proof of calculus theorem used in the Lagrange method
Let f(𝐱) and gi(𝐱),i=0,…,m
be differentiable scalar functions; 𝐱∈Rn.
We will find local extremes of the function f(𝐱) where
∇f=0. This can be proved by contradiction:
∇f≠0 |
⇔∃ϵ0>0,∀ϵ;0<ϵ<ϵ0:f(𝐱-ϵ∇f)<f(𝐱)<f(𝐱+ϵ∇𝐟) |
but then f(𝐱) is not a local extreme.
Now we put up some conditions, such that we should find the 𝐱∈S⊂Rn that gives a local extreme of f. Let S=⋂mi=1Si, and let Si be defined so that gi(𝐱)=0∀𝐱∈Si.
Any vector 𝐱∈Rn can have one component perpendicular
to
the subset Si (for visualization, think n=3 and let
Si be a flat surface). ∇gi will be perpendicular to
Si, because:
∃ϵ0>0,∀ϵ;0<ϵ<ϵ0:gi(𝐱-ϵ∇gi)<gi(𝐱)<gi(𝐱+ϵ∇gi) |
But gi(𝐱)=0, so any vector 𝐱+ϵ∇gi must be outside Si, and also outside S. (todo: I have proved that there might exist a component perpendicular to each subset Si, but not that there exists only one; this should be done)
By the argument above, ∇f must be zero - but now we can ignore
all components of ∇f perpendicular to S. (todo: this should be expressed more formally and proved)
So we will have a local extreme within Si if there exists a λi such that
∇f=λi∇gi |
We will have local extreme(s) within S where there exists a set λi,i=1,…,m such that
∇f=∑λi∇gi |
Title | proof of calculus theorem used in the Lagrange method |
---|---|
Canonical name | ProofOfCalculusTheoremUsedInTheLagrangeMethod |
Date of creation | 2013-03-22 13:29:51 |
Last modified on | 2013-03-22 13:29:51 |
Owner | mathcam (2727) |
Last modified by | mathcam (2727) |
Numerical id | 6 |
Author | mathcam (2727) |
Entry type | Proof |
Classification | msc 15A18 |
Classification | msc 15A42 |