square root spiral grows by roughly an additive increment of Pi each revolution.
It is very misleading for them to show just one revolution, so it looks like a reasonable model, but I don't think this error is commonly made. Do you know other places where the Spiral of Theodorus is used as a model of the nautilus?
George http://georgehart.com/
The very name Schneckenkonstante! --rwg
Stephen Finch has given this "Constant of Theodorus" (good name, but not to be confused with Mathworld's Theodorus's Constant = √3) his usual superb treatment in http://www.people.fas.harvard.edu/~sfinch/csolve/th.pdf , but the rapid numerical method I give below may be new. It seems to me this umbral notation trick should apply more generally to the Euler-Maclaurin formula (perhaps even accommodating my prescription for Bernpolys over plain Berns), streamlining both notation and computation. A further, sickening answer to George's request for published spirals confusion: http://www.ehow.com/how_8671584_decorate-pythagorean-spiral.html . It reminds me of stories I heard of a Simpsons episode wherein all the girls get vectored into a touchy-feely "math" course. --rwg I said, re high school math texts resembling "hardcover comics"
It must come as a shock when today's college freshmen first open their texts and find only text.
Gaa, Neil just showed me his >$120 Linear Algebra text--loaded with color pictures! I think more for anæsthesia than education. [I have no idea how or why GMail sent two copies of this.] On Thu, Aug 30, 2012 at 1:35 AM, Bill Gosper <billgosper@gmail.com> wrote:
http://en.wikipedia.org/wiki/Spiral_of_Theodorus gives just two terms of the expansion Sum[ArcTan[1/Sqrt[n]], {n, 1, k}] == 1411/(12672*k^(9/2)) - 1147/(8064*k^(7/2)) + 167/(840*k^(5/2)) - 41/(120*k^(3/2)) + 7/(6*Sqrt[k]) + H + 2*Sqrt[k]
with H, Hlawka's Schneckenkonstante ( A105459 - The On-Line Encyclopedia of Integer Sequences™ (OEIS *...* <http://oeis.org/A105459/internal>
), given as thirteen digits of -2.1577829966594462209291427868295777235041...
I'm surprised that this constant term (which is not the leading term) is the only difficult one to compute. With encouragement from Corey, Neil, and Mozart,
H==-Sqrt[-B + k] + (-1 + B - k)*ArcCot[Sqrt[-B + k]] + Sum[ArcTan[1/Sqrt[n]], {n, 1, k}]
where k is any positive integer and B is Bernoulli's umbra. I.e., pick, say, k=1, expand around B=0, then Normal[%]/. B^(p_: 1) -> BernoulliB[p]]. Catch: This process only works to about B^d, where d ~ 2πk, and gives about 2πk/(ln 10) digits. Depending on the relative costs of arctan vs Bernoulli terms, it's cheaper to shoot for higher k and lower d.
This would have involved a messy numerical integration but for the minor miracle that ArcTan[x^rational] integrates in closed form. There's a hint in OEIS that H comes out in half-integer zetas, but no hits from ries nor http://isc.carma.newcastle.edu.au/ .
On Thu, Aug 16, 2012 at 7:41 PM, George Hart <george@georgehart.com>wrote:
On 8/16/2012 4:52 PM, Bill Gosper wrote:
Here's a piecewise linear spiral approximation to a nautilus (shown me by Jack Holloway): http://www.amazon.com/The-**Irrationals-Story-Numbers-** Count/dp/0691143420/ref=pd_**sim_b_25#reader_0691143420<http://www.amazon.com/The-Irrationals-Story-Numbers-Count/dp/0691143420/ref=pd_sim_b_25#reader_0691143420>
Hi Bill,
That's a "Spiral of Theodorus", which closely approximates an Archimedean spiral, not an exponential spiral. See:
http://en.wikipedia.org/wiki/**Spiral_of_Theodorus<http://en.wikipedia.org/wiki/Spiral_of_Theodorus>
If the image on that book cover included another revolution, it would become obvious how it doesn't match the nautilus it is superimposed on. A nautilus grows by a multiplicative factor of roughly three each revolution, while the square root spiral grows by roughly an additive increment of Pi each revolution.
It is very misleading for them to show just one revolution, so it looks like a reasonable model, but I don't think this error is commonly made. Do you know other places where the Spiral of Theodorus is used as a model of the nautilus?
George http://georgehart.com/
The very name Schneckenkonstante! --rwg
On Sat, Sep 1, 2012 at 11:47 AM, Bill Gosper <billgosper@gmail.com> wrote:
Stephen Finch has given this "Constant of Theodorus" (good name, but not to be confused with Mathworld's Theodorus's Constant = √3) his usual superb treatment in http://www.people.fas.harvard.edu/~sfinch/csolve/th.pdf , but the rapid numerical method I give below may be new.
It seems to me this umbral notation trick should apply more generally to the Euler-Maclaurin formula (perhaps even accommodating my prescription for Bernpolys over plain Berns), streamlining both notation and computation.
For Stirling's formula, this works unreasonably well: Out[817]= Log[k!] ->(Log[2 π]-1)/2 + (-B + k + y) Log[-B + k + y] - k (The constant term coming from k! == (2^(-2 k) Sqrt[π] (2 k)!)/(k-1/2)! ) In[818]:= Normal[Series[#,{B,0,5}]&/@%817]/.B^(p_:1)->BernoulliB[p,y] Out[818]= Log[k!]->-(1/2)-k+(1/6-y+y^2)/(2 (k+y))+(y/2-(3 y^2)/2+y^3)/(6 (k+y)^2)+(-(1/30)+y^2-2 y^3+y^4)/(12 (k+y)^3)+(-(y/6)+(5 y^3)/3-(5 y^4)/2+y^5)/(20 (k+y)^4)+1/2 Log[2 π]+(-(1/2)+y) (-1-Log[k+y])+k Log[k+y]+y Log[k+y] In[819]:= TableForm[Table[N[%],{k,1,3},{y,0,1,1/3}]] Out[819]//TableForm= 0.->-0.000505911 0.->5.91272*10^-6 0.->0.0000436355 0.->-0.0000212515 0.693147->0.693126 0.693147->0.693149 0.693147->0.693151 0.693147->0.693144 1.79176->1.79176 1.79176->1.79176 1.79176->1.79176 1.79176->1.79176 In[820]:= TableForm[Table[N[%%,9],{k,3,5},{y,0,1,1/2}]]
Out[820]//TableForm=
1.79175947->1.79175644 1.79175947->1.79176085 1.79175947->1.79175873 3.17805383->3.17805309 3.17805383->3.17805423 3.17805383->3.17805358 4.78749174->4.78749150 4.78749174->4.78749189 4.78749174->4.78749164
Probably even better with y some simple function of k. --rwg
Stephen Finch has given this "Constant of Theodorus" (good name, but not to be confused with Mathworld's Theodorus's Constant = √3) his usual superb treatment in http://www.people.fas.harvard.edu/~sfinch/csolve/th.pdf , but the rapid numerical method I give below may be new. It seems to me this umbral notation trick should apply more generally to the Euler-Maclaurin formula (perhaps even accommodating my prescription for Bernpolys over plain Berns), streamlining both notation and computation. A further, sickening answer to George's request for published spirals confusion: http://www.ehow.com/how_8671584_decorate-pythagorean-spiral.html . It reminds me of stories I heard of a Simpsons episode wherein all the girls get vectored into a touchy-feely "math" course. --rwg I said, re high school math texts resembling "hardcover comics"
It must come as a shock when today's college freshmen first open their texts and find only text.
Gaa, Neil just showed me his >$120 (or did he say $160?) Linear Algebra text--loaded with color pictures! I think more for anæsthesia than education. The beauty of hack commercial art replaces the beauty of mathematics. [I have no idea how or why GMail sent two copies of this.] On Thu, Aug 30, 2012 at 1:35 AM, Bill Gosper <billgosper@gmail.com> wrote:
http://en.wikipedia.org/wiki/Spiral_of_Theodorus gives just two terms of the expansion Sum[ArcTan[1/Sqrt[n]], {n, 1, k}] == 1411/(12672*k^(9/2)) - 1147/(8064*k^(7/2)) + 167/(840*k^(5/2)) - 41/(120*k^(3/2)) + 7/(6*Sqrt[k]) + H + 2*Sqrt[k]
with H, Hlawka's Schneckenkonstante ( A105459 - The On-Line Encyclopedia of Integer Sequences™ (OEIS *...* <http://oeis.org/A105459/internal>
), given as thirteen digits of -2.1577829966594462209291427868295777235041...
I'm surprised that this constant term (which is not the leading term) is the only difficult one to compute. With encouragement from Corey, Neil, and Mozart,
H==-Sqrt[-B + k] + (-1 + B - k)*ArcCot[Sqrt[-B + k]] + Sum[ArcTan[1/Sqrt[n]], {n, 1, k}]
where k is any positive integer and B is Bernoulli's umbra. I.e., pick, say, k=1, expand around B=0, then Normal[%]/. B^(p_: 1) -> BernoulliB[p]]. Catch: This process only works to about B^d, where d ~ 2πk, and gives about 2πk/(ln 10) digits. Depending on the relative costs of arctan vs Bernoulli terms, it's cheaper to shoot for higher k and lower d.
This would have involved a messy numerical integration but for the minor miracle that ArcTan[x^rational] integrates in closed form. There's a hint in OEIS that H comes out in half-integer zetas, but no hits from ries nor http://isc.carma.newcastle.edu.au/ .
On Thu, Aug 16, 2012 at 7:41 PM, George Hart <george@georgehart.com>wrote:
On 8/16/2012 4:52 PM, Bill Gosper wrote:
Here's a piecewise linear spiral approximation to a nautilus (shown me by Jack Holloway): http://www.amazon.com/The-**Irrationals-Story-Numbers-** Count/dp/0691143420/ref=pd_**sim_b_25#reader_0691143420<http://www.amazon.com/The-Irrationals-Story-Numbers-Count/dp/0691143420/ref=pd_sim_b_25#reader_0691143420>
Hi Bill,
That's a "Spiral of Theodorus", which closely approximates an Archimedean spiral, not an exponential spiral. See:
http://en.wikipedia.org/wiki/**Spiral_of_Theodorus<http://en.wikipedia.org/wiki/Spiral_of_Theodorus>
If the image on that book cover included another revolution, it would become obvious how it doesn't match the nautilus it is superimposed on. A nautilus grows by a multiplicative factor of roughly three each revolution, while the square root spiral grows by roughly an additive increment of Pi each revolution.
It is very misleading for them to show just one revolution, so it looks like a reasonable model, but I don't think this error is commonly made. Do you know other places where the Spiral of Theodorus is used as a model of the nautilus?
George http://georgehart.com/
The very name Schneckenkonstante! --rwg
http://en.wikipedia.org/wiki/Spiral_of_Theodorus gives just two terms of the expansion Sum[ArcTan[1/Sqrt[n]], {n, 1, k}] == 1411/(12672*k^(9/2)) - 1147/(8064*k^(7/2)) + 167/(840*k^(5/2)) - 41/(120*k^(3/2)) + 7/(6*Sqrt[k]) + H + 2*Sqrt[k] with H, Hlawka's Schneckenkonstante ( A105459 - The On-Line Encyclopedia of Integer Sequences(tm) (OEIS *...* <http://oeis.org/A105459/internal> ), given as thirteen digits of -2.1577829966594462209291427868295777235041... I'm surprised that this constant term (which is not the leading term) is the only difficult one to compute. With encouragement from Corey, Neil, and Mozart, H==-Sqrt[-B + k] + (-1 + B - k)*ArcCot[Sqrt[-B + k]] + Sum[ArcTan[1/Sqrt[n]], {n, 1, k}] where k is any positive integer and B is Bernoulli's umbra. I.e., pick, say, k=1, expand around B=0, then Normal[%]/. B^(p_: 1) -> BernoulliB[p]]. Catch: This process only works to about B^d, where d ~ 2πk, and gives about 2πk/(ln 10) digits. Depending on the relative costs of arctan vs Bernoulli terms, it's cheaper to shoot for higher k and lower d. This would have involved a messy numerical integration but for the minor miracle that ArcTan[x^rational] integrates in closed form. There's a hint in OEIS that H comes out in half-integer zetas, but no hits from ries nor http://isc.carma.newcastle.edu.au/ . On Thu, Aug 16, 2012 at 7:41 PM, George Hart <george@georgehart.com> wrote:
On 8/16/2012 4:52 PM, Bill Gosper wrote:
Here's a piecewise linear spiral approximation to a nautilus (shown me by Jack Holloway): http://www.amazon.com/The-**Irrationals-Story-Numbers-** Count/dp/0691143420/ref=pd_**sim_b_25#reader_0691143420<http://www.amazon.com/The-Irrationals-Story-Numbers-Count/dp/0691143420/ref=pd_sim_b_25#reader_0691143420>
Hi Bill,
That's a "Spiral of Theodorus", which closely approximates an Archimedean spiral, not an exponential spiral. See:
http://en.wikipedia.org/wiki/**Spiral_of_Theodorus<http://en.wikipedia.org/wiki/Spiral_of_Theodorus>
If the image on that book cover included another revolution, it would become obvious how it doesn't match the nautilus it is superimposed on. A nautilus grows by a multiplicative factor of roughly three each revolution, while the square root spiral grows by roughly an additive increment of Pi each revolution.
It is very misleading for them to show just one revolution, so it looks like a reasonable model, but I don't think this error is commonly made. Do you know other places where the Spiral of Theodorus is used as a model of the nautilus?
George http://georgehart.com/
The very name Schneckenkonstante! --rwg
On Thu, Aug 30, 2012 at 1:40 AM, Bill Gosper <billgosper@gmail.com> wrote:
http://en.wikipedia.org/wiki/Spiral_of_Theodorus gives just two terms of the expansion Sum[ArcTan[1/Sqrt[n]], {n, 1, k}] == 1411/(12672*k^(9/2)) - 1147/(8064*k^(7/2)) + 167/(840*k^(5/2)) - 41/(120*k^(3/2)) + 7/(6*Sqrt[k]) + H + 2*Sqrt[k]
with H, Hlawka's Schneckenkonstante ( A105459 - The On-Line Encyclopedia of Integer Sequences™ (OEIS *...* <http://oeis.org/A105459/internal>
), given as thirteen digits of -2.1577829966594462209291427868295777235041...
=-2.1577829966594462209291427868295777235041395986075624551548955508588696467966064814966942989464 ...
I'm surprised that this constant term (which is not the leading term) is the only difficult one to compute.
Actually, this is analogous to the √2π in Stirling's formula.
With encouragement from Corey, Neil, and Mozart,
H==-Sqrt[-B + k] + (-1 + B - k)*ArcCot[Sqrt[-B + k]] + Sum[ArcTan[1/Sqrt[n]], {n, 1, k}]
where k is any positive integer and B is Bernoulli's umbra. I.e., pick, say, k=1, expand around B=0, then Normal[%]/. B^(p_: 1) -> BernoulliB[p]]. Catch: This process only works to about B^d, where d ~ 2πk, and gives about 2πk/(ln 10) digits. Depending on the relative costs of arctan vs Bernoulli terms, it's cheaper to shoot for higher k and lower d.
This would have involved a messy numerical integration but for the minor miracle that ArcTan[x^rational] integrates in closed form. There's a hint in OEIS that H comes out in half-integer zetas, but no hits from ries nor http://isc.carma.newcastle.edu.au/ .
Foo, H==Sum[((-1)^i*Zeta[1/2 + i])/(1 + 2*i),{i, 1, ∞}] + Zeta[1/2] . (This is Finch's K inhttp://www.people.fas.harvard.edu/~sfinch/csolve/th.pdf.) This is the first simple formula I've seen, and is good for 1 bit/term if rewritten -1 + π/4 + Sum[((-1)^n*(-1 + Zeta[(1/2)*(1 + 2*n)]))/ (1 + 2*n), {n, 1,∞}] + Zeta[1/2] --rwg
On Thu, Aug 16, 2012 at 7:41 PM, George Hart <george@georgehart.com>wrote:
On 8/16/2012 4:52 PM, Bill Gosper wrote:
Here's a piecewise linear spiral approximation to a nautilus (shown me by Jack Holloway): http://www.amazon.com/The-**Irrationals-Story-Numbers-** Count/dp/0691143420/ref=pd_**sim_b_25#reader_0691143420<http://www.amazon.com/The-Irrationals-Story-Numbers-Count/dp/0691143420/ref=pd_sim_b_25#reader_0691143420>
Hi Bill,
That's a "Spiral of Theodorus", which closely approximates an Archimedean spiral, not an exponential spiral. See:
http://en.wikipedia.org/wiki/**Spiral_of_Theodorus<http://en.wikipedia.org/wiki/Spiral_of_Theodorus>
If the image on that book cover included another revolution, it would become obvious how it doesn't match the nautilus it is superimposed on. A nautilus grows by a multiplicative factor of roughly three each revolution, while the square root spiral grows by roughly an additive increment of Pi each revolution.
It is very misleading for them to show just one revolution, so it looks like a reasonable model, but I don't think this error is commonly made. Do you know other places where the Spiral of Theodorus is used as a model of the nautilus?
George http://georgehart.com/
The very name Schneckenkonstante! --rwg
Bill Gosper:
with H, Hlawka's Schneckenkonstante ( A105459 - The On-Line Encyclopedia of Integer Sequences™ (OEIS *...* <http://oeis.org/A105459/internal>
), given as thirteen digits of -2.1577829966594462209291427868295777235041...
= -2.1577829966594462209291427868295777235041395986075624551548955508588696467966064814966942989464 ...
David Brink has contributed 125 terms of the continued fraction expansion: http://oeis.org/A185051
participants (3)
-
Bill Gosper -
George Hart -
Hans Havermann