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	<title>integration Archives - rweber.net</title>
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		<title>Power series special cases</title>
		<link>https://www.rweber.net/mathematics/calculus/power-series-special-cases/</link>
					<comments>https://www.rweber.net/mathematics/calculus/power-series-special-cases/#respond</comments>
		
		<dc:creator><![CDATA[Rebecca]]></dc:creator>
		<pubDate>Mon, 19 May 2014 12:00:42 +0000</pubDate>
				<category><![CDATA[calculus]]></category>
		<category><![CDATA[derivative]]></category>
		<category><![CDATA[integration]]></category>
		<category><![CDATA[series]]></category>
		<guid isPermaLink="false">http://www.rweber.net/?p=5525</guid>

					<description><![CDATA[<p>Using the ease of moving between geometric series and their sums, plus the fact that for power series the derivative of a series is the sum of the derivatives of each term (and likewise for integrals), we can find a wider range of power series without too much trouble. I like making diagrams so here [&#8230;]</p>
<p>The post <a href="https://www.rweber.net/mathematics/calculus/power-series-special-cases/">Power series special cases</a> appeared first on <a href="https://www.rweber.net">rweber.net</a>.</p>
]]></description>
										<content:encoded><![CDATA[<p>Using the ease of moving between geometric series and their sums, plus the fact that for power series the derivative of a series is the sum of the derivatives of each term (and likewise for integrals), we can find a wider range of power series without too much trouble. I like making diagrams so here we go.</p>
<p>1. Find a power series representation of <img decoding="async" src="https://s0.wp.com/latex.php?latex=%5Cln%281%2Bx%29.&#038;bg=ffffff&#038;fg=000&#038;s=0&#038;c=20201002" alt="&#92;ln(1+x)." class="latex" /></p>
<p>This function differentiates to a fraction we can interpret as the sum of a geometric series.</p>
<p><a href="https://www.rweber.net/wp-content/uploads/2014/03/powerseries1.jpg"><img fetchpriority="high" decoding="async" src="https://www.rweber.net/wp-content/uploads/2014/03/powerseries1.jpg" alt="power series 1" width="625" height="221" class="aligncenter size-full wp-image-5548" srcset="https://www.rweber.net/wp-content/uploads/2014/03/powerseries1.jpg 625w, https://www.rweber.net/wp-content/uploads/2014/03/powerseries1-300x106.jpg 300w" sizes="(max-width: 625px) 100vw, 625px" /></a></p>
<p>To complete Step 4, note that for x=0, <img decoding="async" src="https://s0.wp.com/latex.php?latex=%5Cln%281%2Bx%29&#038;bg=ffffff&#038;fg=000&#038;s=0&#038;c=20201002" alt="&#92;ln(1+x)" class="latex" /> = 0 and the series also equals 0, so C must be 0. The radius of convergence for the series obtained in Step 2 is 1, with an interval of (-1, 1). When you integrate the interval of convergence can gain or lose endpoints, but that&#8217;s a small thing to check after this fairly quick conversion.</p>
<p>2. Find a power series representation of <img decoding="async" src="https://s0.wp.com/latex.php?latex=%5Cdisplaystyle%7B+%5Cfrac%7B5%7D%7B%283-x%29%5E2%7D+%7D+.&#038;bg=ffffff&#038;fg=000&#038;s=0&#038;c=20201002" alt="&#92;displaystyle{ &#92;frac{5}{(3-x)^2} } ." class="latex" /></p>
<p>Instead of differentiating, this time we integrate to get the sum of a geometric series.</p>
<p><a href="https://www.rweber.net/wp-content/uploads/2014/03/powerseries2.jpg"><img decoding="async" src="https://www.rweber.net/wp-content/uploads/2014/03/powerseries2.jpg" alt="power series 2" width="625" height="237" class="aligncenter size-full wp-image-5551" srcset="https://www.rweber.net/wp-content/uploads/2014/03/powerseries2.jpg 625w, https://www.rweber.net/wp-content/uploads/2014/03/powerseries2-300x113.jpg 300w" sizes="(max-width: 625px) 100vw, 625px" /></a></p>
<p>In Step 3 note that from the perspective of the derivative operator <img decoding="async" src="https://s0.wp.com/latex.php?latex=3%5E%7Bn%2B1%7D&#038;bg=ffffff&#038;fg=000&#038;s=0&#038;c=20201002" alt="3^{n+1}" class="latex" /> is constant. There&#8217;s no constant of integration to find so once you&#8217;ve rounded the bases you&#8217;re done &#8211; well, except for thinking about the radius and interval of convergence.</p>
<p>One more after the jump. <span id="more-5525"></span></p>
<p>3. Find a power series representation for <img decoding="async" src="https://s0.wp.com/latex.php?latex=%5Cdisplaystyle%7B%5Cint+%5Ctan%5E%7B-1%7D%28x%29+%5C%3Adx%7D.&#038;bg=ffffff&#038;fg=000&#038;s=0&#038;c=20201002" alt="&#92;displaystyle{&#92;int &#92;tan^{-1}(x) &#92;:dx}." class="latex" /></p>
<p>This time we differentiate twice.</p>
<p><a href="https://www.rweber.net/wp-content/uploads/2014/03/powerseries3.jpg"><img decoding="async" src="https://www.rweber.net/wp-content/uploads/2014/03/powerseries3.jpg" alt="power series 3" width="625" height="302" class="aligncenter size-full wp-image-5552" srcset="https://www.rweber.net/wp-content/uploads/2014/03/powerseries3.jpg 625w, https://www.rweber.net/wp-content/uploads/2014/03/powerseries3-300x144.jpg 300w" sizes="(max-width: 625px) 100vw, 625px" /></a></p>
<p>Once again you do not have to find C at the end, but in this case it is because C stays undefined: the series is intended to equal an indefinite integral, not a specific function. C does not affect the radius or interval of convergence of the series.</p>
<p>We can use the series above to find a power series for the definite integral <img decoding="async" src="https://s0.wp.com/latex.php?latex=%5Cdisplaystyle%7B%5Cint_0%5E1+%5Ctan%5E%7B-1%7D%28x%29+%5C%3Adx%7D.&#038;bg=ffffff&#038;fg=000&#038;s=0&#038;c=20201002" alt="&#92;displaystyle{&#92;int_0^1 &#92;tan^{-1}(x) &#92;:dx}." class="latex" /> Since plugging 0 into the series obtained in Step 6 gives 0, the answer is the result of plugging in 1: <img decoding="async" src="https://s0.wp.com/latex.php?latex=%5Cdisplaystyle%7B+%5Csum_%7Bn%3D0%7D%5E%5Cinfty+%5Cfrac%7B%28-1%29%5En%7D%7B%282n%2B1%29%282n%2B2%29%7D+%7D.&#038;bg=ffffff&#038;fg=000&#038;s=0&#038;c=20201002" alt="&#92;displaystyle{ &#92;sum_{n=0}^&#92;infty &#92;frac{(-1)^n}{(2n+1)(2n+2)} }." class="latex" /></p>
<p>The post <a href="https://www.rweber.net/mathematics/calculus/power-series-special-cases/">Power series special cases</a> appeared first on <a href="https://www.rweber.net">rweber.net</a>.</p>
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		<post-id xmlns="com-wordpress:feed-additions:1">5525</post-id>	</item>
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		<title>Integration overview</title>
		<link>https://www.rweber.net/mathematics/calculus/integration-overview/</link>
					<comments>https://www.rweber.net/mathematics/calculus/integration-overview/#respond</comments>
		
		<dc:creator><![CDATA[Rebecca]]></dc:creator>
		<pubDate>Mon, 14 Apr 2014 12:00:04 +0000</pubDate>
				<category><![CDATA[calculus]]></category>
		<category><![CDATA[integration]]></category>
		<guid isPermaLink="false">http://www.rweber.net/?p=5562</guid>

					<description><![CDATA[<p>There are basically 4 techniques for solving an indefinite integral. Directly/by rules. After algebraic manipulation. Using substitution. By parts. Using limits for improper bounds or internal discontinuities should be mentioned here, but it&#8217;s not really in the same category since you don&#8217;t use the limit to find an antiderivative. is the method you learn first. [&#8230;]</p>
<p>The post <a href="https://www.rweber.net/mathematics/calculus/integration-overview/">Integration overview</a> appeared first on <a href="https://www.rweber.net">rweber.net</a>.</p>
]]></description>
										<content:encoded><![CDATA[<p>There are basically 4 techniques for solving an indefinite integral.</p>
<ol>
<li>Directly/by rules.</li>
<li>After algebraic manipulation.</li>
<li>Using substitution.</li>
<li>By parts.</li>
</ol>
<p>Using limits for improper bounds or internal discontinuities should be mentioned here, but it&#8217;s not really in the same category since you don&#8217;t use the limit to find an antiderivative.</p>
<ol>
<li>is the method you learn first.</li>
<li>includes completing the square, partial fractions, trigonometric integrals where you apply trig identities to whittle down powers or make substitution possible. In a (strong) sense it also includes integration by Taylor series.</li>
<li>comes from the chain rule for differentiation, and includes u-substitution and inverse substitution, such as trig substitution.</li>
<li>comes from the product rule for differentiation, and is one of the most powerful techniques you learn in calc 2.</li>
</ol>
<p>When should you apply each technique? There are rules of thumb, though it is also possible to solve many integrals in more than one way.</p>
<p>Completing the square comes in when you have a quadratic polynomial in an awkward spot such as the denominator of a fraction or under a square root. In the latter case it is often followed by trig substitution. In the former case it may lead to a u-substitution where the expression that becomes u has a constant derivative.</p>
<p>Partial fractions are for rational functions, to try to get down to simpler fractions which are amenable to u-substitution or trig substitution.</p>
<p>Trig integrals are easy to identify; they are powers and products of trig functions: sine with cosine, tangent with secant, cotangent with cosecant. They appear frequently after making a trig substitution.</p>
<p>Inverse substitution is generally used when there is an ugly function of x such as a square root, but the rest of the integrand is not simply a constant times the derivative of that function. You set u as a function of x, but solve for x in order to complete the substitution. Square roots of linear functions are a common place for this: if the integrand were <img decoding="async" src="https://s0.wp.com/latex.php?latex=x+%5Csqrt%7Bx%2B1%7D%5C%3B+dx&#038;bg=ffffff&#038;fg=000&#038;s=0&#038;c=20201002" alt="x &#92;sqrt{x+1}&#92;; dx" class="latex" />, you could let <img decoding="async" src="https://s0.wp.com/latex.php?latex=u+%3D+%5Csqrt%7Bx%2B1%7D&#038;bg=ffffff&#038;fg=000&#038;s=0&#038;c=20201002" alt="u = &#92;sqrt{x+1}" class="latex" />, and then let x = u<sup>2</sup> &#8211; 1 and dx = 2u du. The integrand becomes 2u<sup>4</sup> &#8211; 2u<sup>2</sup>. For trig substitution the traditional variable is theta, and here we exploit similarities to trig identities to collapse binomials into monomials and eliminate problematic square roots or fractions: 1 &#8211; sin<sup>2</sup> x collapses to cos<sup>2</sup> x, 1 + tan<sup>2</sup> x to sec<sup>2</sup> x, and sec<sup>2</sup> x &#8211; 1 to tan<sup>2</sup> x. [Why introduce sine into the binomial instead of cosine? Because the derivative of your chosen function comes into play in the substitution, and using sine means no negatives are introduced. That is, no reason except it makes for simpler bookkeeping.]</p>
<p>Integration by parts is essential when your integrand is the product of two functions of different &#8220;types&#8221;: polynomial, trigonometric, exponential, logarithmic.</p>
<p>These techniques can be chained together, of course. Completing the square leading to trig substitution leading to a trig integral is common. Making a substitution to make partial fractions possible is easy to come by (e.g., <img decoding="async" src="https://s0.wp.com/latex.php?latex=%5Csqrt%7Bx%2B1%7D%2Fx&#038;bg=ffffff&#038;fg=000&#038;s=0&#038;c=20201002" alt="&#92;sqrt{x+1}/x" class="latex" />).</p>
<p>The post <a href="https://www.rweber.net/mathematics/calculus/integration-overview/">Integration overview</a> appeared first on <a href="https://www.rweber.net">rweber.net</a>.</p>
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		<post-id xmlns="com-wordpress:feed-additions:1">5562</post-id>	</item>
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		<title>Calculus handouts</title>
		<link>https://www.rweber.net/mathematics/calculus/calculus-handouts/</link>
					<comments>https://www.rweber.net/mathematics/calculus/calculus-handouts/#respond</comments>
		
		<dc:creator><![CDATA[Rebecca]]></dc:creator>
		<pubDate>Mon, 26 Nov 2012 13:00:54 +0000</pubDate>
				<category><![CDATA[calculus]]></category>
		<category><![CDATA[graphing]]></category>
		<category><![CDATA[integration]]></category>
		<guid isPermaLink="false">http://www.rweber.net/?p=88</guid>

					<description><![CDATA[<p>Here are a few handouts I&#8217;ve used in the past to teach parts of calculus. Relating latitude and longitude to spherical coordinates. Bonus appendix about great circles. Trigonometry review for calculus students. A guide to curve sketching. Painfully detailed examples of surface integration.</p>
<p>The post <a href="https://www.rweber.net/mathematics/calculus/calculus-handouts/">Calculus handouts</a> appeared first on <a href="https://www.rweber.net">rweber.net</a>.</p>
]]></description>
										<content:encoded><![CDATA[<p>Here are a few handouts I&#8217;ve used in the past to teach parts of calculus.</p>
<ul>
<li>Relating <a href="https://www.rweber.net/calculus/lat_long.pdf">latitude and longitude</a> to spherical coordinates.  Bonus appendix about great circles.</li>
<li><a href="https://www.rweber.net/calculus/trigreview.pdf">Trigonometry review</a> for calculus students.</li>
<li>A guide to <a href="https://www.rweber.net/calculus/curvesketch.pdf">curve sketching</a>.</li>
<li>Painfully detailed examples of <a href="https://www.rweber.net/calculus/surfaces.pdf">surface integration</a>.</li>
</ul>
<p>The post <a href="https://www.rweber.net/mathematics/calculus/calculus-handouts/">Calculus handouts</a> appeared first on <a href="https://www.rweber.net">rweber.net</a>.</p>
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