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**{7/3}****Member**- Registered: 2013-02-11
- Posts: 210

The p-norm of a vector(in 2d) is (|x|^p+|y|^p)^(1/p) my question is can p be 0,fraction or nagetive?

There are 10 kinds of people in the world,people who understand binary and people who don't.

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**zetafunc.****Guest**

If I recall correctly, p can be any real number apart from 0.

**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 14,885

Actually, p must be a real number greater or equal to 1.

Here lies the reader who will never open this book. He is forever dead.

Taking a new step, uttering a new word, is what people fear most. ― Fyodor Dostoyevsky, Crime and Punishment

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**{7/3}****Member**- Registered: 2013-02-11
- Posts: 210

Why isn't nagetive number or fractions<1 allowed?

There are 10 kinds of people in the world,people who understand binary and people who don't.

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 14,885

Because it wouldn't be a norm (it would give zero values if any of the coordinates is 0).

Here lies the reader who will never open this book. He is forever dead.

Taking a new step, uttering a new word, is what people fear most. ― Fyodor Dostoyevsky, Crime and Punishment

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**{7/3}****Member**- Registered: 2013-02-11
- Posts: 210

Oh,thanks.

There are 10 kinds of people in the world,people who understand binary and people who don't.

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