Calculate the gradient of the function f(x1,x2,x3) = 2x51x32x43 +x1x2 +x2x3 at (1,1,1), homework help

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  1. Calculate the gradient of the function f(x1,x2,x3) = 2x51x32x43 +x1x2 +x2x3 at (1,1,1).

  2. Let f(x1,x2) = (x1 +x2)4 8(x1 +x2)2. Find all the local minimas and local maximas

    of this function. Guess what the graph of this function looks like.

  3. Let x R2. Consider the function f(x) = xTAx. If A is an invertible matrix, then prove that this function has only one stationary point at 0(stationary points are the points at which the gradient is zero). Give an A for which 0 is the minimizer of f(x). Give an A for which 0 is neither a maximizer nor a minimizer of f(x).

  4. Complete the proof of Lemma 7 in Lecture note 2 (posted on our course website).

  5. Let x Rp and find the gradient of the following functions:

    (a) f1(x) = (xT Ax)2, where A is an n × n matrix. (b) f2(x) = (xT Ax)n, where A is an n × n matrix.

  6. Let A Rn×p denote a fat matrix, i.e., n < p. Explain why we should expect the equation y = Ax to have infinitely many solutions. Among all those solutions we would like to find the one with minimum Euclidean norm, i.e., we want to find the solution with the smallest xT x. Find that solution and prove your answer.

  7. Plotthefunctionf(x1,x2)=(x11)2+(x21)2+x1x2 inR.Assumethatx1 [2,2] and x2 [2, 2]. Before doing this problem you may want to study the commands “outer” and “persp” in R. 

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Explanation & Answer

see attached

1.

The gradient of the function f would be calculated as:
∇𝑓 (𝑥1, 𝑥2, 𝑥3) =

𝛿𝑓(𝑥1, 𝑥2, 𝑥3)
𝛿𝑓(𝑥1, 𝑥2, 𝑥3)
𝛿𝑓(𝑥1, 𝑥2, 𝑥3)
𝑖+
𝑗+
𝑘
𝛿𝑥1
𝛿𝑥2
𝛿𝑥3

Taking this into account, if we calculate the partial derivatives of the function we find:
𝛿𝑓(𝑥1, 𝑥2, 𝑥3)
= 10𝑥14 𝑥23 𝑥34 + 𝑥2
𝛿𝑥1
𝛿𝑓(𝑥1, 𝑥2, 𝑥3)
= 6𝑥15 𝑥22 𝑥34 + 𝑥1 + 𝑥3
𝛿𝑥2
𝛿𝑓(𝑥1, 𝑥2, 𝑥3)
= 8𝑥15 𝑥23 𝑥33 + 𝑥2
𝛿𝑥3
Such that the gradient of f is:
∇𝑓(𝑥1, 𝑥2, 𝑥3) = (10𝑥14 𝑥23 𝑥34 + 𝑥2)𝑖 + (6𝑥15 𝑥22 𝑥34 + 𝑥1 + 𝑥3)𝑗 + (8𝑥15 𝑥23 𝑥33 + 𝑥2)𝑘
Finally, to calculate the gradient at the point (1,1,1) we simply need to substitute x1=1, x2=1 and x3=1 in
the above formula, such that:
∇𝑓(1,1,1) = (10 + 1)𝑖 + (6 + 1 + 1)𝑗 + (8 + 1)𝑘 = 11𝑖 + 8𝑗 + 9𝑘

2.

A local minimum point must satisfy f 1 = f 2 = 0, f 11 0
A local maximum point must satisfy f 1 = f 2 = 0, f 11 >0, f 11*f 22-f12 2 > 0
In case the test f 11*f22 -f122 = 0, it will be inconclu...


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