proof of calculus theorem used in the Lagrange method
Let and be differentiable scalar functions; .
We will find local extremes of the function where . This can be proved by contradiction:
but then is not a local extreme.
Now we put up some conditions, such that we should find the that gives a local extreme of . Let , and let be defined so that .
Any vector can have one component perpendicular to the subset (for visualization, think and let be a flat surface). will be perpendicular to , because:
But , so any vector must be outside , and also outside . (todo: I have proved that there might exist a component perpendicular to each subset , but not that there exists only one; this should be done)
By the argument above, must be zero - but now we can ignore all components of perpendicular to . (todo: this should be expressed more formally and proved)
So we will have a local extreme within if there exists a such that
We will have local extreme(s) within where there exists a set such that
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 |