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cateater2000
Oct19-04, 08:15 PM
I'm stuck on this question

Show that f(x1,x2) has a strict local minimizer at t=0 along every line

{ x1=at
{ x2=bt


through (0,0).


Any hints or tips would be great thanks

HallsofIvy
Oct20-04, 08:06 AM
What exactly do you mean by "minimizer"? I'm not familiar with that term.

At first I thought you meant "minimum" but in that case the theorem is not true.

Suppose f(x,y)= (x-1)2+ (y-1)2.

Saying that f has a strict local minimum at t= 0 on x= at, y= bt would simply mean that f has a strict local minimum at (0,0) but that is not true- the only minimum of f is at (1,1).

cateater2000
Oct20-04, 03:43 PM
The minimizer is the point t where the minimum is. THat's why I'm a bit confused with the question. The wording I was given in my book is a bit awkward.

I think what it means is. For every f(x1,x2) given that x1 and x2 are lines.

There is a minimum at t=0

Those that make sense?

HallsofIvy
Oct21-04, 08:08 AM
The minimizer is the point t where the minimum is. THat's why I'm a bit confused with the question. The wording I was given in my book is a bit awkward.

I think what it means is. For every f(x1,x2) given that x1 and x2 are lines.

There is a minimum at t=0

Those that make sense?

No! In particular, the sentence "For every f(x1,x2) given that x1 and x2 are lines." makes no sense at all. x1, x2 are variables that I presume are numbers, not lines.

Aren't there some conditions on the function f(x1, x2)? The statement is certainly NOT true for general f.

cateater2000
Oct21-04, 01:22 PM
I think you're right. The wording of the question is not very good. A strict local minimizer is in fact the same as a strict local minimum.

I geuss I'll have to ask the prof