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Posers and Puzzles

Posers and Puzzles

  1. Standard member royalchicken
    CHAOS GHOST!!!
    28 Jan '04 04:32 / 1 edit
    Let f:R^2 --> R be such that for all positive integers N, f(1/N,0) = 1 and f(0,1/N) = -1. Does there exist real L such that f(x,y) --> L as (x,y) --> (0,0)? Prove it.
  2. Standard member flexmore
    Quack Quack Quack !
    28 Jan '04 05:07 / 1 edit
    Originally posted by royalchicken
    Let f:R^2 --> R be such that for all positive integers N, f(1/N,0) = 1 and f(0,1/N) = -1. Does there exist real L such that f(x,y) --> L as (x,y) --> (0,0)? Prove it.
    you don't seem to have defined f off the axes - why is this?

    as for proof: consider integer N --> big positive.

    then f(0,1/N) --> -1

    so L must be -1 (if such L exists)

    also f(1/N,0) --> 1

    so L = 1, contradiction

    so no such L,

    so f is naughty at (0,0)

    (does this have a bigger significance?)

    edit for non-maths folks: you get a topographical map. there is a place on this particular map where as you walk towards it from one direction you go up high in the clouds, if you walk towards that same place from a different direction you end up under the sea. a bit like a cliff edge.
  3. Standard member royalchicken
    CHAOS GHOST!!!
    28 Jan '04 05:09
    Originally posted by flexmore
    you don't seem to have defined f off the axes - why is this?

    as for proof: consider integer N --> big positive.

    then f(0,1/N) --> -1

    so L must be -1 (if such L exists)

    also f(1/N,0) --> 1

    so L = 1, contradiction

    so no such L,

    so f is naughty at (0,0)

    (does this have a bigger significance?)
    It doesn't need to be defined off the axes--it is not completely defined ON the axes, but you are correct.
  4. Standard member StarValleyWy
    BentnevolentDictater
    31 Jan '04 21:56
    Originally posted by royalchicken
    Let f:R^2 --> R be such that for all positive integers N, f(1/N,0) = 1 and f(0,1/N) = -1. Does there exist real L such that f(x,y) --> L as (x,y) --> (0,0)? Prove it.
    Ok. If a function of L goes to lunch, it pays. Therefore it is broke!

    How'd I do?