F has an uniquely defined global minimum

  • #1
mathmari
Gold Member
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Hey! :eek:

Let and defined by with the euclidean norm .
Show that f has an uniquely defined global minimum and calculate it. I have done the following:
The partial derivative in respect to is

The gradient is

We get the critical point if we set the gradient equal to :

So, we have an extremum at .

The partial derivatives of second order are


So the Hessian-Matrix is:


Since we have a diagonal matrix, we get the eigenvalue from the diagonal: .

Since is positive, it follows that at the critical point we have a local minimum. Is everything correct? How can we show that this is a global minimum?
By uniquely defined is it mean that we have just one minimum?

(Wondering)
 
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  • #2
mathmari said:
Hey! :eek:

Let and defined by with the euclidean norm .
Show that f has an uniquely defined global minimum and calculate it. I have done the following:

Hey mathmari!

Shouldn't that be:

(Thinking)

mathmari said:
The partial derivative in respect to is

That also applies here, and wasn't a factor 2 dropped?
That is, shouldn't it be:

(Thinking)
 
  • #3
I like Serena said:
Shouldn't that be:

(Thinking)
That also applies here, and wasn't a factor 2 dropped?
That is, shouldn't it be:

(Thinking)

Oh yes, you're right! (Tmi) The gradient is then equal to

For the ccritical point we set the gradient equal to :

So, we have an extremum at .

The partial derivatives of second order are:


The Hessian matrix is equal to


The only eigenvalue is . So, it follows that at the critical point we have a local minimum.
Is now everything correct so far? (Wondering)
 
  • #4
mathmari said:
Is now everything correct so far? (Wondering)

Yep. All correct now. (Happy)
 
  • #5
I like Serena said:
Yep. All correct now. (Happy)

Great! (Happy)

By uniquely defined is it mean that we have just one minimum?

How can we show that the local minimum that we found is a global minimum?

(Wondering)
 
  • #6
mathmari said:
Great! (Happy)

By uniquely defined is it mean that we have just one minimum?

How can we show that the local minimum that we found is a global minimum?

(Wondering)

Yes. Since there is only 1 critical point, there is only 1 internal local extremum, which is a local minimum.
It is a global minimum if there is no lower minimum on the boundary, or, as in this case, the value of is greater than the minimum as .
What happens to if ? (Wondering)
 
  • #7
I like Serena said:
Yes. Since there is only 1 critical point, there is only 1 internal local extremum, which is a local minimum.
It is a global minimum if there is no lower minimum on the boundary, or, as in this case, the value of is greater than the minimum as .
What happens to if ? (Wondering)

Ah ok!

Does tend to infinity if goes to infinity? (Wondering)
 
  • #8
mathmari said:
Ah ok!

Does tend to infinity if goes to infinity?

Yes. Can you tell why? (Wondering)
 
  • #9
I like Serena said:
Yes. Can you tell why? (Wondering)

The function must tend to so that we can tell that the loca minimum that we found is also a global one, or not?

We have that If then and so the whole right side tends to because of the square, or not? (Wondering)
 
  • #10
mathmari said:
The function must tend to so that we can tell that the loca minimum that we found is also a global one, or not?

We have that If then and so the whole right side tends to because of the square, or not? (Wondering)

Yep. All correct. (Nod)
 
  • #11
I like Serena said:
Yep. All correct. (Nod)

Great! (Yes) The minimum of the function is therefore equal to Can we simplify this expression further? I am a bit confused with the two 's. Only the first one is in the inner sum, isn't it? But both are in the outer sum? (Wondering)
 
  • #12
mathmari said:
Great! (Yes) The minimum of the function is therefore equal to Can we simplify this expression further? I am a bit confused with the two 's. Only the first one is in the inner sum, isn't it? But both are in the outer sum? (Wondering)

Indeed. Let's use a different index variable before substituting, say .
Then we get:

where is the Kronecker delta, don't we? (Wondering)
 
  • #13
I like Serena said:


Indeed. Let's use a different index variable before substituting, say .
Then we get:

where is the Kronecker delta, don't we? (Wondering)

Ah ok! Thank you! (Mmm)
 
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