009C Sample Final 3, Problem 5 - Revision history

Kayla Murray at 23:20, 3 December 2017

2017-12-03T23:20:26Z

Kayla Murray at 01:47, 3 December 2017

2017-12-03T01:47:18Z

Kayla Murray at 21:56, 12 March 2017

2017-03-12T21:56:30Z

Kayla Murray at 21:30, 5 March 2017

2017-03-05T21:30:57Z

Kayla Murray at 21:30, 5 March 2017

2017-03-05T21:30:04Z

Kayla Murray at 21:20, 5 March 2017

2017-03-05T21:20:13Z

Kayla Murray at 21:12, 5 March 2017

2017-03-05T21:12:48Z

Kayla Murray at 01:19, 5 March 2017

2017-03-05T01:19:40Z

Kayla Murray at 00:56, 5 March 2017

2017-03-05T00:56:24Z

Kayla Murray at 18:52, 18 February 2017

2017-02-18T18:52:48Z

@@ Line 1: / Line 1: @@
 <span class="exam"> Consider the function
-::<math>f(x)=e^{-\frac{1}{3}x}</math>
+::<math>f(x)=e^{-\frac{1}{3}x}.</math>
 <span class="exam">(a) Find a formula for the &nbsp;<math>n</math>th derivative &nbsp;<math style="vertical-align: -5px">f^{(n)}(x)</math>&nbsp; of &nbsp;<math style="vertical-align: -5px">f</math>&nbsp; and then find &nbsp;<math style="vertical-align: -5px">f'(3).</math>

@@ Line 9: / Line 9: @@
 ::<math>f(x)=\sum_{n=0}^\infty a_n(x-3)^n.</math>
-{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
+<hr>
-!Foundations: &nbsp;
+[[009C Sample Final 3, Problem 5 Solution|'''<u>Solution</u>''']]
-'''Solution:'''
+[[009C Sample Final 3, Problem 5 Detailed Solution|'''<u>Detailed Solution</u>''']]
 [[009C_Sample_Final_3|'''<u>Return to Sample Exam</u>''']]

@@ Line 28: / Line 28: @@
 |We have
 |-
-|&nbsp; &nbsp; &nbsp; &nbsp;<math>f'(x)=\bigg(\frac{-1}{3}\bigg)e^{-\frac{1}{3}x},</math>
+|&nbsp; &nbsp; &nbsp; &nbsp;<math>f'(x)=\bigg(-\frac{1}{3}\bigg)e^{-\frac{1}{3}x},</math>
 |-
-|&nbsp; &nbsp; &nbsp; &nbsp;<math>f''(x)=\bigg(\frac{-1}{3}\bigg)^2 e^{-\frac{1}{3}x},</math>
+|&nbsp; &nbsp; &nbsp; &nbsp;<math>f''(x)=\bigg(-\frac{1}{3}\bigg)^2 e^{-\frac{1}{3}x},</math>
 |-
 |and
 |-
-|&nbsp; &nbsp; &nbsp; &nbsp;<math>f^{(3)}(x)=\bigg(\frac{-1}{3}\bigg)^3e^{-\frac{1}{3}x}.</math>
+|&nbsp; &nbsp; &nbsp; &nbsp;<math>f^{(3)}(x)=\bigg(-\frac{1}{3}\bigg)^3e^{-\frac{1}{3}x}.</math>
 |-
 |If we compare these three equations, we notice a pattern.
@@ Line 40: / Line 40: @@
 |Thus,
 |-
-|&nbsp; &nbsp; &nbsp; &nbsp;<math>f^{(n)}(x)=\bigg(\frac{-1}{3}\bigg)^ne^{-\frac{1}{3}x}.</math>
+|&nbsp; &nbsp; &nbsp; &nbsp;<math>f^{(n)}(x)=\bigg(-\frac{1}{3}\bigg)^ne^{-\frac{1}{3}x}.</math>
 |}
@@ Line 48: / Line 48: @@
 |Since
 |-
-|&nbsp; &nbsp; &nbsp; &nbsp;<math>f'(x)=\bigg(\frac{-1}{3}\bigg)e^{-\frac{1}{3}x},</math>
+|&nbsp; &nbsp; &nbsp; &nbsp;<math>f'(x)=\bigg(-\frac{1}{3}\bigg)e^{-\frac{1}{3}x},</math>
 |-
 |we have
 |-
-|&nbsp; &nbsp; &nbsp; &nbsp;<math>f'(3)=\bigg(\frac{-1}{3}\bigg)e^{-1}.</math>
+|&nbsp; &nbsp; &nbsp; &nbsp;<math>f'(3)=\bigg(-\frac{1}{3}\bigg)e^{-1}.</math>
 |}
@@ Line 62: / Line 62: @@
 |Since
 |-
-|&nbsp; &nbsp; &nbsp; &nbsp;<math>f^{(n)}(x)=\bigg(\frac{-1}{3}\bigg)^3e^{-\frac{1}{3}x},</math>
+|&nbsp; &nbsp; &nbsp; &nbsp;<math>f^{(n)}(x)=\bigg(-\frac{1}{3}\bigg)^3e^{-\frac{1}{3}x},</math>
 |-
 |we have
 |-
-|&nbsp; &nbsp; &nbsp; &nbsp;<math>f^{(n)}(3)=\bigg(\frac{-1}{3}\bigg)^ne^{-1}.</math>
+|&nbsp; &nbsp; &nbsp; &nbsp;<math>f^{(n)}(3)=\bigg(-\frac{1}{3}\bigg)^ne^{-1}.</math>
 |-
 |Therefore, the coefficients of the Taylor series are
 |-
-|&nbsp; &nbsp; &nbsp; &nbsp;<math>c_n=\frac{\bigg(\frac{-1}{3}\bigg)^ne^{-1}}{n!}.</math>
+|&nbsp; &nbsp; &nbsp; &nbsp;<math>c_n=\frac{\bigg(-\frac{1}{3}\bigg)^ne^{-1}}{n!}.</math>
 |}
@@ Line 78: / Line 78: @@
 |Therefore, the Taylor series for &nbsp;<math style="vertical-align: -5px">f(x)</math>&nbsp; at &nbsp;<math style="vertical-align: -3px">x_0=3</math>&nbsp; is
 |-
-|&nbsp; &nbsp; &nbsp; &nbsp;<math>\sum_{n=0}^\infty \bigg(\frac{-1}{3}\bigg)^n\frac{1}{e (n!)}(x-3)^n.</math>
+|&nbsp; &nbsp; &nbsp; &nbsp;<math>\sum_{n=0}^\infty \bigg(-\frac{1}{3}\bigg)^n\frac{1}{e (n!)}(x-3)^n.</math>
 |}
@@ Line 85: / Line 85: @@
 !Final Answer: &nbsp;
 |-
-|&nbsp; &nbsp; '''(a)'''&nbsp; &nbsp; <math>f^{(n)}(x)=\bigg(\frac{-1}{3}\bigg)^ne^{-\frac{1}{3}x},~f'(3)=\bigg(\frac{-1}{3}\bigg)e^{-1}</math>
+|&nbsp; &nbsp; '''(a)'''&nbsp; &nbsp; <math>f^{(n)}(x)=\bigg(-\frac{1}{3}\bigg)^ne^{-\frac{1}{3}x},~f'(3)=\bigg(-\frac{1}{3}\bigg)e^{-1}</math>
 |-
-|&nbsp; &nbsp; '''(b)'''&nbsp; &nbsp; <math>\sum_{n=0}^\infty \bigg(\frac{-1}{3}\bigg)^n\frac{1}{e (n!)}(x-3)^n</math>
+|&nbsp; &nbsp; '''(b)'''&nbsp; &nbsp; <math>\sum_{n=0}^\infty \bigg(-\frac{1}{3}\bigg)^n\frac{1}{e (n!)}(x-3)^n</math>
 |}
 [[009C_Sample_Final_3|'''<u>Return to Sample Exam</u>''']]

@@ Line 38: / Line 38: @@
 |If we compare these three equations, we notice a pattern.
 |-
-|We have
+|Thus,
 |-
 |&nbsp; &nbsp; &nbsp; &nbsp;<math>f^{(n)}(x)=\bigg(\frac{-1}{3}\bigg)^ne^{-\frac{1}{3}x}.</math>

@@ Line 40: / Line 40: @@
 |We have
 |-
-|&nbsp; &nbsp; &nbsp; &nbsp;<math>f^{(n)}(x)=\bigg(\frac{-1}{3}\bigg)^3e^{-\frac{1}{3}x}.</math>
+|&nbsp; &nbsp; &nbsp; &nbsp;<math>f^{(n)}(x)=\bigg(\frac{-1}{3}\bigg)^ne^{-\frac{1}{3}x}.</math>
 |}
@@ Line 60: / Line 60: @@
 !Step 1: &nbsp;
 |-
+|Since
 |-
+|Therefore, the coefficients of the Taylor series are
 |-
+|&nbsp; &nbsp; &nbsp; &nbsp;<math>c_n=\frac{\bigg(\frac{-1}{3}\bigg)^ne^{-1}}{n!}.</math>
 |}
@@ Line 70: / Line 76: @@
 !Step 2: &nbsp;
 |-
+|Therefore, the Taylor series for &nbsp;<math style="vertical-align: -5px">f(x)</math>&nbsp; at &nbsp;<math style="vertical-align: -3px">x_0=3</math>&nbsp; is
 |-
+|&nbsp; &nbsp; &nbsp; &nbsp;<math>\sum_{n=0}^\infty \bigg(\frac{-1}{3}\bigg)^n\frac{1}{e (n!)}(x-3)^n.</math>
 |}
@@ Line 79: / Line 85: @@
 !Final Answer: &nbsp;
 |-
-|&nbsp; &nbsp; '''(a)'''&nbsp; &nbsp; <math>f^{(n)}(x)=\bigg(\frac{-1}{3}\bigg)^3e^{-\frac{1}{3}x},~f'(3)=\bigg(\frac{-1}{3}\bigg)e^{-1}</math>
+|&nbsp; &nbsp; '''(a)'''&nbsp; &nbsp; <math>f^{(n)}(x)=\bigg(\frac{-1}{3}\bigg)^ne^{-\frac{1}{3}x},~f'(3)=\bigg(\frac{-1}{3}\bigg)e^{-1}</math>
 |-
-|&nbsp;&nbsp; '''(b)'''
+|&nbsp; &nbsp; '''(b)'''&nbsp; &nbsp; <math>\sum_{n=0}^\infty \bigg(\frac{-1}{3}\bigg)^n\frac{1}{e (n!)}(x-3)^n</math>
 |}
 [[009C_Sample_Final_3|'''<u>Return to Sample Exam</u>''']]

@@ Line 12: / Line 12: @@
 !Foundations: &nbsp;
 |-
+|The Taylor polynomial of  &nbsp; <math style="vertical-align: -5px">f(x)</math> &nbsp; at &nbsp; <math style="vertical-align: -1px">a</math> &nbsp; is
 |-
+|
 |}

@@ Line 3: / Line 3: @@
 ::<math>f(x)=e^{-\frac{1}{3}x}</math>
-<span class="exam">(a) Find a formula for the &nbsp;<math>n</math>th derivative &nbsp;<math>f^{(n)}(x)</math>&nbsp; of &nbsp;<math>f</math>&nbsp; and then find &nbsp;<math>f'(3).</math>
+<span class="exam">(a) Find a formula for the &nbsp;<math>n</math>th derivative &nbsp;<math style="vertical-align: -5px">f^{(n)}(x)</math>&nbsp; of &nbsp;<math style="vertical-align: -5px">f</math>&nbsp; and then find &nbsp;<math style="vertical-align: -5px">f'(3).</math>
-<span class="exam">(b) Find the Taylor series for &nbsp;<math>f(x)</math>&nbsp; at &nbsp;<math>x_0=3,</math>&nbsp; i.e. write &nbsp;<math>f(x)</math>&nbsp; in the form
+<span class="exam">(b) Find the Taylor series for &nbsp;<math style="vertical-align: -5px">f(x)</math>&nbsp; at &nbsp;<math style="vertical-align: -5px">x_0=3,</math>&nbsp; i.e. write &nbsp;<math style="vertical-align: -5px">f(x)</math>&nbsp; in the form
 ::<math>f(x)=\sum_{n=0}^\infty a_n(x-3)^n.</math>

@@ Line 1: / Line 1: @@
-<span class="exam">Compute
+<span class="exam"> Consider the function
-::<span class="exam">a) <math style="vertical-align: -12px">\lim_{n\rightarrow \infty} \frac{3-2n^2}{5n^2+n+1}</math>
+::::<math>f(x)=e^{-\frac{1}{3}x}</math>
-::<span class="exam">b) <math style="vertical-align: -12px">\lim_{n\rightarrow \infty} \frac{\ln n}{\ln 3n}</math>
+::<span class="exam">a) Find a formula for the <math>n</math>th derivative <math>f^{(n)}(x)</math> of <math>f</math> and then find <math>f'(3).</math>
 {| class="mw-collapsible mw-collapsed" style = "text-align:left;"