{"id":70,"date":"2017-02-04T11:31:56","date_gmt":"2017-02-04T11:31:56","guid":{"rendered":"https:\/\/petermc.net\/blog\/?p=70"},"modified":"2017-02-04T11:31:56","modified_gmt":"2017-02-04T11:31:56","slug":"pi-is-trancendental","status":"publish","type":"post","link":"https:\/\/petermc.net\/blog\/2017\/02\/04\/pi-is-trancendental\/","title":{"rendered":"Pi is trancendental"},"content":{"rendered":"

First the rabbit. The introduction of this function is the part which I don’t know how to motivate. Let \"f\" be a polynomial and define<\/p>\n

\"\displaystyle <\/p>\n

Integration by parts gives the recursion<\/p>\n

\"\displaystyle <\/p>\n

and therefore we have the formula<\/p>\n

\"\displaystyle <\/p>\n

Now suppose (for want of a contradition) that \"\pi. Let the set of Galois conjugates of \"\pi be \"\{a_1,\ldots,a_k\}.\" Then we have \"\prod_{i=1}^k, expand this as \"\sum_{J\subset[k]}e^{\sum_{j\in and rewrite as<\/p>\n

\"\displaystyle <\/p>\n

where the \"\theta_i\" are the nonzero exponents.<\/p>\n

Now consider<\/p>\n

\"\displaystyle <\/p>\n

Let \"N\" be an integer such that \"N\pi is an algebraic integer. Let \"p\" be a (large) prime and we choose to take<\/p>\n

\"\displaystyle <\/p>\n

There are absolute constants \"A\" and \"B\" (independent of \"p\") such that<\/p>\n

\"\displaystyle <\/p>\n

(look at the integral definition of \"I(z,f)\" and apply the naive estimate).<\/p>\n

Now consider \"\sum_{i=1}^d. It is a Galois-invariant algebraic integer, hence an integer. We have<\/p>\n

\"\displaystyle <\/p>\n

Here higher order means at least \"p\" derivatives appearing. Each of these higher order terms is divisible by \"p!\", hence by \"p\". Since \"p\" is prime, for large enough \"p\", \"\sum_{i=1}^d is not divisible by \"p\", hence nonzero.<\/p>\n

Now every term is divisible by \"(p-1)!\", so we get the lower bound<\/p>\n

\"\displaystyle <\/p>\n

As there are infinitely many primes, we can send choose \"p\" large enough to get a contradiction, QED.<\/p>\n","protected":false},"excerpt":{"rendered":"

First the rabbit. The introduction of this function is the part which I don’t know how to motivate. Let be a polynomial and define Integration by parts gives the recursion and therefore we have the formula Now suppose (for want … Continue reading →<\/span><\/a><\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_jetpack_newsletter_access":"","_jetpack_dont_email_post_to_subs":false,"_jetpack_newsletter_tier_id":0,"_jetpack_memberships_contains_paywalled_content":false,"_jetpack_memberships_contains_paid_content":false,"footnotes":"","jetpack_publicize_message":"","jetpack_publicize_feature_enabled":true,"jetpack_social_post_already_shared":true,"jetpack_social_options":{"image_generator_settings":{"template":"highway","default_image_id":0,"font":"","enabled":false},"version":2},"jetpack_post_was_ever_published":false},"categories":[6],"tags":[],"class_list":["post-70","post","type-post","status-publish","format-standard","hentry","category-maths"],"jetpack_publicize_connections":[],"jetpack_featured_media_url":"","jetpack_sharing_enabled":false,"jetpack_shortlink":"https:\/\/wp.me\/p7V6a7-18","_links":{"self":[{"href":"https:\/\/petermc.net\/blog\/wp-json\/wp\/v2\/posts\/70","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/petermc.net\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/petermc.net\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/petermc.net\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/petermc.net\/blog\/wp-json\/wp\/v2\/comments?post=70"}],"version-history":[{"count":10,"href":"https:\/\/petermc.net\/blog\/wp-json\/wp\/v2\/posts\/70\/revisions"}],"predecessor-version":[{"id":84,"href":"https:\/\/petermc.net\/blog\/wp-json\/wp\/v2\/posts\/70\/revisions\/84"}],"wp:attachment":[{"href":"https:\/\/petermc.net\/blog\/wp-json\/wp\/v2\/media?parent=70"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/petermc.net\/blog\/wp-json\/wp\/v2\/categories?post=70"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/petermc.net\/blog\/wp-json\/wp\/v2\/tags?post=70"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}