- #1

Zurtex

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[tex]|| \text{v} ||_1 \quad{and} \quad || \text{v} ||_2 \quad{and} \quad || \text{v} ||_\infty[/tex]

Any help would be greatly appreciated.

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- #1

Zurtex

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[tex]|| \text{v} ||_1 \quad{and} \quad || \text{v} ||_2 \quad{and} \quad || \text{v} ||_\infty[/tex]

Any help would be greatly appreciated.

- #2

matt grime

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A gues would be,

|v|_n is the n'th roots of the sum nth powers of the absolute values of the components.

If they were function spaces then that would make sense, in integrals. Infinity ought to mean the abs value of the largest component, or the sup norm.

- #3

HallsofIvy

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Matt! Of course they are standard.

In a finite dimensional space, [itex]||v||\infty= max |v_i|[/itex], the largest component, in absolute value, of v.

In an infinite dimensional, function space, [itex]||v||_\infnty= max(|v(x)|)[/itex] where the max is over whatever compact set v(x) is defined on.

In a finite dimensional space [itex]||v||_1= \Sigma |v_i|[/itex].

In an infinite dimensional, function space, [itex]||v||_1= \int |v(x)|dx[/itex] where the integral is over the set v(x) is defined on.

In a finite dimensioal space [itex]||v||_2= \sqrt{\Sigma (v_i)^2}[/itex].

in an infinite dimensional, function space, [itex]||v||_2= \sqrt{\int (v(x))^2 dx}[/itex]

In a finite dimensional space, [itex]||v||\infty= max |v_i|[/itex], the largest component, in absolute value, of v.

In an infinite dimensional, function space, [itex]||v||_\infnty= max(|v(x)|)[/itex] where the max is over whatever compact set v(x) is defined on.

In a finite dimensional space [itex]||v||_1= \Sigma |v_i|[/itex].

In an infinite dimensional, function space, [itex]||v||_1= \int |v(x)|dx[/itex] where the integral is over the set v(x) is defined on.

In a finite dimensioal space [itex]||v||_2= \sqrt{\Sigma (v_i)^2}[/itex].

in an infinite dimensional, function space, [itex]||v||_2= \sqrt{\int (v(x))^2 dx}[/itex]

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- #4

Zurtex

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Always seem to forget them as most the manipulation needed doesn't require their actual definitions but rather the properties they have. Should help for later today

- #5

matt grime

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HallsofIvy said:Matt! Of course they are standard.

you live and you learn.

In a finite dimensional space, [itex]||v||\infty= max |v_i|[/itex], the largest component, in absolute value, of v.

In an infinite dimensional, function space, [itex]||v||_\infnty= max(|v(x)|)[/itex] where the max is over whatever compact set v(x) is defined on.

In a finite dimensional space [itex]||v||_1= \sum |v_i|[/itex].

In an infinite dimensional, function space, [itex]||v||_1= \int |v(x)|dx[/itex] where the integral is over the set v(x) is defined on.

In a finite dimensioal space [itex]||v||_2= \sqrt{\sum (v_i)^2}[/itex].

in an infinite dimensional, function space, [itex]||v||_2= \sqrt{\int (v(x))^2 dx}[/itex]

So they are what i said they ought to be.

I certainly agree that || ||_n is a standard norm on the space of functions on some space, but that information wasn't given in the post, was it? I personally have never seen || ||_n used on ordinary finite dimensional vector spaces. If the question had said norms on Banach spaces then I'd not've been confused. In anycase, topologically the norms on a finite dimensional vector space are all the same.

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Zurtex

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HallsofIvy

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Since I now understand that that was the sense in which Matt meant they were "not standard", I must say that I agree with him.

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Zurtex

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- #9

matt grime

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In general the 'n' norms will make me think of functional analytic beasts, which is why I got confused about the references to (what I presumed were finite dimensional) vector spaces.

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The [itex]L_2[/itex] norm does not necessarily produce the "best" fit. The [itex]L_1[/itex] norm produces a "better" fit than does the [itex]L_2[/itex] norm for error sources with large outliers because the [itex]L_1[/itex] norm is less sensitive to outliers. In cases where controlling outliers is important, the [itex]L_\infty[/itex] norm produces a "better" fit because this norm is very sensitive to outliers.

By the way, I often am guilty of moth-like behavior. Finding the least-squares fit is easy. Finding some other best fit is much more difficult.

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