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-   -   Continuous function (https://www.askmehelpdesk.com/showthread.php?t=553047)

  • Feb 9, 2011, 05:54 PM
    western50
    continuous function
    1. Consider the function f : R9 to R de fined as follows. For each vector a =
    (a1; a2; : : : ; a9) in R9, let f(a) be the 3 by 3 determinant

    a1 a2 a3
    a4 a5 a6
    a7 a8 a9

    :
    Is f a continuous function from R9 to R; why or why not?

    please explain the concept to me!
  • Feb 9, 2011, 08:34 PM
    jcaron2
    You have a function f which maps points in 9-dimensional real space to 1-dimensional real space (i.e. a scalar). Even though it has more dimensions than you're used to working with, conceptually it's no different than a function which maps, say, two dimensional space. For example

    f(x,y) = 3x - 2y + 6

    maps R2 to R. For any coordinate pair (x,y), the function f(x,y) will have some real value, and it's pretty obvious that it's continuous. Changing x or y by some infinitesimal amount will result in an infinitesimal change in f(x,y).

    Back to your question, since the function is simply the determinant of the matrix formed as indicated from the 9 coordinates, you can just write it out in long form:

    f(a) = a1 * (a5*a9-a6*a8) - a2 * (... etc.

    The determinant function is just some simple (albeit tedious) addition, subtraction, and multiplication. Like my simple 2-dimensional example, those are all continuous operations, so the function f(a) is continuous.

    Does that make sense to you?
  • Feb 10, 2011, 12:00 AM
    western50
    So with what you said, because the determinant will give me some a real numbers, so that I can conclude from it that the function is continuous?
  • Feb 10, 2011, 06:23 AM
    jcaron2
    Comment on western50's post
    Yes, that's right. Not only are the answers real, but they change smoothly (i.e. continuously) as the values for a1 through a9 change.
  • Feb 10, 2011, 11:25 AM
    western50
    So is there any way making f(a) not continuous? Just a after thought
  • Feb 10, 2011, 03:02 PM
    jcaron2
    Sure. For example, if f(a) = 1/a1 + 1/a2 + 1/a3 +... + 1/a9, then it would be discontinuous at any point where one or more of the coordinates was 0. That's because the function blows up when the denominator goes to zero. Of course you could come up with an infinite variety of functions whose denominators go to zero at some point or points.

    Likewise, if it included a discontinuous trig function like tan or csc, or any of countless others, it would map discontinuously.




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