[{"id":247547,"date":"2026-08-05T13:26:25","date_gmt":"2026-08-05T18:26:25","guid":{"rendered":"https:\/\/www.johndcook.com\/blog\/?p=247547"},"modified":"2026-08-05T13:26:25","modified_gmt":"2026-08-05T18:26:25","slug":"math-log","status":"publish","type":"post","link":"https:\/\/www.johndcook.com\/blog\/2026\/08\/05\/math-log\/","title":{"rendered":"The code that didn’t break"},"content":{"rendered":"

Last week I wrote a post<\/a> on hiding cryptographic keys in decks of cards. I wrote some code for that post that shouldn’t work, but before fixing I noticed that it in fact did work.<\/p>\n

The code computes logarithms for integers larger than the largest representable float. For example, the largest float is on the order of 10308<\/sup>, and yet the following code works.<\/p>\n

>>> import math\r\n>>> math.log10(10**400)\r\n400.0\r\n<\/pre>\n
The log<\/code>, log2<\/code>, and log10<\/code> functions have some code inside that handles large integers specially. It doesn’t simply convert the integers to floats before taking the logarithm. If it did, it would overflow.<\/p>\n
While playing around with this I also noticed that you can define floats larger than the largest float without warnings.<\/p>\n
>>> math.log(1e308)\r\n709.1962086421661\r\n>>> math.log(1e309)\r\ninf\r\n<\/pre>\n
This isn’t a feature of math.log<\/code> but of how Python handles scientific notation. The expression 1e308<\/code> is the floating point representation of 10308<\/sup>. It is a float, not an int.<\/p>\n
>>> type(1e308)\r\n<class 'float'>\r\n<\/pre>\n
The expression 1e309<\/code> is also a float. But since it’s larger than is possible for a float, Python interprets it as inf<\/code>. The code<\/p>\n
math.log(1e309)<\/pre>\n
returns inf<\/code> based on the reasoning that log(\u221e) = \u221e.<\/p>\n
That explains the following behavior:<\/p>\n
>>> 1e309 == 1e310\r\nTrue\r\n<\/pre>\n
The expressions 1e309<\/code> and 1e310<\/code> are equal because both are alternate ways of writing inf<\/code>.<\/p>\n","protected":false},"excerpt":{"rendered":"
Last week I wrote a post on hiding cryptographic keys in decks of cards. I wrote some code for that post that shouldn’t work, but before fixing I noticed that it in fact did work. The code computes logarithms for integers larger than the largest representable float. For example, the largest float is on the […]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"_acf_changed":false,"footnotes":""},"categories":[9],"tags":[169],"class_list":["post-247547","post","type-post","status-publish","format-standard","hentry","category-math","tag-python"],"acf":[],"aioseo_notices":[],"aioseo_head":"\n\t\t\n\t\n\t\n\t\n\t\n\t\n\t\n\t\t\n\t\t\n\t\t\n\t\t\n\t\t\n\t\t\n\t\t\n\t\t\n\t\t\n\t\t\n\t\t\n\t\t\n\t\t\n\n","aioseo_head_json":{"title":"The code that didn\u2019t break","description":"A couple oddities in math with very large numbers in Python","canonical_url":"https:\/\/www.johndcook.com\/blog\/2026\/08\/05\/math-log\/","robots":"max-image-preview:large","keywords":"python","webmasterTools":{"miscellaneous":""},"schema":null,"og:locale":"en_US","og:site_name":"John D. Cook | Applied Mathematics Consulting","og:type":"article","og:title":"The code that didn\u2019t break","og:description":"A couple oddities in math with very large numbers in Python","og:url":"https:\/\/www.johndcook.com\/blog\/2026\/08\/05\/math-log\/","article:published_time":"2026-08-05T18:26:25+00:00","article:modified_time":"2026-08-05T18:26:25+00:00","twitter:card":"summary","twitter:title":"The code that didn\u2019t break","twitter:description":"A couple oddities in math with very large numbers in Python","twitter:image":"https:\/\/www.johndcook.com\/blog\/wp-content\/uploads\/2022\/05\/twittercard.png"},"aioseo_meta_data":{"post_id":"247547","title":null,"description":"A couple oddities in math with very large numbers in Python","keywords":null,"keyphrases":{"focus":{"keyphrase":"","score":0,"analysis":{"keyphraseInTitle":{"score":0,"maxScore":9,"error":1}}},"additional":[]},"primary_term":null,"canonical_url":null,"og_title":null,"og_description":null,"og_object_type":"default","og_image_type":"default","og_image_url":null,"og_image_width":null,"og_image_height":null,"og_image_custom_url":null,"og_image_custom_fields":null,"og_video":"","og_custom_url":null,"og_article_section":null,"og_article_tags":null,"twitter_use_og":false,"twitter_card":"default","twitter_image_type":"default","twitter_image_url":null,"twitter_image_custom_url":null,"twitter_image_custom_fields":null,"twitter_title":null,"twitter_description":null,"schema":{"blockGraphs":[],"customGraphs":[],"default":{"data":{"Article":[],"Course":[],"Dataset":[],"FAQPage":[],"Movie":[],"Person":[],"Product":[],"ProductReview":[],"Car":[],"Recipe":[],"Service":[],"SoftwareApplication":[],"WebPage":[]},"graphName":"Article","isEnabled":true},"graphs":[]},"schema_type":"default","schema_type_options":null,"pillar_content":false,"robots_default":true,"robots_noindex":false,"robots_noarchive":false,"robots_nosnippet":false,"robots_nofollow":false,"robots_noimageindex":false,"robots_noodp":false,"robots_notranslate":false,"robots_max_snippet":"-1","robots_max_videopreview":"-1","robots_max_imagepreview":"large","priority":null,"frequency":"default","location":null,"local_seo":null,"breadcrumb_settings":null,"limit_modified_date":false,"created":"2026-08-05 17:47:04","updated":"2026-08-05 18:26:25","ai":{"faqs":[],"keyPoints":[],"schemas":[],"titles":[],"descriptions":[],"socialPosts":{"email":{"subject":"","preview":"","content":""},"linkedin":[],"twitter":[],"facebook":[],"instagram":[]}},"seo_analyzer_scan_date":null},"aioseo_breadcrumb":"
\n\t\t\tHome<\/a>\n\t\t<\/span>»<\/span>\n\t\t\tMath<\/a>\n\t\t<\/span>»<\/span>\n\t\t\tThe code that didn\u2019t break\n\t\t<\/span><\/div>","aioseo_breadcrumb_json":[{"label":"Home","link":"https:\/\/www.johndcook.com\/blog"},{"label":"Math","link":"https:\/\/www.johndcook.com\/blog\/category\/math\/"},{"label":"The code that didn’t break","link":"https:\/\/www.johndcook.com\/blog\/2026\/08\/05\/math-log\/"}],"_links":{"self":[{"href":"https:\/\/www.johndcook.com\/blog\/wp-json\/wp\/v2\/posts\/247547","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.johndcook.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.johndcook.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.johndcook.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.johndcook.com\/blog\/wp-json\/wp\/v2\/comments?post=247547"}],"version-history":[{"count":2,"href":"https:\/\/www.johndcook.com\/blog\/wp-json\/wp\/v2\/posts\/247547\/revisions"}],"predecessor-version":[{"id":247549,"href":"https:\/\/www.johndcook.com\/blog\/wp-json\/wp\/v2\/posts\/247547\/revisions\/247549"}],"wp:attachment":[{"href":"https:\/\/www.johndcook.com\/blog\/wp-json\/wp\/v2\/media?parent=247547"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.johndcook.com\/blog\/wp-json\/wp\/v2\/categories?post=247547"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.johndcook.com\/blog\/wp-json\/wp\/v2\/tags?post=247547"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}},{"id":247540,"date":"2026-08-05T09:49:35","date_gmt":"2026-08-05T14:49:35","guid":{"rendered":"https:\/\/www.johndcook.com\/blog\/?p=247540"},"modified":"2026-08-05T10:51:33","modified_gmt":"2026-08-05T15:51:33","slug":"enumerating-trees-and-circles","status":"publish","type":"post","link":"https:\/\/www.johndcook.com\/blog\/2026\/08\/05\/enumerating-trees-and-circles\/","title":{"rendered":"Enumerating trees and circles"},"content":{"rendered":"
A few days ago I wrote a post on counting rooted trees<\/a>. That post looked at the sequence c<\/em>(n<\/em>) which counts the number of rooted trees with n<\/em> nodes. Here one node is distinguished as the root, but the nodes below the root are not distinguished from each other; all that matters is how the nodes are connected.<\/p>\n
<\/p>\n
The number of rooted trees with n<\/em> nodes is the same as the number of ways to configure n<\/em> \u2212 1 non-overlapping circles. Not only are the counts the same, there is a natural correspondence between the trees and the circles. It’s not obvious that there should be such a correspondence, with the right notation the correspondence is sort of a pun.<\/p>\n
The standard way to represent unlabeled trees is as a multiset<\/a> of their children. We use a multiset, not a set, because some elements will be repeated. We represent a leaf as a pair of parentheses: ()<\/code>.<\/p>\n
There is only one rooted tree with one node: ()<\/code>.<\/p>\n
There is only one rooted tree with one two nodes: (())<\/code>. Here the outer parentheses represent the root node and the inner parentheses represent its child.<\/p>\n
There are two rooted trees with three nodes, and we can represent them as ((()))<\/code> and ((),())<\/code>. The first is the straight line tree: a node that has a single child node that has a single child node. The second is a node that branches to two nodes. (Here’s where we need multisets.)<\/p>\n
The four rooted trees with four nodes can be represented as (((())))<\/code>, ((((),()))<\/code>, ((),(()))<\/code>, and ((),(),(),())<\/code>.<\/p>\n
Here are the nine rooted trees with five nodes:<\/p>\n
((((()))))\r\n((((),())))\r\n(((),(())))\r\n(((),(),()))\r\n((()),(()))\r\n((),((())))\r\n((),((),()))\r\n((),(),(()))\r\n((),(),(),())\r\n<\/pre>\n
The correspondence with non-overlapping circles removes the outer parentheses then joins the rest to form circles, with nested parentheses corresponding to concentric circles. A more geometric way to see the correspondence is to start at the bottom of the tree, replace leaves with circles, then work your way up circling connected components.<\/p>\n
<\/p>\n","protected":false},"excerpt":{"rendered":"
A few days ago I wrote a post on counting rooted trees. That post looked at the sequence c(n) which counts the number of rooted trees with n nodes. Here one node is distinguished as the root, but the nodes below the root are not distinguished from each other; all that matters is how the […]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"_acf_changed":false,"footnotes":""},"categories":[1],"tags":[],"class_list":["post-247540","post","type-post","status-publish","format-standard","hentry","category-uncategorized"],"acf":[],"aioseo_notices":[],"aioseo_head":"\n\t\t\n\t\n\t\n\t\n\t\n\t\n\t\t\n\t\t\n\t\t\n\t\t\n\t\t\n\t\t\n\t\t\n\t\t\n\t\t\n\t\t\n\t\t\n\t\t\n\t\t\n\n","aioseo_head_json":{"title":"Enumerating trees and circles","description":"There's a one-to-one correspondence between rooted trees and non-overlapping circles","canonical_url":"https:\/\/www.johndcook.com\/blog\/2026\/08\/05\/enumerating-trees-and-circles\/","robots":"max-image-preview:large","keywords":"","webmasterTools":{"miscellaneous":""},"schema":null,"og:locale":"en_US","og:site_name":"John D. Cook | Applied Mathematics Consulting","og:type":"article","og:title":"Enumerating trees and circles","og:description":"There's a one-to-one correspondence between rooted trees and non-overlapping circles","og:url":"https:\/\/www.johndcook.com\/blog\/2026\/08\/05\/enumerating-trees-and-circles\/","article:published_time":"2026-08-05T14:49:35+00:00","article:modified_time":"2026-08-05T15:51:33+00:00","twitter:card":"summary","twitter:title":"Enumerating trees and circles","twitter:description":"There's a one-to-one correspondence between rooted trees and non-overlapping circles","twitter:image":"https:\/\/www.johndcook.com\/blog\/wp-content\/uploads\/2022\/05\/twittercard.png"},"aioseo_meta_data":{"post_id":"247540","title":null,"description":"There's a one-to-one correspondence between rooted trees and non-overlapping circles","keywords":null,"keyphrases":{"focus":[],"additional":[]},"primary_term":null,"canonical_url":null,"og_title":null,"og_description":null,"og_object_type":"default","og_image_type":"default","og_image_url":null,"og_image_width":null,"og_image_height":null,"og_image_custom_url":null,"og_image_custom_fields":null,"og_video":"","og_custom_url":null,"og_article_section":null,"og_article_tags":null,"twitter_use_og":false,"twitter_card":"default","twitter_image_type":"default","twitter_image_url":null,"twitter_image_custom_url":null,"twitter_image_custom_fields":null,"twitter_title":null,"twitter_description":null,"schema":{"blockGraphs":[],"customGraphs":[],"default":{"data":{"Article":[],"Course":[],"Dataset":[],"FAQPage":[],"Movie":[],"Person":[],"Product":[],"ProductReview":[],"Car":[],"Recipe":[],"Service":[],"SoftwareApplication":[],"WebPage":[]},"graphName":"Article","isEnabled":true},"graphs":[]},"schema_type":"default","schema_type_options":null,"pillar_content":false,"robots_default":true,"robots_noindex":false,"robots_noarchive":false,"robots_nosnippet":false,"robots_nofollow":false,"robots_noimageindex":false,"robots_noodp":false,"robots_notranslate":false,"robots_max_snippet":"-1","robots_max_videopreview":"-1","robots_max_imagepreview":"large","priority":null,"frequency":"default","location":null,"local_seo":null,"breadcrumb_settings":null,"limit_modified_date":false,"created":"2026-08-05 12:48:37","updated":"2026-08-05 17:40:38","ai":{"faqs":[],"keyPoints":[],"schemas":[],"titles":[],"descriptions":[],"socialPosts":{"email":{"subject":"","preview":"","content":""},"linkedin":[],"twitter":[],"facebook":[],"instagram":[]}},"seo_analyzer_scan_date":null},"aioseo_breadcrumb":"
\n\t\t\tHome<\/a>\n\t\t<\/span>»<\/span>\n\t\t\tUncategorized<\/a>\n\t\t<\/span>»<\/span>\n\t\t\tEnumerating trees and circles\n\t\t<\/span><\/div>","aioseo_breadcrumb_json":[{"label":"Home","link":"https:\/\/www.johndcook.com\/blog"},{"label":"Uncategorized","link":"https:\/\/www.johndcook.com\/blog\/category\/uncategorized\/"},{"label":"Enumerating trees and circles","link":"https:\/\/www.johndcook.com\/blog\/2026\/08\/05\/enumerating-trees-and-circles\/"}],"_links":{"self":[{"href":"https:\/\/www.johndcook.com\/blog\/wp-json\/wp\/v2\/posts\/247540","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.johndcook.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.johndcook.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.johndcook.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.johndcook.com\/blog\/wp-json\/wp\/v2\/comments?post=247540"}],"version-history":[{"count":3,"href":"https:\/\/www.johndcook.com\/blog\/wp-json\/wp\/v2\/posts\/247540\/revisions"}],"predecessor-version":[{"id":247544,"href":"https:\/\/www.johndcook.com\/blog\/wp-json\/wp\/v2\/posts\/247540\/revisions\/247544"}],"wp:attachment":[{"href":"https:\/\/www.johndcook.com\/blog\/wp-json\/wp\/v2\/media?parent=247540"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.johndcook.com\/blog\/wp-json\/wp\/v2\/categories?post=247540"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.johndcook.com\/blog\/wp-json\/wp\/v2\/tags?post=247540"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}},{"id":247536,"date":"2026-08-04T08:18:58","date_gmt":"2026-08-04T13:18:58","guid":{"rendered":"https:\/\/www.johndcook.com\/blog\/?p=247536"},"modified":"2026-08-04T20:20:01","modified_gmt":"2026-08-05T01:20:01","slug":"metallic-alchemy","status":"publish","type":"post","link":"https:\/\/www.johndcook.com\/blog\/2026\/08\/04\/metallic-alchemy\/","title":{"rendered":"Mathematical alchemy"},"content":{"rendered":"
After writing the previous post<\/a> about metallic ratios, I thought about the analogy to alchemy and the attempt to make precious metals out of base metals.<\/p>\n
When can you make one metallic ratio out of another? Can you make the golden ratio out of the lead ratio?<\/p>\n
Before we can make gold out of lead, we have to say what lead is.<\/p>\n
Defining metallic ratios<\/h2>\n
The metallic ratios\u00a0M<\/em>(n<\/em>) can be defined several ways. The most interesting definition is the number whose continued fraction representation contains all n<\/em>s. A more prosaic but more convenient definition is the larger number that equals its reciprocal plus n<\/em>, which can be found using the quadratic formula.<\/p>\n
\"M(n)<\/p>\n
The golden ratio is\u00a0M<\/em>(1), the silver ratio is\u00a0M<\/em>(2), and the bronze ratio is\u00a0M<\/em>(3).<\/p>\n
Gold from silver and bronze?<\/h2>\n
Can you make the golden ratio out of the silver and bronze ratios? Not by integer arithmetic. The golden ratio involves \u221a5, the silver ratio \u221a2 and the bronze ratio \u221a13. No integer operations on the latter two radicals will produce the former, though you can come arbitrarily close.<\/p>\n
Gold from lead<\/h2>\n
The metallic ratios for n<\/em> > 3 don’t have standard names, but let’s call M<\/em>(4) the lead ratio. Can you make the golden ratio out of the lead ratio? Yes you can:<\/p>\n
M<\/em>(1) = (M<\/em>(4) \u2212 1)\/2.<\/p>\n
General solution<\/h2>\n
In general, when can you make\u00a0M<\/em>(n<\/em>) out of\u00a0M<\/em>(m<\/em>)? In abstract terms the question is when the fields<\/p>\n
\u211a(\u221a(n<\/em>\u00b2 + 4))<\/p>\n
and<\/p>\n
\u211a(\u221a(m<\/em>\u00b2 + 4))<\/p>\n
are the same, i.e. when adjoining \u221a(n<\/em>\u00b2 + 4) to the rational numbers gives the same field as adjoining \u221a(m<\/em>\u00b2 + 4) to the rational numbers. This occurs if and only if<\/p>\n
(n<\/em>\u00b2 + 4)\/(m\u00b2<\/em> + 4)<\/p>\n
is the square of a rational number.<\/p>\n
Bronze from copper and tin<\/h2>\n
Can you make bronze out of copper and tin? Yes, if you define\u00a0M<\/em>(36) to be the copper ratio and\u00a0M<\/em>(393) to be the tin ratio, because<\/p>\n
(3\u00b2 + 4)\/(36\u00b2 + 4) = (1\/10)\u00b2<\/p>\n
and<\/p>\n
(3\u00b2 + 4)\/(292\u00b2 + 4) = (1\/109)\u00b2.<\/p>\n","protected":false},"excerpt":{"rendered":"
After writing the previous post about metallic ratios, I thought about the analogy to alchemy and the attempt to make precious metals out of base metals. When can you make one metallic ratio out of another? Can you make the golden ratio out of the lead ratio? Before we can make gold out of lead, […]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"_acf_changed":false,"footnotes":""},"categories":[9],"tags":[94],"class_list":["post-247536","post","type-post","status-publish","format-standard","hentry","category-math","tag-number-theory"],"acf":[],"aioseo_notices":[],"aioseo_head":"\n\t\t\n\t\n\t\n\t\n\t\n\t\n\t\n\t\t\n\t\t\n\t\t\n\t\t\n\t\t\n\t\t\n\t\t\n\t\t\n\t\t\n\t\t\n\t\t\n\t\t\n\t\t\n\n","aioseo_head_json":{"title":"Metallic alchemy: making one metallic ratio from another","description":"By analogy with alchemy, can you make gold from lead? i.e. can you make the golden ratio by integer operations on the lead ratio?","canonical_url":"https:\/\/www.johndcook.com\/blog\/2026\/08\/04\/metallic-alchemy\/","robots":"max-image-preview:large","keywords":"number theory","webmasterTools":{"miscellaneous":""},"schema":null,"og:locale":"en_US","og:site_name":"John D. Cook | Applied Mathematics Consulting","og:type":"article","og:title":"Metallic alchemy: making one metallic ratio from another","og:description":"By analogy with alchemy, can you make gold from lead? i.e. can you make the golden ratio by integer operations on the lead ratio?","og:url":"https:\/\/www.johndcook.com\/blog\/2026\/08\/04\/metallic-alchemy\/","article:published_time":"2026-08-04T13:18:58+00:00","article:modified_time":"2026-08-05T01:20:01+00:00","twitter:card":"summary","twitter:title":"Metallic alchemy: making one metallic ratio from another","twitter:description":"By analogy with alchemy, can you make gold from lead? i.e. can you make the golden ratio by integer operations on the lead ratio?","twitter:image":"https:\/\/www.johndcook.com\/blog\/wp-content\/uploads\/2022\/05\/twittercard.png"},"aioseo_meta_data":{"post_id":"247536","title":"Metallic alchemy: making one metallic ratio from another","description":"By analogy with alchemy, can you make gold from lead? i.e. can you make the golden ratio by integer operations on the lead ratio?","keywords":null,"keyphrases":{"focus":[],"additional":[]},"primary_term":null,"canonical_url":null,"og_title":null,"og_description":null,"og_object_type":"default","og_image_type":"default","og_image_url":null,"og_image_width":null,"og_image_height":null,"og_image_custom_url":null,"og_image_custom_fields":null,"og_video":"","og_custom_url":null,"og_article_section":null,"og_article_tags":null,"twitter_use_og":false,"twitter_card":"default","twitter_image_type":"default","twitter_image_url":null,"twitter_image_custom_url":null,"twitter_image_custom_fields":null,"twitter_title":null,"twitter_description":null,"schema":{"blockGraphs":[],"customGraphs":[],"default":{"data":{"Article":[],"Course":[],"Dataset":[],"FAQPage":[],"Movie":[],"Person":[],"Product":[],"ProductReview":[],"Car":[],"Recipe":[],"Service":[],"SoftwareApplication":[],"WebPage":[]},"graphName":"Article","isEnabled":true},"graphs":[]},"schema_type":"default","schema_type_options":null,"pillar_content":false,"robots_default":true,"robots_noindex":false,"robots_noarchive":false,"robots_nosnippet":false,"robots_nofollow":false,"robots_noimageindex":false,"robots_noodp":false,"robots_notranslate":false,"robots_max_snippet":"-1","robots_max_videopreview":"-1","robots_max_imagepreview":"large","priority":null,"frequency":"default","location":null,"local_seo":null,"breadcrumb_settings":null,"limit_modified_date":false,"created":"2026-08-04 12:23:52","updated":"2026-08-05 17:40:38","ai":{"faqs":[],"keyPoints":[],"schemas":[],"titles":[],"descriptions":[],"socialPosts":{"email":{"subject":"","preview":"","content":""},"linkedin":[],"twitter":[],"facebook":[],"instagram":[]}},"seo_analyzer_scan_date":null},"aioseo_breadcrumb":"
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The golden ratio is the first and best known of the metallic ratios. I’ve written about the silver ratio a few times, most recently here<\/a>. And I’ve mentioned the bronze ratio<\/a> a couple times. The metallic ratios after bronze don’t have standard names.<\/p>\n
The n<\/em>th metallic ratio M<\/em>(n<\/em>) is the number whose continued fraction representation contains all n<\/em>s.<\/p>\n
\"n<\/p>\n
When n<\/em> = 1, 2, and 3 we get the gold, silver, and bronze ratios.<\/p>\n
You can approximate any positive real number as a ratio of metallic ratios. To see this, note that for large\u00a0n<\/em>, M<\/i>(n<\/em>)\u00a0is approximately\u00a0n<\/em>. For any positive rational number a<\/em>\/b<\/em>,<\/p>\n
\"\\lim_{n\\to\\infty}<\/p>\n
and so you can make\u00a0M<\/em>(na<\/em>) \/\u00a0M<\/em>(nb<\/em>) as close to\u00a0a<\/em>\/b<\/em> as you like by taking\u00a0n<\/em> large enough. And since the rationals are dense in the reals, you can approximate any positive real number as close as you’d like.<\/p>\n
Let’s look for metallic ratios whose ratios approximate \u03c0 to within 0.001 with the following Python code.<\/p>\n
from math import pi, sqrt\r\n\r\nM = lambda n: 0.5*(n + sqrt(n**2 + 4))\r\n\r\nfor n in range(1, 100):\r\n    a = round(pi*n)\r\n    b = n\r\n    r = M(a)\/M(b)\r\n    if abs(r - pi) < 0.001:\r\n        print(a, b, r)\r\n<\/pre>\n
This shows<\/p>\n
\u03c0 \u2248\u00a0M<\/em>(132) \/\u00a0M<\/em>(42) = 3.1412\u2026<\/p>\n
Could we find smaller numbers that work? The following code shows the answer is no.<\/p>\n
k = 132 + 42\r\n# loop over numbers whose sum is less than k\r\nfor n in range(1, k):\r\n    for a in range(1, n):\r\n        b = n - a\r\n        r = M(a)\/M(b)\r\n        if abs(r - pi) < 0.001:\r\n            print(a, b, r)\r\n            exit()\r\n<\/pre>\n
Related posts<\/h2>\n