{"id":1641,"date":"2017-12-05T00:28:48","date_gmt":"2017-12-05T05:28:48","guid":{"rendered":"http:\/\/blog.richmond.edu\/math320\/?p=1641"},"modified":"2017-12-06T19:16:53","modified_gmt":"2017-12-07T00:16:53","slug":"muddiest-point-1130","status":"publish","type":"post","link":"https:\/\/blog.richmond.edu\/math320\/2017\/12\/05\/muddiest-point-1130\/","title":{"rendered":"Muddiest Point 11\/30"},"content":{"rendered":"<p>I thought that the muddiest point from last Thursday&#8217;s class was when we went over proving that <img decoding=\"async\" src=\"https:\/\/s0.wp.com\/latex.php?latex=%5Cint+kf+%3D+k%5Cint+f&#038;bg=ffffff&#038;fg=000&#038;s=0&#038;c=20201002\" alt=\"&#92;int kf = k&#92;int f\" class=\"latex\" \/>. The reason why I thought that this was the muddiest is because the proof required to cases. A case in which k was negative and a case that case was positive . This is often a detail that is easy to look over as it seems so trivial which is why it is important to go over it. To start the proof we assume that f is integrable and that <img decoding=\"async\" src=\"https:\/\/s0.wp.com\/latex.php?latex=k%3E0&#038;bg=ffffff&#038;fg=000&#038;s=0&#038;c=20201002\" alt=\"k&gt;0\" class=\"latex\" \/>. Then we see that <img decoding=\"async\" src=\"https:\/\/s0.wp.com\/latex.php?latex=U%28kf%2CP%29+%3D%5Csum+sup%5C%7Bkf%28x%29%3A+x+%5Cin+%5Bx_%7Bk-1%7D%2Cx_k%5D+%5C%7D&#038;bg=ffffff&#038;fg=000&#038;s=0&#038;c=20201002\" alt=\"U(kf,P) =&#92;sum sup&#92;{kf(x): x &#92;in [x_{k-1},x_k] &#92;}\" class=\"latex\" \/>. We already know from earlier this semester that <img decoding=\"async\" src=\"https:\/\/s0.wp.com\/latex.php?latex=csup%28A%29+%3D+sup%28cA%29&#038;bg=ffffff&#038;fg=000&#038;s=0&#038;c=20201002\" alt=\"csup(A) = sup(cA)\" class=\"latex\" \/> therefore we can say that $latex\u00a0sup\\{kf(x): x \\in [x_{k-1},x_k] \\} = ksup\\{f(x): x \\in [x_{k-1},x_k] \\}$ and in turn <img decoding=\"async\" src=\"https:\/\/s0.wp.com\/latex.php?latex=U%28kf%2CP%29+%3D+kU%28f%2CP%29&#038;bg=ffffff&#038;fg=000&#038;s=0&#038;c=20201002\" alt=\"U(kf,P) = kU(f,P)\" class=\"latex\" \/>. we can use the same logic for the lower sums as well. Since\u00a0<img decoding=\"async\" src=\"https:\/\/s0.wp.com\/latex.php?latex=L%28kf%2CP%29+%3D+inf%5C%7Bkf%28x%29%3A+x+%5Cin+%5Bx_%7Bk-1%7D%2Cx_k%5D+%5C%7D&#038;bg=ffffff&#038;fg=000&#038;s=0&#038;c=20201002\" alt=\"L(kf,P) = inf&#92;{kf(x): x &#92;in [x_{k-1},x_k] &#92;}\" class=\"latex\" \/> we can once again use our knowledge about infinums and supremums from earlier in the semester that <img decoding=\"async\" src=\"https:\/\/s0.wp.com\/latex.php?latex=kinf%28A%29+%3D+inf%28kA%29&#038;bg=ffffff&#038;fg=000&#038;s=0&#038;c=20201002\" alt=\"kinf(A) = inf(kA)\" class=\"latex\" \/> to justify the statement that\u00a0<img decoding=\"async\" src=\"https:\/\/s0.wp.com\/latex.php?latex=inf%5C%7Bkf%28x%29%3A+x+%5Cin+%5Bx_%7Bk-1%7D%2Cx_k%5D+%5C%7D+%3D+kinf%5C%7Bf%28x%29%3A+x+%5Cin+%5Bx_%7Bk-1%7D%2Cx_k%5D+%5C%7D&#038;bg=ffffff&#038;fg=000&#038;s=0&#038;c=20201002\" alt=\"inf&#92;{kf(x): x &#92;in [x_{k-1},x_k] &#92;} = kinf&#92;{f(x): x &#92;in [x_{k-1},x_k] &#92;}\" class=\"latex\" \/> and <img decoding=\"async\" src=\"https:\/\/s0.wp.com\/latex.php?latex=L%28kf%2CP%29+%3D+kL%28f%2CP%29&#038;bg=ffffff&#038;fg=000&#038;s=0&#038;c=20201002\" alt=\"L(kf,P) = kL(f,P)\" class=\"latex\" \/>. Since the lower sums and Upper sums are unchanged by the location of the <img decoding=\"async\" src=\"https:\/\/s0.wp.com\/latex.php?latex=k&#038;bg=ffffff&#038;fg=000&#038;s=0&#038;c=20201002\" alt=\"k\" class=\"latex\" \/> we can conclude that if <img decoding=\"async\" src=\"https:\/\/s0.wp.com\/latex.php?latex=k&#038;bg=ffffff&#038;fg=000&#038;s=0&#038;c=20201002\" alt=\"k\" class=\"latex\" \/> is positive then\u00a0<img decoding=\"async\" src=\"https:\/\/s0.wp.com\/latex.php?latex=%5Cint+kf+%3D+k%5Cint+f&#038;bg=ffffff&#038;fg=000&#038;s=0&#038;c=20201002\" alt=\"&#92;int kf = k&#92;int f\" class=\"latex\" \/> and that there exists a partition <img decoding=\"async\" src=\"https:\/\/s0.wp.com\/latex.php?latex=P_n&#038;bg=ffffff&#038;fg=000&#038;s=0&#038;c=20201002\" alt=\"P_n\" class=\"latex\" \/> such that $ \\lim_{n \\to \\infty} [U(kf, P_n) \u2212 L(kf, P_n)] = \\lim_{n \\to \\infty} k[U(f,P_n) \u2212 L(f,P_n)] = 0$. Now we must consider the case that <img decoding=\"async\" src=\"https:\/\/s0.wp.com\/latex.php?latex=k&#038;bg=ffffff&#038;fg=000&#038;s=0&#038;c=20201002\" alt=\"k\" class=\"latex\" \/> is negative. If k is negative <img decoding=\"async\" src=\"https:\/\/s0.wp.com\/latex.php?latex=U%28f%2Cp%29&#038;bg=ffffff&#038;fg=000&#038;s=0&#038;c=20201002\" alt=\"U(f,p)\" class=\"latex\" \/> will be flipped when multiplied by k and in turn <img decoding=\"async\" src=\"https:\/\/s0.wp.com\/latex.php?latex=L%28f%2CP%29&#038;bg=ffffff&#038;fg=000&#038;s=0&#038;c=20201002\" alt=\"L(f,P)\" class=\"latex\" \/> will be flipped when multiplied by <img decoding=\"async\" src=\"https:\/\/s0.wp.com\/latex.php?latex=k&#038;bg=ffffff&#038;fg=000&#038;s=0&#038;c=20201002\" alt=\"k\" class=\"latex\" \/>. This distinction is important to make and is why the second case is needed. With the lower and upper sums flipped by the multiplication by <img decoding=\"async\" src=\"https:\/\/s0.wp.com\/latex.php?latex=k&#038;bg=ffffff&#038;fg=000&#038;s=0&#038;c=20201002\" alt=\"k\" class=\"latex\" \/> we now have that <img decoding=\"async\" src=\"https:\/\/s0.wp.com\/latex.php?latex=U%28kf%2CP%29+%3D+kL%28f%2CP%29&#038;bg=ffffff&#038;fg=000&#038;s=0&#038;c=20201002\" alt=\"U(kf,P) = kL(f,P)\" class=\"latex\" \/> and <img decoding=\"async\" src=\"https:\/\/s0.wp.com\/latex.php?latex=kU%28f%2CP%29+%3D+L%28kf%2CP%29&#038;bg=ffffff&#038;fg=000&#038;s=0&#038;c=20201002\" alt=\"kU(f,P) = L(kf,P)\" class=\"latex\" \/>. Therefore it is still the case that\u00a0$ \\lim_{n \\rightarrow \\infty} [U(kf, P_n) \u2212 L(kf, P_n)] = \\lim_{n \\rightarrow \\infty} k[U(f,P_n) \u2212 L(f,P_n)] = 0$. So we can conclude that\u00a0<img decoding=\"async\" src=\"https:\/\/s0.wp.com\/latex.php?latex=%5Cint+kf+%3D+k%5Cint+f&#038;bg=ffffff&#038;fg=000&#038;s=0&#038;c=20201002\" alt=\"&#92;int kf = k&#92;int f\" class=\"latex\" \/> regardless of the value of <img decoding=\"async\" src=\"https:\/\/s0.wp.com\/latex.php?latex=k&#038;bg=ffffff&#038;fg=000&#038;s=0&#038;c=20201002\" alt=\"k\" class=\"latex\" \/>.<\/p>\n","protected":false},"excerpt":{"rendered":"<p>I thought that the muddiest point from last Thursday&#8217;s class was when we went over proving that . The reason why I thought that this was the muddiest is because the proof required to cases. A case in which k was negative and a case that case was positive . This is often a detail [&hellip;]<\/p>\n","protected":false},"author":3524,"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":[58822],"tags":[],"class_list":["post-1641","post","type-post","status-publish","format-standard","hentry","category-muddiest-point"],"jetpack_publicize_connections":[],"jetpack_featured_media_url":"","jetpack_sharing_enabled":true,"jetpack_shortlink":"https:\/\/wp.me\/p7L4E1-qt","jetpack-related-posts":[],"_links":{"self":[{"href":"https:\/\/blog.richmond.edu\/math320\/wp-json\/wp\/v2\/posts\/1641","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\/3524"}],"replies":[{"embeddable":true,"href":"https:\/\/blog.richmond.edu\/math320\/wp-json\/wp\/v2\/comments?post=1641"}],"version-history":[{"count":0,"href":"https:\/\/blog.richmond.edu\/math320\/wp-json\/wp\/v2\/posts\/1641\/revisions"}],"wp:attachment":[{"href":"https:\/\/blog.richmond.edu\/math320\/wp-json\/wp\/v2\/media?parent=1641"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/blog.richmond.edu\/math320\/wp-json\/wp\/v2\/categories?post=1641"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/blog.richmond.edu\/math320\/wp-json\/wp\/v2\/tags?post=1641"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}