009C Sample Final 2, Problem 8 - Revision history

Kayla Murray at 22:57, 3 December 2017

2017-12-03T22:57:19Z

Kayla Murray at 01:34, 3 December 2017

2017-12-03T01:34:42Z

Grad at 23:56, 10 March 2017

2017-03-10T23:56:48Z

Kayla Murray at 19:49, 5 March 2017

2017-03-05T19:49:01Z

Kayla Murray at 19:48, 5 March 2017

2017-03-05T19:48:32Z

Kayla Murray at 19:37, 5 March 2017

2017-03-05T19:37:45Z

Kayla Murray at 19:29, 5 March 2017

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Kayla Murray at 01:15, 5 March 2017

2017-03-05T01:15:43Z

Kayla Murray at 00:49, 5 March 2017

2017-03-05T00:49:43Z

Kayla Murray at 00:44, 5 March 2017

2017-03-05T00:44:38Z

← Older revision		Revision as of 22:57, 3 December 2017
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−	<span class="exam">Find  <math>n</math>  such that the Maclaurin polynomial of degree  <math>n</math>  of  <math style="vertical-align: -5px">f(x)=\cos(x)</math>  approximates  <math style="vertical-align: -13px">\cos \frac{\pi}{3}</math>  within 0.0001 of the actual value.	+	<span class="exam">Find  <math>n</math>  such that the Maclaurin polynomial of degree  <math>n</math>  of  <math style="vertical-align: -5px">f(x)=\cos(x)</math>  approximates  <math style="vertical-align: -13px">\cos \bigg(\frac{\pi}{3}\bigg)</math>  within 0.0001 of the actual value.

	<hr>		<hr>

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 <span class="exam">Find &nbsp;<math>n</math>&nbsp; such that the Maclaurin polynomial of degree &nbsp;<math>n</math>&nbsp; of &nbsp;<math style="vertical-align: -5px">f(x)=\cos(x)</math>&nbsp; approximates &nbsp;<math style="vertical-align: -13px">\cos \frac{\pi}{3}</math>&nbsp; within 0.0001 of the actual value.
-{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
+<hr>
-!Foundations: &nbsp;
+[[009C Sample Final 2, Problem 8 Solution|'''<u>Solution</u>''']]
-'''Solution:'''
+[[009C Sample Final 2, Problem 8 Detailed Solution|'''<u>Detailed Solution</u>''']]
 [[009C_Sample_Final_2|'''<u>Return to Sample Exam</u>''']]

@@ Line 6: / Line 6: @@
 |'''Taylor's Theorem'''
 |-
-|&nbsp; &nbsp; &nbsp; &nbsp; Let &nbsp;<math style="vertical-align: -5px">f</math>&nbsp; be a function whose &nbsp;<math style="vertical-align: -2px">n+1</math>th derivative exists on an interval &nbsp;<math style="vertical-align: 0px">I</math>&nbsp; and let &nbsp;<math style="vertical-align: 0px">c</math>&nbsp; be in &nbsp;<math style="vertical-align: 0px">I.</math>
+|&nbsp; &nbsp; &nbsp; &nbsp; Let &nbsp;<math style="vertical-align: -5px">f</math>&nbsp; be a function whose &nbsp;<math style="vertical-align: -4px">(n+1)^{\mathrm{th}}</math> derivative exists on an interval &nbsp;<math style="vertical-align: 0px">I</math>,&nbsp; and let &nbsp;<math style="vertical-align: 0px">c</math>&nbsp; be in &nbsp;<math style="vertical-align: 0px">I.</math>
 |-
 |&nbsp; &nbsp; &nbsp; &nbsp; Then, for each &nbsp;<math style="vertical-align: 0px">x</math>&nbsp; in &nbsp;<math style="vertical-align: -4px">I,</math>&nbsp; there exists &nbsp;<math style="vertical-align: -3px">z_x</math>&nbsp; between &nbsp;<math style="vertical-align: 0px">x</math>&nbsp; and &nbsp;<math style="vertical-align: 0px">c</math>&nbsp; such that

@@ Line 37: / Line 37: @@
 |We note that
 |-
-|&nbsp; &nbsp; &nbsp; &nbsp;<math style="vertical-align: -5px">|f^{n+1}(z)|=|\cos(z)|\le 1</math>&nbsp; or &nbsp;<math style="vertical-align: -5px">|f^{n+1}(z)|=|\cos(z)|\le 1.</math>
+|&nbsp; &nbsp; &nbsp; &nbsp;<math style="vertical-align: -5px">|f^{n+1}(z)|=|\cos(z)|\le 1</math>&nbsp; or &nbsp;<math style="vertical-align: -5px">|f^{n+1}(z)|=|\sin(z)|\le 1.</math>
 |-
 |Therefore, we have

@@ Line 14: / Line 14: @@
 |&nbsp; &nbsp; &nbsp; &nbsp; where &nbsp;<math style="vertical-align: -18px">R_n(x)=\frac{f^{n+1}(z_x)}{(n+1)!}(x-c)^{n+1}.</math>
 |-
-|&nbsp; &nbsp; &nbsp; &nbsp; Also, <math style="vertical-align: -17px">|R_n(x)|\le \frac{\max |f^{n+1}(z)|}{(n+1)!}|(x-c)^{n+1}|.</math>
+|&nbsp; &nbsp; &nbsp; &nbsp; Also, &nbsp;<math style="vertical-align: -17px">|R_n(x)|\le \frac{\max |f^{n+1}(z)|}{(n+1)!}|(x-c)^{n+1}|.</math>
 |}
@@ Line 23: / Line 23: @@
 !Step 1: &nbsp;
 |-
-|Using Taylor's Theorem, we have that the error in approximating &nbsp;<math>\cos \frac{\pi}{3}</math>&nbsp; with
+|Using Taylor's Theorem, we have that the error in approximating &nbsp;<math style="vertical-align: -13px">\cos \frac{\pi}{3}</math>&nbsp; with
 |-
-|the Maclaurin polynomial of degree &nbsp;<math>n</math>&nbsp; is &nbsp;<math>R_n\bigg(\frac{\pi}{3}\bigg)</math>&nbsp; where
+|the Maclaurin polynomial of degree &nbsp;<math style="vertical-align: 0px">n</math>&nbsp; is &nbsp;<math style="vertical-align: -16px">R_n\bigg(\frac{\pi}{3}\bigg)</math>&nbsp; where
 |-
 |&nbsp; &nbsp; &nbsp; &nbsp; <math>\bigg|R_n\bigg(\frac{\pi}{3}\bigg)\bigg|\le \frac{\max |f^{n+1}(z)|}{(n+1)!}\bigg|\bigg(\frac{\pi}{3}-0\bigg)^{n+1}\bigg|.</math>
@@ Line 35: / Line 35: @@
 !Step 2: &nbsp;
 |-
-|We note that &nbsp;<math>|f^{n+1}(z)|=|\cos(z)|\le 1</math>&nbsp; or &nbsp;<math>|f^{n+1}(z)|=|\cos(z)|\le 1.</math>
+|We note that
 |-
 |Therefore, we have
@@ Line 78: / Line 80: @@
 </table>
 |-
-|So, &nbsp;<math>n=7</math>&nbsp; is the smallest value of &nbsp;<math>n</math>&nbsp; where the error is less than or equal to 0.0001.
+|So, &nbsp;<math style="vertical-align: 0px">n=7</math>&nbsp; is the smallest value of &nbsp;<math style="vertical-align: 0px">n</math>&nbsp; where the error is less than or equal to 0.0001.
 |-
-|Therefore, for &nbsp;<math>n=7</math>&nbsp; the Maclaurin polynomial approximates &nbsp;<math>\cos \frac{\pi}{3}</math>&nbsp; within 0.0001 of the actual value.
+|Therefore, for &nbsp;<math style="vertical-align: 0px">n=7</math>&nbsp; the Maclaurin polynomial approximates &nbsp;<math style="vertical-align: -13px">\cos \frac{\pi}{3}</math>&nbsp; within 0.0001 of the actual value.
 |}
@@ Line 87: / Line 89: @@
 !Final Answer: &nbsp;
 |-
-|&nbsp; &nbsp; &nbsp; &nbsp;<math>n=7.</math>
+|&nbsp; &nbsp; &nbsp; &nbsp;<math>n=7</math>
 |}
 [[009C_Sample_Final_2|'''<u>Return to Sample Exam</u>''']]

← Older revision		Revision as of 01:15, 5 March 2017
Line 1:		Line 1:
−	<span class="exam">Find  <math>n</math>  such that the Maclaurin polynomial of degree  <math>n</math>  of  <math>f(x)=\cos(x)</math>  approximates  <math>\cos \frac{\pi}{3}</math>  within 0.0001 of the actual value.	+	<span class="exam">Find  <math>n</math>  such that the Maclaurin polynomial of degree  <math>n</math>  of  <math style="vertical-align: -5px">f(x)=\cos(x)</math>  approximates  <math style="vertical-align: -13px">\cos \frac{\pi}{3}</math>  within 0.0001 of the actual value.

	{\| class="mw-collapsible mw-collapsed" style = "text-align:left;"		{\| class="mw-collapsible mw-collapsed" style = "text-align:left;"

← Older revision		Revision as of 00:44, 5 March 2017
Line 1:		Line 1:
−	<span class="exam">Find <math>n</math> such that the Maclaurin polynomial of degree <math>n</math> of <math>f(x)=\cos(x)</math> approximates <math>\cos \frac{\pi}{3}</math> within 0.0001 of the actual value.	+	<span class="exam">Find  <math>n</math>  such that the Maclaurin polynomial of degree  <math>n</math>  of  <math>f(x)=\cos(x)</math>  approximates  <math>\cos \frac{\pi}{3}</math>  within 0.0001 of the actual value.

	{\| class="mw-collapsible mw-collapsed" style = "text-align:left;"		{\| class="mw-collapsible mw-collapsed" style = "text-align:left;"