{"id":1033,"date":"2014-10-31T23:47:46","date_gmt":"2014-11-01T03:47:46","guid":{"rendered":"http:\/\/www.curtisbright.com\/bln\/?p=1033"},"modified":"2014-10-31T23:47:46","modified_gmt":"2014-11-01T03:47:46","slug":"revisiting-a-lemma","status":"publish","type":"post","link":"http:\/\/localhost\/blog\/index.php\/2014\/10\/31\/revisiting-a-lemma\/","title":{"rendered":"Revisiting a lemma"},"content":{"rendered":"

We’ve discussed before the “trigonometric harmonic” series $\\sum_{n=1}^\\infty\\cos n\/n$. In particular, we\u00a0showed<\/a> that the series converges (conditionally<\/a>). The argument involved the partial sums\u00a0of the sequence\u00a0$\\{\\cos n\\}_{n=1}^\\infty$, and\u00a0we denoted these by $C(m)$. The closed-form expression we found for $C(m)$ involved the quantity $\\cos m-\\cos(m+1)$; in this post we show that this expression can also be written in the alternative form $2\\sin(1\/2)\\sin(m+1\/2)$.<\/p>\n

<\/p>\n

The proof is a straightforward application of the complex exponential expression for\u00a0sine and cosine:<\/p>\n

\\begin{align*}
\n2\\sin(1\/2)\\sin(m+1\/2) &= 2\\frac{e^{i\/2}-e^{-i\/2}}{2i}\\cdot\\frac{e^{i(m+1\/2)}-e^{-i(m+1\/2)}}{2i} \\\\
\n&= \\frac{e^{i(m+1)}-e^{im}-e^{-im}+e^{-i(m+1)}}{-2} \\\\
\n&= \\cos m-\\cos(m+1)
\n\\end{align*}<\/p>\n

Applying this with $m:=0$ yields\u00a0$2\\sin(1\/2)^2=1-\\cos 1$, and we have<\/p>\n

\\begin{align*}
\nC(m) &= \\frac{\\cos m-\\cos(m+1)}{2(1-\\cos 1)}-\\frac{1}{2} \\\\
\n&= \\frac{2\\sin(1\/2)\\sin(m+1\/2)}{4\\sin(1\/2)^2}-\\frac{1}{2} \\\\
\n&= \\frac{\\sin(m+1\/2)}{2\\sin(1\/2)}-\\frac{1}{2} .
\n\\end{align*}<\/p>\n

Written in this form, the naive upper bound on $\\lvert C(m)\\rvert$ becomes<\/p>\n

\\[ \\frac{1}{2\\sin(1\/2)}+\\frac{1}{2} \\approx\u00a01.54\u00a0, \\]<\/p>\n

which is actually slightly better than the previous upper bound we gave of<\/p>\n

\\[ \\frac{1}{1-\\cos1}+\\frac{1}{2}\u00a0\\approx 2.68\u00a0. \\]<\/p>\n

In fact, it is easy to see that our new upper bound the best possible, since it is reached at for example $m:=(3\\pi-1)\/2$.<\/p>\n","protected":false},"excerpt":{"rendered":"

We’ve discussed before the “trigonometric harmonic” series $\\sum_{n=1}^\\infty\\cos n\/n$. In particular, we\u00a0showed that the series converges (conditionally). The argument involved the partial sums\u00a0of the sequence\u00a0$\\{\\cos n\\}_{n=1}^\\infty$, and\u00a0we denoted these by $C(m)$. The closed-form expression we found for $C(m)$ involved the … Continue reading →<\/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\/1033"}],"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=1033"}],"version-history":[{"count":0,"href":"http:\/\/localhost\/blog\/index.php\/wp-json\/wp\/v2\/posts\/1033\/revisions"}],"wp:attachment":[{"href":"http:\/\/localhost\/blog\/index.php\/wp-json\/wp\/v2\/media?parent=1033"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"http:\/\/localhost\/blog\/index.php\/wp-json\/wp\/v2\/categories?post=1033"},{"taxonomy":"post_tag","embeddable":true,"href":"http:\/\/localhost\/blog\/index.php\/wp-json\/wp\/v2\/tags?post=1033"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}