{"id":1599,"date":"2017-11-29T22:25:19","date_gmt":"2017-11-30T03:25:19","guid":{"rendered":"http:\/\/blog.richmond.edu\/math320\/?p=1599"},"modified":"2017-11-29T22:25:19","modified_gmt":"2017-11-30T03:25:19","slug":"muddiest-point-1128","status":"publish","type":"post","link":"https:\/\/blog.richmond.edu\/math320\/2017\/11\/29\/muddiest-point-1128\/","title":{"rendered":"Muddiest Point 11\/28"},"content":{"rendered":"

The muddiest point for me from the lecture on Tuesday is visualizing a proposition we proved. The proposition states that if a function g from [0,1] to [0,1] is bijective and increasing, then the sum of its integral from 0 to 1 and its inverse’s integral from 0 to 1 is 1. It’s not essentially related to our topic but since I had a hard time making sense of that graphically, I thought I’d use this opportunity to clear this up.<\/p>\n


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In the graph above, I used y=x^2 where x is from 0 to 1 as an example. To find the inverse of this function, we simply flip it along the y=x line. Observe that the integral of y=x^2 from 0 to 1 is the area of D. Because the function is from [0,1] to [0,1], we can draw a 1×1 square, and it is symmetric along the y=x line. Thus, we can map D to A because their areas are the same. Then, we know that the integral of the inverse function is the area of B,C,D. Adding them together, we get the area of A,B,C,D, which is just the whole square. And since the square is 1×1, we have that the sum of the integrals is 1. This fact is not limited to y=x^2. Any function from [0,1] to [0,1] would work because of the symmetry.<\/p>\n","protected":false},"excerpt":{"rendered":"

The muddiest point for me from the lecture on Tuesday is visualizing a proposition we proved. The proposition states that if a function g from [0,1] to [0,1] is bijective and increasing, then the sum of its integral from 0 to 1 and its inverse’s integral from 0 to 1 is 1. It’s not essentially […]<\/p>\n","protected":false},"author":3539,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"jetpack_post_was_ever_published":false,"_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}},"categories":[58820,58822],"tags":[],"class_list":["post-1599","post","type-post","status-publish","format-standard","hentry","category-daily-blogs","category-muddiest-point"],"jetpack_publicize_connections":[],"jetpack_featured_media_url":"","jetpack_sharing_enabled":true,"jetpack_shortlink":"https:\/\/wp.me\/p7L4E1-pN","jetpack-related-posts":[],"_links":{"self":[{"href":"https:\/\/blog.richmond.edu\/math320\/wp-json\/wp\/v2\/posts\/1599","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/blog.richmond.edu\/math320\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/blog.richmond.edu\/math320\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/blog.richmond.edu\/math320\/wp-json\/wp\/v2\/users\/3539"}],"replies":[{"embeddable":true,"href":"https:\/\/blog.richmond.edu\/math320\/wp-json\/wp\/v2\/comments?post=1599"}],"version-history":[{"count":0,"href":"https:\/\/blog.richmond.edu\/math320\/wp-json\/wp\/v2\/posts\/1599\/revisions"}],"wp:attachment":[{"href":"https:\/\/blog.richmond.edu\/math320\/wp-json\/wp\/v2\/media?parent=1599"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/blog.richmond.edu\/math320\/wp-json\/wp\/v2\/categories?post=1599"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/blog.richmond.edu\/math320\/wp-json\/wp\/v2\/tags?post=1599"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}