Question

Let $f(x) = \frac{1}{2}x^TQx - c^Tx$, where $Q = \begin{pmatrix} 2 & 0 \ 0 & 2 \end{pmatrix}$, $c = \begin{bmatrix} 1 \ 0 \end{bmatrix}$, and $x_0 = \begin{bmatrix} 1 \ 1 \end{bmatrix}$. Use Dogleg method to derive the value of Cauchy point $P_U$?

          Let $f(x) = \frac{1}{2}x^TQx - c^Tx$, where $Q = \begin{pmatrix} 2 & 0 \ 0 & 2 \end{pmatrix}$, $c = \begin{bmatrix} 1 \ 0 \end{bmatrix}$, and $x_0 = \begin{bmatrix} 1 \ 1 \end{bmatrix}$. Use Dogleg method to derive the value of Cauchy point $P_U$?
        
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Let f(x) = (1)/(2)x^TQx - c^Tx, where Q = 
    < p m a t r i x >, c = 
    < b m a t r i x >, and x0 = 
    < b m a t r i x >. Use Dogleg method to derive the value of Cauchy point PU?

Added by Carla T.

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Calculus: Early Transcendentals
Calculus: Early Transcendentals
James Stewart 8th Edition
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1 ʒ(2)==-z¹Qz-c¹z, where Q=(32) _._D] [ f(x) Let derive the value of Cauchy point PU? ON O [=0.5] O [05] O [9] O O [-0.5] O [05] ā—‹ 0.5 -0.5 сх and o TH . Use Dogleg method to Le--where([]and-[] Use Dogleg method to derive the value of Cauchy point Pu? o[H] O[-0:] 0[] O[] O[] O[] 0[] 0[]
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Transcript

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00:01 So, here it is given that the heat equation can be infinite to meant when u of t would be equals to the beta times of u of x x where x belongs to the real and t will be greater than to zero and also the u of x comma zero is equals to the x square where x belongs to the real.
00:25 So, here we are given del u by del x is equals to let's say the beta times of sorry this is not beta this is k here.
00:36 So, let's do del u by del x so that will comes out k times of del x square u by del x square let's suppose this as equation number a.
00:48 Similarly if we try to see the given that is u of x comma zero so that is given x square let's say this as the second function that is u of x and let's suppose this as equation number b.
01:02 Now, let the solution of a is in the form of u of x comma t which should be equals to the f of x dot t of t let's say this as equation number first.
01:16 So then by using the above relation u of x x comma t that will be equals to the f of x of x dot t of t where u x x is equals to the f x x dot t right and similarly the u of t will be equals to the f dot t of t.
01:42 Now, by using the derivative in equation a we get f dt by dt this should be equals to k d square f by the dx square.
01:57 So 1 over f into del square f by del x square this term is going to be equals to 1 over k times of t dt by dt.
02:12 Now if you remember so phi square or minus phi square separation constant is here.
02:17 So now we get the three sets of two ordinary differential equations.
02:23 The first set will be the del square f by del x square that should be equals to 0 and the same way del t by del t is equals to 0.
02:36 If lambda will be equals to 0 that is the separation constant you will say and the second we will get del square by del x square minus of p square f this will be equals to 0 and del t by del t is equals to p square c square t.
02:57 If the value of separation constant lambda will be greater than to 0 or lambda would be equals to p square.
03:06 And similarly the third case would be the del square f by the del x square or you will say the square that does not matter then plus p square f is equals to 0 for del t by del t is equals to minus of p square c square t and the separation coefficient sorry constant lambda less than to 0 or lambda is equals to minus of p square...
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