Difference between revisions of "009C Sample Final 3, Problem 5"

Revision as of 17:47, 2 December 2017

Consider the function

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f(x)=e^{-\frac{1}{3}x}}

(a) Find a formula for the Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle n} th derivative Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f^{(n)}(x)} of Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f} and then find Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f'(3).}

(b) Find the Taylor series for Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f(x)} at Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x_0=3,} i.e. write Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f(x)} in the form

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f(x)=\sum_{n=0}^\infty a_n(x-3)^n.}

Solution

Detailed Solution

Return to Sample Exam

@@ 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>''']]
-|-
-|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
-|-
-|
-&nbsp; &nbsp; &nbsp; &nbsp;<math>\sum_{n=0}^{\infty}c_n(x-a)^n</math> where <math style="vertical-align: -14px">c_n=\frac{f^{(n)}(a)}{n!}.</math>
-|}
-'''Solution:'''
+[[009C Sample Final 3, Problem 5 Detailed Solution|'''<u>Detailed Solution</u>''']]
-'''(a)'''
-{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
-!Step 1: &nbsp;
-|-
-|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)^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>
-|-
-|If we compare these three equations, we notice a pattern.
-|-
-|Thus,
-|-
-|&nbsp; &nbsp; &nbsp; &nbsp;<math>f^{(n)}(x)=\bigg(-\frac{1}{3}\bigg)^ne^{-\frac{1}{3}x}.</math>
-|}
-{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
-!Step 2: &nbsp;
-|-
-|Since
-|-
-|&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>
-|}
-'''(b)'''
-{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
-!Step 1: &nbsp;
-|-
-|Since
-|-
-|&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>
-|-
-|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>
-|}
-{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
-!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>
-|}
-{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
-!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; '''(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>''']]

Difference between revisions of "009C Sample Final 3, Problem 5"

Revision as of 17:47, 2 December 2017

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