{"id":3061,"date":"2020-04-19T13:50:21","date_gmt":"2020-04-19T13:50:21","guid":{"rendered":"https:\/\/students.pingry.org\/record\/?p=3061"},"modified":"2020-04-19T13:51:04","modified_gmt":"2020-04-19T13:51:04","slug":"fractals-math-problem","status":"publish","type":"post","link":"https:\/\/students.pingry.org\/record\/2020\/04\/19\/fractals-math-problem\/","title":{"rendered":"Fractals Math Problem"},"content":{"rendered":"\n\n[et_pb_section fb_built=&#8221;1&#8243; _builder_version=&#8221;3.18.9&#8243;][et_pb_row _builder_version=&#8221;3.18.9&#8243;][et_pb_column type=&#8221;4_4&#8243; _builder_version=&#8221;3.18.9&#8243; parallax=&#8221;off&#8221; parallax_method=&#8221;on&#8221;][et_pb_image src=&#8221;https:\/\/students.pingry.org\/record\/wp-content\/uploads\/sites\/3\/2020\/04\/fractalsMath.png&#8221; _builder_version=&#8221;3.18.9&#8243;][\/et_pb_image][et_pb_text text_line_height=&#8221;1.2em&#8221; _builder_version=&#8221;3.18.9&#8243; inline_fonts=&#8221;Inconsolata&#8221; text_font=&#8221;||||||||&#8221;]<p><span style=\"font-weight: normal;font-family: Inconsolata;font-size: small\">Fractals are self-similar.\u00a0<\/span><span style=\"font-family: Inconsolata;font-size: small\">The Sierpinski Triangle is an interesting fractal pattern. You can build one like this:<\/span><\/p>\n<ul>\n<li style=\"font-weight: 400\"><span style=\"font-weight: normal;font-family: Inconsolata;font-size: small\">1. Start with an \u201cupright\u201d equilateral triangle (as seen in stage zero)<\/span><\/li>\n<li style=\"font-weight: 400\"><span style=\"font-weight: normal;font-family: Inconsolata;font-size: small\">2. Place an upside-down triangle in the center of any upright triangle<\/span><\/li>\n<li style=\"font-weight: 400\"><span style=\"font-weight: normal;font-family: Inconsolata;font-size: small\">3. Repeat Step 2 infinitely\u00a0<\/span><\/li>\n<\/ul>\n<p><span style=\"font-weight: normal;font-family: Inconsolata;font-size: small\">Write a recursive formula that can model how many similar triangles you can count in the figure in a given stage. The figure above will be helpful in finding the pattern.<\/span><\/p>\n<p><span style=\"font-family: Inconsolata;font-weight: normal;font-size: small\"><b>Extra Challenge<\/b>: Turn that recursive formula into an explicit formula\u00a0<span>(i.e.\u00a0<\/span><span>find S(n), where n is the \u201cstage\u201d and S(n) is the number of triangles)<\/span><\/span><\/p>\n<p><span style=\"font-family: Inconsolata;font-weight: normal;font-size: small\"><\/span><\/p>[\/et_pb_text][\/et_pb_column][\/et_pb_row][et_pb_row _builder_version=&#8221;3.18.9&#8243;][et_pb_column type=&#8221;4_4&#8243; _builder_version=&#8221;3.18.9&#8243; parallax=&#8221;off&#8221; parallax_method=&#8221;on&#8221;][et_pb_toggle title=&#8221;Click for Solutions!&#8221; open_toggle_text_color=&#8221;#ffffff&#8221; open_toggle_background_color=&#8221;#0063a7&#8243; closed_toggle_text_color=&#8221;#ffffff&#8221; closed_toggle_background_color=&#8221;#0063a7&#8243; _builder_version=&#8221;3.18.9&#8243; title_level=&#8221;h6&#8243; title_font=&#8221;|600|||||||&#8221; title_text_align=&#8221;center&#8221; title_font_size=&#8221;16px&#8221; body_font=&#8221;||||||||&#8221; body_text_color=&#8221;#ffffff&#8221; body_line_height=&#8221;1.1em&#8221;]<p><strong>S(0) = 1<\/strong><\/p>\n<p><strong>S(n) = 3S(n-1) + 2<\/strong><\/p>\n<p><span style=\"font-weight: 400\">Base case is 1 because there is only 1 triangle in Stage 0.<\/span><\/p>\n<p><span style=\"font-weight: 400\">Look at the jump from stage 0 to 1. When you add the white triangle, you essentially have 3 smaller versions of Stage 0, along with the central triangle and the entire triangle. That\u2019s 3*1+1+1.\u00a0<\/span><\/p>\n<p><span style=\"font-weight: 400\">Stage 1 to 2 follows the same pattern. 3(4) + 1 + 1.<\/span><\/p>\n<p><span style=\"font-weight: 400\">Generalize the pattern to obtain your recursive step, S(n) = 3S(n-1) + 2.<\/span><\/p>\n<p><span style=\"font-weight: 400\">&#8212;<\/span><\/p>\n<p><span style=\"font-weight: 400\">In order to make an explicit formula, try expanding the recursive step.\u00a0<\/span><\/p>\n<p><span style=\"font-weight: 400\">S(n) = 3S(n-1) + 2<\/span><\/p>\n<p><span style=\"font-weight: 400\">Replace S(n-1)<\/span><\/p>\n<p><span style=\"font-weight: 400\">S(n) = 3( 3S(n-2) + 2 ) + 2<\/span><\/p>\n<p><span style=\"font-weight: 400\">Replace S(n-2)<\/span><\/p>\n<p><span style=\"font-weight: 400\">S(n) = 3( 3( 3S(n-3) + 2) + 2 ) + 2<\/span><\/p>\n<p><span style=\"font-weight: 400\">Simplify to make the pattern more clear.<\/span><\/p>\n<p><span style=\"font-weight: 400\">S(n) = 3^<\/span><span style=\"font-weight: 400\">3 *\u00a0<\/span><span style=\"font-weight: 400\">S(n-3) + 18 + 6\u00a0 + 2<\/span><\/p>\n<p><span style=\"font-weight: 400\">If you keep on replacing k times you find the pattern:<\/span><\/p>\n<p><span style=\"font-weight: 400\">S(n) = 3^<\/span><span style=\"font-weight: 400\">k *\u00a0<\/span><span style=\"font-weight: 400\">S(n-k) + 2 + 6 + 18 + \u2026\u00a0 2*3<\/span><span style=\"font-weight: 400\">k-1<\/span><\/p>\n<p><span style=\"font-weight: 400\">We can get rid of S(n-k) by using our base case of S(0). When k=n, S(n-k) is 0. Thus\u2026<\/span><\/p>\n<p><span style=\"font-weight: 400\">S(n) = 3^<\/span><span style=\"font-weight: 400\">n *\u00a0<\/span><span style=\"font-weight: 400\">S(0) + 2 + 6 + 18 + \u2026\u00a0 2*3<\/span><span style=\"font-weight: 400\">n-1<\/span><\/p>\n<p><span style=\"font-weight: 400\">Final Answer:<\/span><\/p>\n<p><span style=\"font-weight: 400\">S(n) =<\/span><span style=\"font-weight: 400\">3^<\/span><span style=\"font-weight: 400\">n<\/span><span style=\"font-weight: 400\"> + ( 2+6 + 18 + &#8230; 2*<\/span><span style=\"font-weight: 400\">3^(<\/span><span style=\"font-weight: 400\">n-1<\/span><span style=\"font-weight: 400\">) )<\/span><\/p>\n<p><span style=\"font-weight: 400\">Or more formally, using sigma notation:<\/span><\/p>\n<p><span style=\"font-weight: 400\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/students.pingry.org\/record\/wp-content\/uploads\/sites\/3\/2020\/04\/Screen-Shot-2020-04-19-at-9.51.00-AM.png\" width=\"178\" height=\"49\" alt=\"\" class=\"wp-image-3065 alignnone size-full\" \/><\/span><\/p>[\/et_pb_toggle][\/et_pb_column][\/et_pb_row][\/et_pb_section]\n\n","protected":false},"excerpt":{"rendered":"<p>Fractals are self-similar.\u00a0The Sierpinski Triangle is an interesting fractal pattern. You can build one like this: 1. Start with an \u201cupright\u201d equilateral triangle (as seen in stage zero) 2. Place an upside-down triangle in the center of any upright triangle 3. Repeat Step 2 infinitely\u00a0 Write a recursive formula that can model how many similar [&hellip;]<\/p>\n","protected":false},"author":32,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_et_pb_use_builder":"on","_et_pb_old_content":"","_et_gb_content_width":"","footnotes":""},"categories":[220,205],"tags":[],"class_list":["post-3061","post","type-post","status-publish","format-standard","hentry","category-fun","category-noah"],"_links":{"self":[{"href":"https:\/\/students.pingry.org\/record\/wp-json\/wp\/v2\/posts\/3061","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/students.pingry.org\/record\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/students.pingry.org\/record\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/students.pingry.org\/record\/wp-json\/wp\/v2\/users\/32"}],"replies":[{"embeddable":true,"href":"https:\/\/students.pingry.org\/record\/wp-json\/wp\/v2\/comments?post=3061"}],"version-history":[{"count":3,"href":"https:\/\/students.pingry.org\/record\/wp-json\/wp\/v2\/posts\/3061\/revisions"}],"predecessor-version":[{"id":3067,"href":"https:\/\/students.pingry.org\/record\/wp-json\/wp\/v2\/posts\/3061\/revisions\/3067"}],"wp:attachment":[{"href":"https:\/\/students.pingry.org\/record\/wp-json\/wp\/v2\/media?parent=3061"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/students.pingry.org\/record\/wp-json\/wp\/v2\/categories?post=3061"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/students.pingry.org\/record\/wp-json\/wp\/v2\/tags?post=3061"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}