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	<title>taylor polynomial &#8211; Mathemafrica</title>
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		<title>UCT MAM1000 lecture notes part 20 &#8211; part ii &#8211; Monday 17th August</title>
		<link>http://www.mathemafrica.org/?p=11189</link>
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		<pubDate>Sun, 16 Aug 2015 08:20:08 +0000</pubDate>
		<dc:creator><![CDATA[Jonathan Shock]]></dc:creator>
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		<category><![CDATA[Maclaurin polynomial]]></category>
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		<category><![CDATA[taylor polynomial]]></category>
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		<guid isPermaLink="false">http://www.mathemafrica.org/?p=11189</guid>
		<description><![CDATA[<p>So, we saw in the last section that we could work out the polynomial expression for <img src="//s0.wp.com/latex.php?latex=%281%2Bx%29%5E5&#038;bg=ffffff&#038;fg=000&#038;s=0" alt="(1+x)^5" title="(1+x)^5" class="latex" /> both using combinatorics as well as using calculus. We had also found previously that for small <img src="//s0.wp.com/latex.php?latex=x&#038;bg=ffffff&#038;fg=000&#038;s=0" alt="x" title="x" class="latex" /> we could just take the first couple of terms of the polynomial and it was a good approximation of the function itself, depending on how accurate we needed the answer. For instance, for small <img src="//s0.wp.com/latex.php?latex=x&#038;bg=ffffff&#038;fg=000&#038;s=0" alt="x" title="x" class="latex" />, <img src="//s0.wp.com/latex.php?latex=1%2B5x&#038;bg=ffffff&#038;fg=000&#038;s=0" alt="1+5x" title="1+5x" class="latex" /> is a reasonable approximation. One thing to note is that the value of these two functions is exactly the same at <img src="//s0.wp.com/latex.php?latex=x%3D0&#038;bg=ffffff&#038;fg=000&#038;s=0" alt="x=0" title="x=0" class="latex" /> and the derivative of both functions is the same at <img src="//s0.wp.com/latex.php?latex=x%3D0&#038;bg=ffffff&#038;fg=000&#038;s=0" alt="x=0" title="x=0" class="latex" />. The second derivatives are not the same, but had we taken the next term in the polynomial, the second derivatives would have matched as well.</p>
<p>OK, let&#8217;s take this one step further and repeat everything exactly as before, but now with the power, not as 5 but as 5.2 and see what happens.&#8230;</p>]]></description>
				<content:encoded><![CDATA[<p>So, we saw in the last section that we could work out the polynomial expression for <img src="//s0.wp.com/latex.php?latex=%281%2Bx%29%5E5&#038;bg=ffffff&#038;fg=000&#038;s=0" alt="(1+x)^5" title="(1+x)^5" class="latex" /> both using combinatorics as well as using calculus. We had also found previously that for small <img src="//s0.wp.com/latex.php?latex=x&#038;bg=ffffff&#038;fg=000&#038;s=0" alt="x" title="x" class="latex" /> we could just take the first couple of terms of the polynomial and it was a good approximation of the function itself, depending on how accurate we needed the answer. For instance, for small <img src="//s0.wp.com/latex.php?latex=x&#038;bg=ffffff&#038;fg=000&#038;s=0" alt="x" title="x" class="latex" />, <img src="//s0.wp.com/latex.php?latex=1%2B5x&#038;bg=ffffff&#038;fg=000&#038;s=0" alt="1+5x" title="1+5x" class="latex" /> is a reasonable approximation. One thing to note is that the value of these two functions is exactly the same at <img src="//s0.wp.com/latex.php?latex=x%3D0&#038;bg=ffffff&#038;fg=000&#038;s=0" alt="x=0" title="x=0" class="latex" /> and the derivative of both functions is the same at <img src="//s0.wp.com/latex.php?latex=x%3D0&#038;bg=ffffff&#038;fg=000&#038;s=0" alt="x=0" title="x=0" class="latex" />. The second derivatives are not the same, but had we taken the next term in the polynomial, the second derivatives would have matched as well.</p>
<p>OK, let&#8217;s take this one step further and repeat everything exactly as before, but now with the power, not as 5 but as 5.2 and see what happens.&hellip;</p>]]></content:encoded>
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