{"id":1072,"date":"2014-11-22T12:26:05","date_gmt":"2014-11-22T17:26:05","guid":{"rendered":"http:\/\/www.curtisbright.com\/bln\/?p=1072"},"modified":"2014-11-22T12:26:05","modified_gmt":"2014-11-22T17:26:05","slug":"a-sum-of-sines","status":"publish","type":"post","link":"http:\/\/localhost\/blog\/index.php\/2014\/11\/22\/a-sum-of-sines\/","title":{"rendered":"A sum of sines"},"content":{"rendered":"<p>In this post I want to prove a lemma which gives a closed-form expression for the summation $\\sum_{n=1}^m\\sin(nx)$. The method of proof has <a href=\"http:\/\/www.curtisbright.com\/bln\/2014\/04\/29\/harmonic-series-variant-solution\/\">come up before<\/a>; it uses basic algebra, the complex exponential expression for sine, and the summation of a geometric series formula.<\/p>\n<p><!--more--><\/p>\n<p>\\begin{align*}<br \/>\n\\sum_{n=1}^m\\sin(nx) &amp;= \\sum_{n=1}^m \\frac{e^{inx}-e^{-inx}}{2i} \\\\<br \/>\n&amp;= \\frac{1}{2i}\\Bigl(e^{ix}\\frac{e^{imx}-1}{e^{ix}-1}-e^{-ix}\\frac{e^{-imx}-1}{e^{-ix}-1}\\Bigr) \\\\<br \/>\n&amp;= \\frac{e^{ix}(e^{-ix}-1)(e^{imx}-1)-e^{-ix}(e^{ix}-1)(e^{-mx}-1)}{2i(e^{ix}-1)(e^{-ix}-1)} \\\\<br \/>\n&amp;= \\frac{(1-e^{ix})(e^{imx}-1)-(1-e^{-ix})(e^{-mx}-1)}{2i(e^{ix\/2}-e^{-ix\/2})(e^{-ix\/2}-e^{ix\/2})} \\\\<br \/>\n&amp;= \\frac{e^{imx}-e^{-imx}-(e^{i(m+1)x}-e^{-i(m+1)x})+e^{ix}-e^{-ix}}{-2i(e^{ix\/2}-e^{-ix\/2})^2} \\\\<br \/>\n&amp;= \\frac{\\sin(mx)-\\sin((m+1)x)+\\sin x}{4((e^{ix\/2}-e^{-ix\/2})\/2i)^2} \\\\<br \/>\n&amp;= \\frac{\\sin(mx)-\\sin((m+1)x)+\\sin x}{4\\sin(x\/2)^2}<br \/>\n\\end{align*}<\/p>\n<p>Most of this is just grunt-work algebra, although in the fourth line\u00a0a factor of $e^{ix\/2}$ was creatively moved from one factor in the denominator to the other. This allows us to write the denominator in terms of sine; the denominator would end up being $2-2\\cos x$ if this step was skipped.<\/p>\n","protected":false},"excerpt":{"rendered":"<p>In this post I want to prove a lemma which gives a closed-form expression for the summation $\\sum_{n=1}^m\\sin(nx)$. The method of proof has come up before; it uses basic algebra, the complex exponential expression for sine, and the summation of &hellip; <a href=\"http:\/\/localhost\/blog\/index.php\/2014\/11\/22\/a-sum-of-sines\/\">Continue reading <span class=\"meta-nav\">&rarr;<\/span><\/a><\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[4],"tags":[],"_links":{"self":[{"href":"http:\/\/localhost\/blog\/index.php\/wp-json\/wp\/v2\/posts\/1072"}],"collection":[{"href":"http:\/\/localhost\/blog\/index.php\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"http:\/\/localhost\/blog\/index.php\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"http:\/\/localhost\/blog\/index.php\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"http:\/\/localhost\/blog\/index.php\/wp-json\/wp\/v2\/comments?post=1072"}],"version-history":[{"count":0,"href":"http:\/\/localhost\/blog\/index.php\/wp-json\/wp\/v2\/posts\/1072\/revisions"}],"wp:attachment":[{"href":"http:\/\/localhost\/blog\/index.php\/wp-json\/wp\/v2\/media?parent=1072"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"http:\/\/localhost\/blog\/index.php\/wp-json\/wp\/v2\/categories?post=1072"},{"taxonomy":"post_tag","embeddable":true,"href":"http:\/\/localhost\/blog\/index.php\/wp-json\/wp\/v2\/tags?post=1072"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}