{"id":10718,"date":"2017-10-02T12:43:03","date_gmt":"2017-10-02T19:43:03","guid":{"rendered":"https:\/\/magoosh.com\/hs\/?p=10718"},"modified":"2018-10-26T06:23:33","modified_gmt":"2018-10-26T13:23:33","slug":"ap-calculus-bc-review-radius-interval-convergence","status":"publish","type":"post","link":"https:\/\/magoosh.com\/hs\/ap\/ap-calculus-bc-review-radius-interval-convergence\/","title":{"rendered":"AP Calculus BC Review: Radius and Interval of Convergence"},"content":{"rendered":"<p>Whenever you work with a power series, you have to be careful about its radius and interval of convergence.  This article reviews the definitions and techniques for finding radius and interval of convergence of power series.  And we&#8217;ll also see a few examples similar to those you might find on the AP Calculus BC exam.<\/p>\n<h2>The Radius and Interval of Convergence<\/h2>\n<p>Recall that a <strong>power series<\/strong>, with center <em>c<\/em>, is a series of functions of the following form.<\/p>\n<p><img decoding=\"async\" src=\"https:\/\/magoosh.com\/hs\/files\/2017\/07\/power_series.gif\" alt=\"Power series\" width=\"372\" height=\"77\" class=\"aligncenter size-full wp-image-11009\" \/><\/p>\n<p>Now anytime you have an infinite series (infinitely many terms), you have to worry about issues of convergence.  In fact, the series may converge (have a finite sum) for some values of <em>x<\/em>, but diverge at others.<\/p>\n<p>For more about convergence and divergence in general, check out: <a href=\"https:\/\/magoosh.com\/hs\/ap\/ap-calculus-bc-review-series-convergence\/\">AP Calculus BC Review: Series Convergence<\/a>.<\/p>\n<h3>Three Possibilities<\/h3>\n<p>There are three things that could happen.<\/p>\n<ol>\n<li>The series converges regardless of <em>x<\/em>.<\/li>\n<li>There may be a limited range of <em>x<\/em>-values, called the <strong>interval of convergence<\/strong>, for which the series converges.  And the series diverges outside of that interval.  <\/li>\n<li>The converges only at the center <em>x<\/em> = <em>c<\/em>, diverging at every other <em>x<\/em>-value.<\/li>\n<\/ol>\n<p>Now, that second point is probably the trickiest.  When a series has a finite interval of convergence, then it&#8217;s always centered at&#8230; well&#8230; the <em>center<\/em>!  <\/p>\n<p>In other words, there would be a finite positive number <em>R<\/em>, called the <strong>radius of convergence<\/strong>, such that the series converges for all <em>x<\/em>-values that are within <em>R<\/em> units from the center <em>c<\/em>.  Any <em>x<\/em>-values greater than a distance of <em>R<\/em> from <em>c<\/em> would cause the series to diverge.<\/p>\n<p>But what about those points that are exactly <em>R<\/em> units away?  That is, <em>c<\/em> &#8211; <em>R<\/em> and <em>c<\/em> + <em>R<\/em>?<\/p>\n<p>Here, there are a number of possibilities as well.  The series could converge or diverge at each of the two endpoints.  See the table below for the four combinations.<\/p>\n<table id=\"tablepress-138\" class=\"tablepress tablepress-id-138 tablepress-responsive\">\n<thead>\n<tr class=\"row-1 odd\">\n<th class=\"column-1\">Behavior at left endpoint,<br \/>\n<em>c<\/em> &#8211; <em>R<\/em><\/th>\n<th class=\"column-2\">Behavior at right endpoint,<br \/>\n <em>c<\/em> + <em>R<\/em><\/th>\n<th class=\"column-3\">Interval of convergence<\/th>\n<th class=\"column-4\">Interval of convergence <br \/>\nas an inequality<\/th>\n<\/tr>\n<\/thead>\n<tbody class=\"row-hover\">\n<tr class=\"row-2 even\">\n<td class=\"column-1\">Converges<\/td>\n<td class=\"column-2\">Converges<\/td>\n<td class=\"column-3\">[<em>c<\/em> &#8211; <em>R<\/em>, <em>c<\/em> + <em>R<\/em>]<\/td>\n<td class=\"column-4\"><em>c<\/em> &#8211; <em>R<\/em> &le; <em>x<\/em> &le; <em>c<\/em> + <em>R<\/em><\/td>\n<\/tr>\n<tr class=\"row-3 odd\">\n<td class=\"column-1\">Converges<\/td>\n<td class=\"column-2\">Diverges<\/td>\n<td class=\"column-3\">[<em>c<\/em> &#8211; <em>R<\/em>, <em>c<\/em> + <em>R<\/em>)<\/td>\n<td class=\"column-4\"><em>c<\/em> &#8211; <em>R<\/em> &le; <em>x<\/em> &lt; <em>c<\/em> + <em>R<\/em><\/td>\n<\/tr>\n<tr class=\"row-4 even\">\n<td class=\"column-1\">Diverges<\/td>\n<td class=\"column-2\">Converges<\/td>\n<td class=\"column-3\">(<em>c<\/em> &#8211; <em>R<\/em>, <em>c<\/em> + <em>R<\/em>]<\/td>\n<td class=\"column-4\"><em>c<\/em> &#8211; <em>R<\/em> &lt; <em>x<\/em> &le; <em>c<\/em> + <em>R<\/em><\/td>\n<\/tr>\n<tr class=\"row-5 odd\">\n<td class=\"column-1\">Diverges<\/td>\n<td class=\"column-2\">Diverges<\/td>\n<td class=\"column-3\">(<em>c<\/em> &#8211; <em>R<\/em>, <em>c<\/em> + <em>R<\/em>)<\/td>\n<td class=\"column-4\"><em>c<\/em> &#8211; <em>R<\/em> &lt; <em>x<\/em> &lt; <em>c<\/em> + <em>R<\/em><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p><!-- #tablepress-138 from cache --><\/p>\n<h2>Techniques for the Finding Interval of Convergence<\/h2>\n<p>There are typically two phases to finding the interval of convergence.<\/p>\n<ol>\n<li>First, find the radius of convergence, <em>R<\/em>.  Usually the root or ratio test works best for this part.<\/li>\n<li>Second, find out the behavior of the series at each of the two endpoints, <em>c<\/em> &#8211; <em>R<\/em> and <em>c<\/em> + <em>R<\/em>.  This time, <strong>do not<\/strong> use root or ratio tests, as those will almost certainly give you inconclusive results at the endpoints.  Instead, you may use comparison, limit comparison, alternating series test, integral test, <em>p<\/em>-series test, or other appropriate tests.<\/li>\n<\/ol>\n<h3>Example 1<\/h3>\n<p>Determine the interval of convergence for the series.<\/p>\n<p><img decoding=\"async\" src=\"https:\/\/magoosh.com\/hs\/files\/2017\/07\/series_example_1.gif\" alt=\"Example series\" width=\"139\" height=\"51\" class=\"aligncenter size-full wp-image-11011\" srcset=\"https:\/\/magoosh.com\/hs\/files\/2017\/07\/series_example_1.gif 139w, https:\/\/magoosh.com\/hs\/files\/2017\/07\/series_example_1-30x11.gif 30w\" sizes=\"(max-width: 139px) 100vw, 139px\" \/><\/p>\n<h4>Solution<\/h4>\n<p>First, observe that the center of this series is 5.  You find that by looking for the part of the general term that looks like (<em>x<\/em> &#8211; <em>c<\/em>), and looking to see what <em>c<\/em> must be there.<\/p>\n<p>Let&#8217;s use the <strong>root test<\/strong>.  When working with power series, we start by setting the limit less than 1 (for convergence).  Then, after taking limits (as <em>n<\/em> &rarr; &infin;), you&#8217;ll be able to solve for an expression of <em>x<\/em>, giving you the radius of convergence explicitly.<\/p>\n<p>The algebra can get a little hairy&#8230;<\/p>\n<p><img decoding=\"async\" src=\"https:\/\/magoosh.com\/hs\/files\/2017\/07\/radius_convergence_example1.gif\" alt=\"determining the radius of convergence\" width=\"215\" height=\"237\" class=\"aligncenter size-full wp-image-11012\" \/><\/p>\n<p>Thus, the radius of convergence is <em>R<\/em> = 1 (from the right side of the inequality).<\/p>\n<p>Now with center at 5, and radius 1, we can figure out the two endpoints.<\/p>\n<ul>\n<li><em>c<\/em> &#8211; <em>R<\/em> = 5 &#8211; 1 = 4.<\/li>\n<li><em>c<\/em> + <em>R<\/em> = 5 + 1 = 6.<\/li>\n<\/ul>\n<p>So already, we have four possibilities to work with.  We know that the interval of convergence must be from 4 to 6, but we just don&#8217;t know yet whether or not to include any of the endpoints.<\/p>\n<p>Now let&#8217;s plug in each endpoint for <em>x<\/em>, and work out the convergence properties.<\/p>\n<ul>\n<li>At <em>x<\/em> = 4:\n<p><img decoding=\"async\" src=\"https:\/\/magoosh.com\/hs\/files\/2017\/07\/series_example_1A.gif\" alt=\"Series example, continued\" width=\"144\" height=\"169\" class=\"aligncenter size-full wp-image-11013\" \/><\/p>\n<p>The resulting series is a <em>p<\/em>-series with <em>p<\/em> = 1\/2 (because of the square root of <em>n<\/em> in the denominator).<\/p>\n<p>By the <em>p<\/em>-series test, this series diverges.  Therefore, 4 is <strong>not<\/strong> in the interval of convergence.<\/p>\n<\/li>\n<li>At <em>x<\/em> = 6:\n<p><img decoding=\"async\" src=\"https:\/\/magoosh.com\/hs\/files\/2017\/07\/series_example_1B.gif\" alt=\"interval of convergence example, continued\" width=\"139\" height=\"169\" class=\"aligncenter size-full wp-image-11014\" \/><\/p>\n<p>This time, in addition to the square root in the denominator, there is also a factor of (-1)<em><sup>n<\/sup><\/em>.  This makes a huge difference.  Now we&#8217;re looking at an <em>alternating series<\/em>, which is much more likely to converge.<\/p>\n<p>In fact, the series does converge, since the absolute value of the general term limits onto zero.<\/p>\n<p><img decoding=\"async\" src=\"https:\/\/magoosh.com\/hs\/files\/2017\/07\/limit_of_term.gif\" alt=\"limit of general term\" width=\"187\" height=\"66\" class=\"aligncenter size-full wp-image-11015\" \/><\/p>\n<p>Thus by the Alternating Series Test, this series converges.  So <em>x<\/em> = 6 must be included in the interval of convergence.<\/p>\n<p>(For more details about alternating series and the Alternating Series Test, check out: <a href=\"https:\/\/magoosh.com\/hs\/ap\/ap-calculus-bc-review-alternating-series\/\">AP Calculus BC Review: Alternating Series<\/a>.)<\/p>\n<\/li>\n<\/ul>\n<p>Summarizing the above work, we know that 4 is not included, but 6 is.  Therefore, the interval of convergence is: (4, 6], or 4 &lt; <em>x<\/em> &le; 6.<\/p>\n<h3>Example 2<\/h3>\n<p>What is the largest interval on which the Maclaurin series for the exponential function converges?<\/p>\n<p><img decoding=\"async\" src=\"https:\/\/magoosh.com\/hs\/files\/2017\/07\/exp_power_series.gif\" alt=\"power series for e^x\" width=\"91\" height=\"50\" class=\"aligncenter size-full wp-image-11000\" srcset=\"https:\/\/magoosh.com\/hs\/files\/2017\/07\/exp_power_series.gif 91w, https:\/\/magoosh.com\/hs\/files\/2017\/07\/exp_power_series-30x16.gif 30w\" sizes=\"(max-width: 91px) 100vw, 91px\" \/><\/p>\n<h4>Solution<\/h4>\n<p>This time, the ratio test will be best.  Again, we will set the limit less than 1.  But first let&#8217;s figure out what the limit will be.<\/p>\n<p><img decoding=\"async\" src=\"https:\/\/magoosh.com\/hs\/files\/2017\/07\/series_example_2.gif\" alt=\"Series example 2, working out the limit\" width=\"314\" height=\"231\" class=\"aligncenter size-full wp-image-11016\" \/><\/p>\n<p>The reason that we get a limit of zero this time is because the numerator |<em>x<\/em>| must be considered fixed as <em>n<\/em> &rarr; &infin;.  Of course the denominator explicitly involves <em>n<\/em>.  So in the limit, we get <em>constant<\/em>\/&infin;, which is simply 0.<\/p>\n<p>Now, regardless of what value <em>x<\/em> has, the left side of the inequality will always be 0.  Setting 0 &lt; 1 produces a trivial true statement, so no matter what <em>x<\/em> is, the sequence converges!<\/p>\n<p>In this situation, we say that the radius of convergence is infinity (<em>R<\/em> = &infin;), and the interval of convergence is the entire real number line, (-&infin;, &infin;).<\/p>\n<h2>Summary<\/h2>\n<p>Given a power series centered at <em>c<\/em>, <\/p>\n<p><img decoding=\"async\" src=\"https:\/\/magoosh.com\/hs\/files\/2017\/07\/power_series.gif\" alt=\"Power series\" width=\"372\" height=\"77\" class=\"aligncenter size-full wp-image-11009\" \/><\/p>\n<ul>\n<li>The <strong>radius of convergence<\/strong>, <em>R<\/em>, is the largest number such that the series is guaranteed to converge within the interval between <em>c<\/em> &#8211; <em>R<\/em> and <em>c<\/em> + <em>R<\/em>.<\/li>\n<li>The <strong>interval of convergence<\/strong> is the largest interval on which the series converges.<\/li>\n<li>If <em>R<\/em> is finite and nonzero, then there are four combinations for interval of convergence, depending on whether each endpoint is convergent or not in the series.<\/li>\n<\/ul>\n","protected":false},"excerpt":{"rendered":"<p>This article reviews definitions and techniques for finding radius and interval of convergence of power series. <\/p>\n","protected":false},"author":223,"featured_media":0,"comment_status":"open","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[240],"tags":[241],"ppma_author":[24932],"class_list":["post-10718","post","type-post","status-publish","format-standard","hentry","category-ap","tag-ap-calculus"],"acf":[],"yoast_head":"<!-- This site is optimized with the Yoast SEO Premium plugin v21.7 (Yoast SEO v21.7) - https:\/\/yoast.com\/wordpress\/plugins\/seo\/ -->\n<title>AP Calculus BC Review: Radius and Interval of Convergence - Magoosh Blog | High School<\/title>\n<meta name=\"description\" content=\"This article reviews definitions and techniques for finding radius and interval of convergence of power series. 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