https://artofproblemsolving.com/wiki/index.php?title=Taylor_polynomial&feed=atom&action=historyTaylor polynomial - Revision history2024-03-28T21:20:18ZRevision history for this page on the wikiMediaWiki 1.31.1https://artofproblemsolving.com/wiki/index.php?title=Taylor_polynomial&diff=183791&oldid=prevOrange quail 9 at 16:39, 9 December 20222022-12-09T16:39:51Z<p></p>
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<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div><cmath>f(x) = \left\{</div></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div><cmath>f(x) = \left\{</div></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>\begin{array}{lr}  </div></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>\begin{array}{lr}  </div></td></tr>
<tr><td class='diff-marker'>−</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #ffe49c; vertical-align: top; white-space: pre-wrap;"><div>e^{-1/<del class="diffchange diffchange-inline">t</del>^2}  &  x \neq 0 \\</div></td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div>e^{-1/<ins class="diffchange diffchange-inline">x</ins>^2}  &  x \neq 0 \\</div></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>0 & x = 0  </div></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>0 & x = 0  </div></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>\end{array} \right\}</cmath></div></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>\end{array} \right\}</cmath></div></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>are zero at <math>x = 0</math>, so the Maclaurin series of <math>f(x)</math> converges to <math>0</math> for all <math>x</math>. However, <math>f(x) > 0</math> if <math>x \neq 0</math>, so the Maclaurin series converges to a different value than <math>f(x)</math> for all <math>x \neq 0</math>.</div></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>are zero at <math>x = 0</math>, so the Maclaurin series of <math>f(x)</math> converges to <math>0</math> for all <math>x</math>. However, <math>f(x) > 0</math> if <math>x \neq 0</math>, so the Maclaurin series converges to a different value than <math>f(x)</math> for all <math>x \neq 0</math>.</div></td></tr>
</table>Orange quail 9https://artofproblemsolving.com/wiki/index.php?title=Taylor_polynomial&diff=178371&oldid=prevOrange quail 9: Fixed an incorrectly named variable.2022-09-21T20:01:34Z<p>Fixed an incorrectly named variable.</p>
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<td colspan="2" style="background-color: #fff; color: #222; text-align: center;">Revision as of 20:01, 21 September 2022</td>
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<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>We want the Taylor polynomial to have <math>k</math>-th derivative <math>f^{(k)}(a)</math> at <math>x = a</math>. The [[Derivative/Formulas|Power Rule]] for derivatives gives that the derivative of <math>(x-a)^j</math> is <math>j(x-a)^{j-1}</math> for all positive integers <math>j</math>, and <math>0</math> for <math>j = 0</math> (because when <math>j = 0</math> the function is a constant <math>1</math>). Here the [[Chain Rule]] is used implicitly with the fact that <math>x - a</math> has derivative <math>1</math> for all <math>x</math>.</div></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>We want the Taylor polynomial to have <math>k</math>-th derivative <math>f^{(k)}(a)</math> at <math>x = a</math>. The [[Derivative/Formulas|Power Rule]] for derivatives gives that the derivative of <math>(x-a)^j</math> is <math>j(x-a)^{j-1}</math> for all positive integers <math>j</math>, and <math>0</math> for <math>j = 0</math> (because when <math>j = 0</math> the function is a constant <math>1</math>). Here the [[Chain Rule]] is used implicitly with the fact that <math>x - a</math> has derivative <math>1</math> for all <math>x</math>.</div></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td></tr>
<tr><td class='diff-marker'>−</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #ffe49c; vertical-align: top; white-space: pre-wrap;"><div>For <math>m < k</math>, the degree-<math>m</math> term in <math>x - a</math> has <math>k</math>th derivative <math>0</math>, because after <math><del class="diffchange diffchange-inline">k</del></math> differentiations the degree of the term will have reached <math>0</math> and then at least one more differentiation ensures that the term is eliminated.  </div></td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div>For <math>m < k</math>, the degree-<math>m</math> term in <math>x - a</math> has <math>k</math>th derivative <math>0</math>, because after <math><ins class="diffchange diffchange-inline">m</ins></math> differentiations the degree of the term will have reached <math>0</math> and then at least one more differentiation ensures that the term is eliminated.  </div></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>For <math>m > k</math>, the degree-<math>m</math> term in <math>x - a</math> has <math>k</math>th derivative <math>0</math> at <math>x = a</math>, because the <math>k</math> differentiations leave a term with a positive power of <math>(x - a)</math>, which is zero at <math>x = a</math>.</div></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>For <math>m > k</math>, the degree-<math>m</math> term in <math>x - a</math> has <math>k</math>th derivative <math>0</math> at <math>x = a</math>, because the <math>k</math> differentiations leave a term with a positive power of <math>(x - a)</math>, which is zero at <math>x = a</math>.</div></td></tr>
</table>Orange quail 9https://artofproblemsolving.com/wiki/index.php?title=Taylor_polynomial&diff=175472&oldid=prevOrange quail 9: /* Convergence */2022-06-29T21:21:47Z<p><span dir="auto"><span class="autocomment">Convergence</span></span></p>
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<td colspan="2" style="background-color: #fff; color: #222; text-align: center;">Revision as of 21:21, 29 June 2022</td>
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<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>If the limit of the Lagrange error bound, given above, is <math>0</math> as <math>n</math> goes to infinity, then the Taylor series must converge to the function. Because of the rapidly growing factorial term in the denominator of the error bound, Taylor series often converge for all <math>x</math>. For example, all Taylor series for <math>\sin(x)</math>, <math>\cos(x)</math>, and <math>e^x</math> converge to their respective functions for all <math>x</math>.</div></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>If the limit of the Lagrange error bound, given above, is <math>0</math> as <math>n</math> goes to infinity, then the Taylor series must converge to the function. Because of the rapidly growing factorial term in the denominator of the error bound, Taylor series often converge for all <math>x</math>. For example, all Taylor series for <math>\sin(x)</math>, <math>\cos(x)</math>, and <math>e^x</math> converge to their respective functions for all <math>x</math>.</div></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td></tr>
<tr><td class='diff-marker'>−</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #ffe49c; vertical-align: top; white-space: pre-wrap;"><div>However, if the derivatives of successive [[Order (derivative)|orders]] grow fast enough, then Taylor series may only converge for <math>a</math> sufficiently close to <math>x</math>.</div></td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div>However, if the derivatives of successive [[Order (derivative)|orders]] grow fast enough, then <ins class="diffchange diffchange-inline">the </ins>Taylor series may only converge for <math>a</math> sufficiently close to <math>x</math>.</div></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>Consider the Taylor series of <math>\ln(x)</math> about <math>x = 1</math>. By <math>\ln'(x) = \frac{1}{x}</math> and the [[Derivative/Formulas|power rule for derivatives]], <cmath>\ln^{(n)}(x) = (-1)^{n-1}(n-1)!x^{-n}</cmath> for <math>n > 0</math>. Since negative-power functions are strictly decreasing and positive for positive <math>x</math>, the maximum absolute value of <math>x^{-(n+1)}</math> on the interval <math>[1,x]</math> is <math>1</math>, so the maximum absolute value of <math>\ln^{(n+1)}(x)</math> on <math>[1,x]</math> is <math>n!</math>. Therefore, for <math>x \geq 1</math>, the Lagrange error bound evaluates to <cmath>\left| \frac{n!(x-1)^{n+1}}{(n+1)!}\right| = \frac{(x-1)^{n+1}}{n+1}.</cmath> If <math>x \leq 2</math>, then the limit of the above expression as <math>n</math> goes to infinity is <math>0</math>, so the Taylor series must in fact converge to <math>\ln(x)</math>; but for <math>x > 2</math>, the Taylor series is not guaranteed to converge to <math>\ln(x)</math>. In fact, the Taylor series for <math>\ln(x)</math> about <math>x = 1</math> is <cmath>\sum_{k=0}^{\infty} \frac{(-1)^{k-1}(x-1)^k}{k} </cmath> which, by the Ratio Test, does not converge at all for <math>x > 2</math>.</div></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>Consider the Taylor series of <math>\ln(x)</math> about <math>x = 1</math>. By <math>\ln'(x) = \frac{1}{x}</math> and the [[Derivative/Formulas|power rule for derivatives]], <cmath>\ln^{(n)}(x) = (-1)^{n-1}(n-1)!x^{-n}</cmath> for <math>n > 0</math>. Since negative-power functions are strictly decreasing and positive for positive <math>x</math>, the maximum absolute value of <math>x^{-(n+1)}</math> on the interval <math>[1,x]</math> is <math>1</math>, so the maximum absolute value of <math>\ln^{(n+1)}(x)</math> on <math>[1,x]</math> is <math>n!</math>. Therefore, for <math>x \geq 1</math>, the Lagrange error bound evaluates to <cmath>\left| \frac{n!(x-1)^{n+1}}{(n+1)!}\right| = \frac{(x-1)^{n+1}}{n+1}.</cmath> If <math>x \leq 2</math>, then the limit of the above expression as <math>n</math> goes to infinity is <math>0</math>, so the Taylor series must in fact converge to <math>\ln(x)</math>; but for <math>x > 2</math>, the Taylor series is not guaranteed to converge to <math>\ln(x)</math>. In fact, the Taylor series for <math>\ln(x)</math> about <math>x = 1</math> is <cmath>\sum_{k=0}^{\infty} \frac{(-1)^{k-1}(x-1)^k}{k} </cmath> which, by the Ratio Test, does not converge at all for <math>x > 2</math>.</div></td></tr>
</table>Orange quail 9https://artofproblemsolving.com/wiki/index.php?title=Taylor_polynomial&diff=175470&oldid=prevOrange quail 9: /* Convergence */ Edited for concision.2022-06-29T21:21:24Z<p><span dir="auto"><span class="autocomment">Convergence: </span> Edited for concision.</span></p>
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<td colspan="2" style="background-color: #fff; color: #222; text-align: center;">Revision as of 21:21, 29 June 2022</td>
</tr><tr><td colspan="2" class="diff-lineno" id="mw-diff-left-l33" >Line 33:</td>
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<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>If the limit of the Lagrange error bound, given above, is <math>0</math> as <math>n</math> goes to infinity, then the Taylor series must converge to the function. Because of the rapidly growing factorial term in the denominator of the error bound, Taylor series often converge for all <math>x</math>. For example, all Taylor series for <math>\sin(x)</math>, <math>\cos(x)</math>, and <math>e^x</math> converge to their respective functions for all <math>x</math>.</div></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>If the limit of the Lagrange error bound, given above, is <math>0</math> as <math>n</math> goes to infinity, then the Taylor series must converge to the function. Because of the rapidly growing factorial term in the denominator of the error bound, Taylor series often converge for all <math>x</math>. For example, all Taylor series for <math>\sin(x)</math>, <math>\cos(x)</math>, and <math>e^x</math> converge to their respective functions for all <math>x</math>.</div></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td></tr>
<tr><td class='diff-marker'>−</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #ffe49c; vertical-align: top; white-space: pre-wrap;"><div>However, if the derivatives of successive [[Order (derivative)|orders]] grow fast enough, then Taylor series may only converge for <math>a</math> sufficiently close to <math>x<del class="diffchange diffchange-inline"></math>, or may even diverge for all <math>x \neq a</del></math>.</div></td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div>However, if the derivatives of successive [[Order (derivative)|orders]] grow fast enough, then Taylor series may only converge for <math>a</math> sufficiently close to <math>x</math>.</div></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>Consider the Taylor series of <math>\ln(x)</math> about <math>x = 1</math>. By <math>\ln'(x) = \frac{1}{x}</math> and the [[Derivative/Formulas|power rule for derivatives]], <cmath>\ln^{(n)}(x) = (-1)^{n-1}(n-1)!x^{-n}</cmath> for <math>n > 0</math>. Since negative-power functions are strictly decreasing and positive for positive <math>x</math>, the maximum absolute value of <math>x^{-(n+1)}</math> on the interval <math>[1,x]</math> is <math>1</math>, so the maximum absolute value of <math>\ln^{(n+1)}(x)</math> on <math>[1,x]</math> is <math>n!</math>. Therefore, for <math>x \geq 1</math>, the Lagrange error bound evaluates to <cmath>\left| \frac{n!(x-1)^{n+1}}{(n+1)!}\right| = \frac{(x-1)^{n+1}}{n+1}.</cmath> If <math>x \leq 2</math>, then the limit of the above expression as <math>n</math> goes to infinity is <math>0</math>, so the Taylor series must in fact converge to <math>\ln(x)</math>; but for <math>x > 2</math>, the Taylor series is not guaranteed to converge to <math>\ln(x)</math>. In fact, the Taylor series for <math>\ln(x)</math> about <math>x = 1</math> is <cmath>\sum_{k=0}^{\infty} \frac{(-1)^{k-1}(x-1)^k}{k} </cmath> which, by the Ratio Test, does not converge at all for <math>x > 2</math>.</div></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>Consider the Taylor series of <math>\ln(x)</math> about <math>x = 1</math>. By <math>\ln'(x) = \frac{1}{x}</math> and the [[Derivative/Formulas|power rule for derivatives]], <cmath>\ln^{(n)}(x) = (-1)^{n-1}(n-1)!x^{-n}</cmath> for <math>n > 0</math>. Since negative-power functions are strictly decreasing and positive for positive <math>x</math>, the maximum absolute value of <math>x^{-(n+1)}</math> on the interval <math>[1,x]</math> is <math>1</math>, so the maximum absolute value of <math>\ln^{(n+1)}(x)</math> on <math>[1,x]</math> is <math>n!</math>. Therefore, for <math>x \geq 1</math>, the Lagrange error bound evaluates to <cmath>\left| \frac{n!(x-1)^{n+1}}{(n+1)!}\right| = \frac{(x-1)^{n+1}}{n+1}.</cmath> If <math>x \leq 2</math>, then the limit of the above expression as <math>n</math> goes to infinity is <math>0</math>, so the Taylor series must in fact converge to <math>\ln(x)</math>; but for <math>x > 2</math>, the Taylor series is not guaranteed to converge to <math>\ln(x)</math>. In fact, the Taylor series for <math>\ln(x)</math> about <math>x = 1</math> is <cmath>\sum_{k=0}^{\infty} \frac{(-1)^{k-1}(x-1)^k}{k} </cmath> which, by the Ratio Test, does not converge at all for <math>x > 2</math>.</div></td></tr>
</table>Orange quail 9https://artofproblemsolving.com/wiki/index.php?title=Taylor_polynomial&diff=175469&oldid=prevOrange quail 9 at 20:51, 29 June 20222022-06-29T20:51:10Z<p></p>
<table class="diff diff-contentalign-left" data-mw="interface">
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<col class="diff-content" />
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<td colspan="2" style="background-color: #fff; color: #222; text-align: center;">← Older revision</td>
<td colspan="2" style="background-color: #fff; color: #222; text-align: center;">Revision as of 20:51, 29 June 2022</td>
</tr><tr><td colspan="2" class="diff-lineno" id="mw-diff-left-l27" >Line 27:</td>
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<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>==Taylor series==</div></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>==Taylor series==</div></td></tr>
<tr><td class='diff-marker'>−</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #ffe49c; vertical-align: top; white-space: pre-wrap;"><div>The '''Taylor series''' of an infinitely differentiable function <math>f(x)</math> is the [[Sum#Infinite|infinite series]] <cmath>\sum_{k=0}^{\infty} \frac{f^{(k)}(x-a)}{k!} = f(a) + f'(a)(x - a) + \frac{f''(a)(x-a)^2}{2} + \dots.</cmath> The [[partial sums]] of the Taylor series are the Taylor polynomials of <math>f(x)</math> about <math>x = a</math> of each degree.</div></td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div>The '''Taylor series''' of an infinitely differentiable function <math>f(x)</math> is the [[Sum#Infinite|infinite series]] <cmath>\sum_{k=0}^{\infty} \frac{f^{(k)}(x-a)<ins class="diffchange diffchange-inline">^k</ins>}{k!} = f(a) + f'(a)(x - a) + \frac{f''(a)(x-a)^2}{2} + \dots.</cmath> The [[partial sums]] of the Taylor series are the Taylor polynomials of <math>f(x)</math> about <math>x = a</math> of each degree.</div></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>The '''Maclaurin series''' is the Taylor series chosen with <math>a = 0</math>. The partial sums of the Maclaurin series are the Maclaurin polynomials of <math>f(x)</math> of each degree.</div></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>The '''Maclaurin series''' is the Taylor series chosen with <math>a = 0</math>. The partial sums of the Maclaurin series are the Maclaurin polynomials of <math>f(x)</math> of each degree.</div></td></tr>
<tr><td colspan="2" class="diff-lineno" id="mw-diff-left-l35" >Line 35:</td>
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<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>However, if the derivatives of successive [[Order (derivative)|orders]] grow fast enough, then Taylor series may only converge for <math>a</math> sufficiently close to <math>x</math>, or may even diverge for all <math>x \neq a</math>.</div></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>However, if the derivatives of successive [[Order (derivative)|orders]] grow fast enough, then Taylor series may only converge for <math>a</math> sufficiently close to <math>x</math>, or may even diverge for all <math>x \neq a</math>.</div></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td></tr>
<tr><td class='diff-marker'>−</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #ffe49c; vertical-align: top; white-space: pre-wrap;"><div>Consider the Taylor series of <math>\ln(x)</math> about <math>x = 1</math>. By <math>\ln'(x) = \frac{1}{x}</math> and the [[Derivative/Formulas|power rule for derivatives]], <cmath>\ln^{(n)}(x) = (-1)^{n-1}(n-1)!x^{-n}</cmath> for <math>n > 0</math>. Since negative-power functions are strictly decreasing and positive for positive <math>x</math>, the maximum absolute value of <math>x^{-(n+1)}</math> on the interval <math>[1,x]</math> is <math>1</math>, so the maximum absolute value of <math>\ln^{(n+1)}(x)</math> on <math>[1,x]</math> is <math>n!</math>. Therefore, for <math>x \geq 1</math>, the Lagrange error bound evaluates to <cmath>\left| \frac{n!(x-1)^{n+1}}{(n+1)!}\right| = \frac{(x-1)^{n+1}}{n+1}.</cmath> If <math>x \leq 2</math>, then the limit of the above expression as <math>n</math> goes to infinity is <math>0</math>, so the Taylor series must in fact converge to <math>\ln(x)</math>; but for <math>x > 2</math>, the Taylor series is not guaranteed to converge to <math>\ln(x)</math>. In fact, the Taylor series for <math>\ln(x)</math> about <math>x = 1</math> is <cmath>\sum_{<del class="diffchange diffchange-inline">n</del>=0}^{\infty} \frac{(-1)^{<del class="diffchange diffchange-inline">n</del>-1}(x-1)^<del class="diffchange diffchange-inline">n</del>}{<del class="diffchange diffchange-inline">n</del>} </cmath> which, by the Ratio Test, does not converge at all for <math>x > 2</math>.</div></td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div>Consider the Taylor series of <math>\ln(x)</math> about <math>x = 1</math>. By <math>\ln'(x) = \frac{1}{x}</math> and the [[Derivative/Formulas|power rule for derivatives]], <cmath>\ln^{(n)}(x) = (-1)^{n-1}(n-1)!x^{-n}</cmath> for <math>n > 0</math>. Since negative-power functions are strictly decreasing and positive for positive <math>x</math>, the maximum absolute value of <math>x^{-(n+1)}</math> on the interval <math>[1,x]</math> is <math>1</math>, so the maximum absolute value of <math>\ln^{(n+1)}(x)</math> on <math>[1,x]</math> is <math>n!</math>. Therefore, for <math>x \geq 1</math>, the Lagrange error bound evaluates to <cmath>\left| \frac{n!(x-1)^{n+1}}{(n+1)!}\right| = \frac{(x-1)^{n+1}}{n+1}.</cmath> If <math>x \leq 2</math>, then the limit of the above expression as <math>n</math> goes to infinity is <math>0</math>, so the Taylor series must in fact converge to <math>\ln(x)</math>; but for <math>x > 2</math>, the Taylor series is not guaranteed to converge to <math>\ln(x)</math>. In fact, the Taylor series for <math>\ln(x)</math> about <math>x = 1</math> is <cmath>\sum_{<ins class="diffchange diffchange-inline">k</ins>=0}^{\infty} \frac{(-1)^{<ins class="diffchange diffchange-inline">k</ins>-1}(x-1)^<ins class="diffchange diffchange-inline">k</ins>}{<ins class="diffchange diffchange-inline">k</ins>} </cmath> which, by the Ratio Test, does not converge at all for <math>x > 2</math>.</div></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>It is also possible that a Taylor series converges but not to the value of the function it is meant to approximate. For a well-known example, all of the derivatives of the function  </div></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>It is also possible that a Taylor series converges but not to the value of the function it is meant to approximate. For a well-known example, all of the derivatives of the function  </div></td></tr>
</table>Orange quail 9https://artofproblemsolving.com/wiki/index.php?title=Taylor_polynomial&diff=175468&oldid=prevOrange quail 9: /* Convergence */2022-06-29T20:15:27Z<p><span dir="auto"><span class="autocomment">Convergence</span></span></p>
<table class="diff diff-contentalign-left" data-mw="interface">
<col class="diff-marker" />
<col class="diff-content" />
<col class="diff-marker" />
<col class="diff-content" />
<tr class="diff-title" lang="en">
<td colspan="2" style="background-color: #fff; color: #222; text-align: center;">← Older revision</td>
<td colspan="2" style="background-color: #fff; color: #222; text-align: center;">Revision as of 20:15, 29 June 2022</td>
</tr><tr><td colspan="2" class="diff-lineno" id="mw-diff-left-l31" >Line 31:</td>
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<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>The '''Maclaurin series''' is the Taylor series chosen with <math>a = 0</math>. The partial sums of the Maclaurin series are the Maclaurin polynomials of <math>f(x)</math> of each degree.</div></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>The '''Maclaurin series''' is the Taylor series chosen with <math>a = 0</math>. The partial sums of the Maclaurin series are the Maclaurin polynomials of <math>f(x)</math> of each degree.</div></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>===Convergence===</div></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>===Convergence===</div></td></tr>
<tr><td colspan="2"> </td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins style="font-weight: bold; text-decoration: none;">If the limit of the Lagrange error bound, given above, is <math>0</math> as <math>n</math> goes to infinity, then the Taylor series must converge to the function. Because of the rapidly growing factorial term in the denominator of the error bound, Taylor series often converge for all <math>x</math>. For example, all Taylor series for <math>\sin(x)</math>, <math>\cos(x)</math>, and <math>e^x</math> converge to their respective functions for all <math>x</math>.</ins></div></td></tr>
<tr><td colspan="2"> </td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins style="font-weight: bold; text-decoration: none;"></ins></div></td></tr>
<tr><td colspan="2"> </td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins style="font-weight: bold; text-decoration: none;">However, if the derivatives of successive [[Order (derivative)|orders]] grow fast enough, then Taylor series may only converge for <math>a</math> sufficiently close to <math>x</math>, or may even diverge for all <math>x \neq a</math>.</ins></div></td></tr>
<tr><td colspan="2"> </td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins style="font-weight: bold; text-decoration: none;"></ins></div></td></tr>
<tr><td colspan="2"> </td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins style="font-weight: bold; text-decoration: none;">Consider the Taylor series of <math>\ln(x)</math> about <math>x = 1</math>. By <math>\ln'(x) = \frac{1}{x}</math> and the [[Derivative/Formulas|power rule for derivatives]], <cmath>\ln^{(n)}(x) = (-1)^{n-1}(n-1)!x^{-n}</cmath> for <math>n > 0</math>. Since negative-power functions are strictly decreasing and positive for positive <math>x</math>, the maximum absolute value of <math>x^{-(n+1)}</math> on the interval <math>[1,x]</math> is <math>1</math>, so the maximum absolute value of <math>\ln^{(n+1)}(x)</math> on <math>[1,x]</math> is <math>n!</math>. Therefore, for <math>x \geq 1</math>, the Lagrange error bound evaluates to <cmath>\left| \frac{n!(x-1)^{n+1}}{(n+1)!}\right| = \frac{(x-1)^{n+1}}{n+1}.</cmath> If <math>x \leq 2</math>, then the limit of the above expression as <math>n</math> goes to infinity is <math>0</math>, so the Taylor series must in fact converge to <math>\ln(x)</math>; but for <math>x > 2</math>, the Taylor series is not guaranteed to converge to <math>\ln(x)</math>. In fact, the Taylor series for <math>\ln(x)</math> about <math>x = 1</math> is <cmath>\sum_{n=0}^{\infty} \frac{(-1)^{n-1}(x-1)^n}{n} </cmath> which, by the Ratio Test, does not converge at all for <math>x > 2</math>.</ins></div></td></tr>
<tr><td colspan="2"> </td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins style="font-weight: bold; text-decoration: none;"></ins></div></td></tr>
<tr><td colspan="2"> </td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins style="font-weight: bold; text-decoration: none;">It is also possible that a Taylor series converges but not to the value of the function it is meant to approximate. For a well-known example, all of the derivatives of the function </ins></div></td></tr>
<tr><td colspan="2"> </td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins style="font-weight: bold; text-decoration: none;"><cmath>f(x) = \left\{</ins></div></td></tr>
<tr><td colspan="2"> </td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins style="font-weight: bold; text-decoration: none;">\begin{array}{lr} </ins></div></td></tr>
<tr><td colspan="2"> </td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins style="font-weight: bold; text-decoration: none;">e^{-1/t^2}  &  x \neq 0 \\</ins></div></td></tr>
<tr><td colspan="2"> </td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins style="font-weight: bold; text-decoration: none;">0 & x = 0 </ins></div></td></tr>
<tr><td colspan="2"> </td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins style="font-weight: bold; text-decoration: none;">\end{array} \right\}</cmath></ins></div></td></tr>
<tr><td colspan="2"> </td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins style="font-weight: bold; text-decoration: none;">are zero at <math>x = 0</math>, so the Maclaurin series of <math>f(x)</math> converges to <math>0</math> for all <math>x</math>. However, <math>f(x) > 0</math> if <math>x \neq 0</math>, so the Maclaurin series converges to a different value than <math>f(x)</math> for all <math>x \neq 0</math>.</ins></div></td></tr>
</table>Orange quail 9https://artofproblemsolving.com/wiki/index.php?title=Taylor_polynomial&diff=172514&oldid=prevOrange quail 9 at 16:10, 10 March 20222022-03-10T16:10:17Z<p></p>
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<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>The '''Taylor series''' of an infinitely differentiable function <math>f(x)</math> is the [[Sum#Infinite|infinite series]] <cmath>\sum_{k=0}^{\infty} \frac{f^{(k)}(x-a)}{k!} = f(a) + f'(a)(x - a) + \frac{f''(a)(x-a)^2}{2} + \dots.</cmath> The [[partial sums]] of the Taylor series are the Taylor polynomials of <math>f(x)</math> about <math>x = a</math> of each degree.</div></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>The '''Taylor series''' of an infinitely differentiable function <math>f(x)</math> is the [[Sum#Infinite|infinite series]] <cmath>\sum_{k=0}^{\infty} \frac{f^{(k)}(x-a)}{k!} = f(a) + f'(a)(x - a) + \frac{f''(a)(x-a)^2}{2} + \dots.</cmath> The [[partial sums]] of the Taylor series are the Taylor polynomials of <math>f(x)</math> about <math>x = a</math> of each degree.</div></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td></tr>
<tr><td class='diff-marker'>−</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #ffe49c; vertical-align: top; white-space: pre-wrap;"><div>The '''<del class="diffchange diffchange-inline">Taylor </del>series''' <del class="diffchange diffchange-inline">is the Maclaurin series </del>is the Taylor series chosen with <math>a = 0</math>. The partial sums of the Maclaurin series are the Maclaurin polynomials of <math>f(x)</math> of each degree.</div></td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div>The '''<ins class="diffchange diffchange-inline">Maclaurin </ins>series''' is the Taylor series chosen with <math>a = 0</math>. The partial sums of the Maclaurin series are the Maclaurin polynomials of <math>f(x)</math> of each degree.</div></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>===Convergence===</div></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>===Convergence===</div></td></tr>
</table>Orange quail 9https://artofproblemsolving.com/wiki/index.php?title=Taylor_polynomial&diff=172503&oldid=prevOrange quail 9: Added Lagrange error bound2022-03-09T23:33:39Z<p>Added Lagrange error bound</p>
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<td colspan="2" style="background-color: #fff; color: #222; text-align: center;">Revision as of 23:33, 9 March 2022</td>
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<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>A tangent-line approximation is a first-degree Taylor polynomial, given by <math>f(a) + f'(a)(x - a)</math>. The name "tangent-line approximation" comes from the fact that the graph is a line [[tangent]] to the graph of <math>f(x)</math> at <math>x = a</math>. Tangent-line approximations are used in [[Differential equations#Approximations|Euler's method]] and [[Newton's method]].  </div></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>A tangent-line approximation is a first-degree Taylor polynomial, given by <math>f(a) + f'(a)(x - a)</math>. The name "tangent-line approximation" comes from the fact that the graph is a line [[tangent]] to the graph of <math>f(x)</math> at <math>x = a</math>. Tangent-line approximations are used in [[Differential equations#Approximations|Euler's method]] and [[Newton's method]].  </div></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>==Error bound==</div></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>==Error bound==</div></td></tr>
<tr><td colspan="2"> </td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins style="font-weight: bold; text-decoration: none;">Letting <math>P_n</math> be the degree-<math>n</math> Taylor polynomial of <math>f</math> about <math>a</math>, the '''Lagrange Error Bound''' states that <cmath>\lvert f(x) - P_n(x) \rvert \leq \left\lvert \frac{(x-a)^{n+1}M}{(n+1)!} \right\rvert</cmath> if <math>f^{(n+1)}(x)</math> is defined and has [[absolute value]] at most <math>M</math> on the entire interval <math>(a,x)</math> if <math>x > a</math> or <math>(x,a)</math> if <math>x < a</math>. </ins></div></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td></tr>
<tr><td class='diff-marker'>−</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #ffe49c; vertical-align: top; white-space: pre-wrap;"><div> </div></td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins class="diffchange diffchange-inline">The Lagrange Error Bound bounds the true value of <math>f(x)</math> both above and below.</ins></div></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>==Taylor series==</div></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>==Taylor series==</div></td></tr>
</table>Orange quail 9https://artofproblemsolving.com/wiki/index.php?title=Taylor_polynomial&diff=172502&oldid=prevOrange quail 9: Added derivation.2022-03-09T22:53:49Z<p>Added derivation.</p>
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<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>==Derivation of the formula==</div></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>==Derivation of the formula==</div></td></tr>
<tr><td colspan="2"> </td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins style="font-weight: bold; text-decoration: none;">We want the Taylor polynomial to have <math>k</math>-th derivative <math>f^{(k)}(a)</math> at <math>x = a</math>. The [[Derivative/Formulas|Power Rule]] for derivatives gives that the derivative of <math>(x-a)^j</math> is <math>j(x-a)^{j-1}</math> for all positive integers <math>j</math>, and <math>0</math> for <math>j = 0</math> (because when <math>j = 0</math> the function is a constant <math>1</math>). Here the [[Chain Rule]] is used implicitly with the fact that <math>x - a</math> has derivative <math>1</math> for all <math>x</math>.</ins></div></td></tr>
<tr><td colspan="2"> </td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins style="font-weight: bold; text-decoration: none;"></ins></div></td></tr>
<tr><td colspan="2"> </td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins style="font-weight: bold; text-decoration: none;">For <math>m < k</math>, the degree-<math>m</math> term in <math>x - a</math> has <math>k</math>th derivative <math>0</math>, because after <math>k</math> differentiations the degree of the term will have reached <math>0</math> and then at least one more differentiation ensures that the term is eliminated. </ins></div></td></tr>
<tr><td colspan="2"> </td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins style="font-weight: bold; text-decoration: none;"></ins></div></td></tr>
<tr><td colspan="2"> </td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins style="font-weight: bold; text-decoration: none;">For <math>m > k</math>, the degree-<math>m</math> term in <math>x - a</math> has <math>k</math>th derivative <math>0</math> at <math>x = a</math>, because the <math>k</math> differentiations leave a term with a positive power of <math>(x - a)</math>, which is zero at <math>x = a</math>.</ins></div></td></tr>
<tr><td colspan="2"> </td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins style="font-weight: bold; text-decoration: none;"></ins></div></td></tr>
<tr><td colspan="2"> </td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins style="font-weight: bold; text-decoration: none;">The degree-<math>k</math> term undergoes <math>k</math> differentiations, leaving a constant term and accumulating all of the factors <math>j</math> for <math>k \geq j \geq 1</math>. As such, its <math>k</math>th derivative is <math>k!</math> times its original coefficient for all <math>x</math>, so the coefficient of <math>(x-a)^k</math> should be defined as <math>\frac{f^{(k)}(a)}{k!}</math>.</ins></div></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>==Special cases==</div></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>==Special cases==</div></td></tr>
</table>Orange quail 9https://artofproblemsolving.com/wiki/index.php?title=Taylor_polynomial&diff=172500&oldid=prevOrange quail 9: Added details for "Special cases" and Maclaurin series information2022-03-09T22:21:59Z<p>Added details for "Special cases" and Maclaurin series information</p>
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<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>==Special cases==</div></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>==Special cases==</div></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>===Maclaurin polynomial===</div></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>===Maclaurin polynomial===</div></td></tr>
<tr><td colspan="2"> </td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins style="font-weight: bold; text-decoration: none;">A '''Maclaurin polynomial''' is a Taylor series with <math>a = 0</math>. Setting <math>a = 0</math> simplifies the appearance of the polynomial somewhat, since every instance of <math>(x-a)</math> in the formula is replaced with <math>x</math>.</ins></div></td></tr>
<tr><td colspan="2"> </td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins style="font-weight: bold; text-decoration: none;"></ins></div></td></tr>
<tr><td colspan="2"> </td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins style="font-weight: bold; text-decoration: none;">For some functions, like <math>e^x</math> and <math>\sin x</math>, Maclaurin polynomials are generally effective across the [[domain]] (although using a different <math>a</math>-value might allow greater accuracy for the same choice of degree). However, for functions like <math>\ln x</math>, Maclaurin polynomials cannot be defined because the function and its derivatives are undefined at <math>x = 0</math>. For other functions, Maclaurin polynomials can be defined, but do not in general approximate the function well (see [[Taylor polynomial#Taylor series|Taylor series]]), so a value of <math>a</math> closer to the <math>x</math>-value of the desired approximation must be chosen.</ins></div></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>===Tangent-line approximation===</div></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>===Tangent-line approximation===</div></td></tr>
<tr><td colspan="2"> </td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins style="font-weight: bold; text-decoration: none;">A tangent-line approximation is a first-degree Taylor polynomial, given by <math>f(a) + f'(a)(x - a)</math>. The name "tangent-line approximation" comes from the fact that the graph is a line [[tangent]] to the graph of <math>f(x)</math> at <math>x = a</math>. Tangent-line approximations are used in [[Differential equations#Approximations|Euler's method]] and [[Newton's method]]. </ins></div></td></tr>
<tr><td colspan="2"> </td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins style="font-weight: bold; text-decoration: none;">==Error bound==</ins></div></td></tr>
<tr><td colspan="2"> </td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins style="font-weight: bold; text-decoration: none;"></ins></div></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td></tr>
<tr><td class='diff-marker'>−</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #ffe49c; vertical-align: top; white-space: pre-wrap;"><div><del style="font-weight: bold; text-decoration: none;">==Error bound==</del></div></td><td colspan="2"> </td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>==Taylor series==</div></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>==Taylor series==</div></td></tr>
<tr><td class='diff-marker'>−</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #ffe49c; vertical-align: top; white-space: pre-wrap;"><div>The '''Taylor series''' of an infinitely differentiable function <math>f(x)</math> is the [[Sum#Infinite|infinite series]] <cmath>\sum_{k=0}^{\infty} \frac{f^{(k)}(x-a)}{k!} = f(a) + f'(a)(x - a) + \frac{f''(a)(x-a)^2}{2} + \dots.</cmath> The [[partial sums]] of <del class="diffchange diffchange-inline">this </del>series are the Taylor polynomials of <math>f(x)</math> of each degree.</div></td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div>The '''Taylor series''' of an infinitely differentiable function <math>f(x)</math> is the [[Sum#Infinite|infinite series]] <cmath>\sum_{k=0}^{\infty} \frac{f^{(k)}(x-a)}{k!} = f(a) + f'(a)(x - a) + \frac{f''(a)(x-a)^2}{2} + \dots.</cmath> The [[partial sums]] of <ins class="diffchange diffchange-inline">the Taylor </ins>series are the Taylor <ins class="diffchange diffchange-inline">polynomials of <math>f(x)</math> about <math>x = a</math> of each degree.</ins></div></td></tr>
<tr><td colspan="2"> </td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div> </div></td></tr>
<tr><td colspan="2"> </td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins class="diffchange diffchange-inline">The '''Taylor series''' is the Maclaurin series is the Taylor series chosen with <math>a = 0</math>. The partial sums of the Maclaurin series are the Maclaurin </ins>polynomials of <math>f(x)</math> of each degree.</div></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>===Convergence===</div></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>===Convergence===</div></td></tr>
</table>Orange quail 9