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so that <math>g(t_1,t_2)=\dfrac{1}{t_1^2+t_2^2+3}|t_1=0,t_2=0</math> would have some convex property<br>  
 
so that <math>g(t_1,t_2)=\dfrac{1}{t_1^2+t_2^2+3}|t_1=0,t_2=0</math> would have some convex property<br>  
 
with <math>f(x_1,x_2)=\dfrac{1}{(x_1-2)^2+(x_2+1)^2+3}|x_1=2,x_1=-1</math><br>
 
with <math>f(x_1,x_2)=\dfrac{1}{(x_1-2)^2+(x_2+1)^2+3}|x_1=2,x_1=-1</math><br>
 +
<math>D^2g(x)=\dfrac{1}{(t_1^2+t_2^2+3)^3}\begin{bmatrix} 6(t_1)^2-2(t_2)^3-6 & 8t_1t_2 \\ 8t_1t_2 & 6(t_2)^2-2(t_1)^3-6 \end{bmatrix}</math>
 
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[[QE2016_AC-3_ECE580|Back to QE AC question 3, August 2016]]
 
[[QE2016_AC-3_ECE580|Back to QE AC question 3, August 2016]]
  
 
[[ECE_PhD_Qualifying_Exams|Back to ECE Qualifying Exams (QE) page]]
 
[[ECE_PhD_Qualifying_Exams|Back to ECE Qualifying Exams (QE) page]]

Revision as of 22:08, 18 February 2019


ECE Ph.D. Qualifying Exam

Automatic Control (AC)

Question 3: Optimization

August 2016 Problem 3


Solution

Let $ t_1=x_1-2 $, $ t_2=x_2+1 $
so that $ g(t_1,t_2)=\dfrac{1}{t_1^2+t_2^2+3}|t_1=0,t_2=0 $ would have some convex property
with $ f(x_1,x_2)=\dfrac{1}{(x_1-2)^2+(x_2+1)^2+3}|x_1=2,x_1=-1 $
$ D^2g(x)=\dfrac{1}{(t_1^2+t_2^2+3)^3}\begin{bmatrix} 6(t_1)^2-2(t_2)^3-6 & 8t_1t_2 \\ 8t_1t_2 & 6(t_2)^2-2(t_1)^3-6 \end{bmatrix} $


Back to QE AC question 3, August 2016

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