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Thread: Pointwise sum and product

  1. #1
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    Pointwise sum and product

    If A and B are sets where B has the folowing binary operation defined on it.

    BxB -> B
    (b,b') -> b*b'

    Then given 2 maps, f,g contained in Map(A,B) we get another map f*g contained in Map(A,B) by setting

    Map(A,B)xMap(A,B) -> Map(A,B)
    (f,g) -> f*g

    Of particular importance is the case A=B=R (reals) and it is usual to refer to f and g as functions in this case. Then the addition and multiplication on R induce pointwise addition and multiplication on Map(R,R):

    f+g: R -> R
    a -> f(a) + g(a)

    f.g: R -> R
    a -> f(a)g(a)


    Can anyone explain whats going on here? I don't understand how maps can be contained in other maps nor the whole pointwise thing.

    PS does anyone happen to know if you can write greek characters on a keyboard using alt+ number codes as you can with other european characters?
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  2. #2
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    Quote Originally Posted by Obstacle1 View Post
    If A and B are sets where B has the following binary operation defined on it.
    BxB -> B
    (b,b') -> b*b'

    Then given 2 maps, f,g contained in Map(A,B) we get another map f*g contained in Map(A,B) by setting Map(A,B)xMap(A,B) -> Map(A,B) (f,g) -> f*g

    Can anyone explain what’s going on here? I don't understand how maps can be contained in other maps nor the whole pointwise thing.
    PS does anyone happen to know if you can write greek characters on a keyboard using alt+ number codes as you can with other european characters?
    This appears to be a very interesting question. However, I am clueless about the notation. Is Map(A,B) the set of all mapping(functions) from A to B. If so the usual notations is $\displaystyle B^A$. What is the operation f*g? That doesn’t make any sense without a complete set of definitions( I do not have your textbook). Please give us more information.

    BTW. You should try to learn LaTeX.
    If you type [math]\lambda[/tex] we can see $\displaystyle \lambda$
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  3. #3
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    Quote Originally Posted by Plato View Post
    This appears to be a very interesting question. However, I am clueless about the notation. Is Map(A,B) the set of all mapping(functions) from A to B. If so the usual notations is $\displaystyle B^A$. What is the operation f*g? That doesn’t make any sense without a complete set of definitions( I do not have your textbook). Please give us more information.

    BTW. You should try to learn LaTeX.
    If you type $\displaystyle \lambda$ we can see $\displaystyle \lambda$

    Yea, Map(A,B) is the set of all maps from A to B. Sorry, my textbook gave this as the standard notation and said that $\displaystyle B^A$ is now seldom used. Although I won't be offended if you use it . I don't exactly understand what f*g is either, but I thought it was defined in some way by the initial binary operation given. i.e.
    $\displaystyle \beta$: BxB -> B where (b,b') -> b*b' (ordinary multiplication)
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  4. #4
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    Quote Originally Posted by Obstacle1 View Post
    I don't exactly understand what f*g is either, but I thought it was defined in some way by the initial binary operation given. i.e.
    $\displaystyle \beta$: BxB -> B where (b,b') -> b*b' (ordinary multiplication)
    Because each of f & g is a mapping of A to B, if we define f*g(x) as f(x)*g(x) where * is the binary operation on B, then f*g is a mapping from A to B also.

    But I still am not clear as to what the question is asking.
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  5. #5
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    Quote Originally Posted by Plato View Post
    Because each of f & g is a mapping of A to B, if we define f*g(x) as f(x)*g(x) where * is the binary operation on B, then f*g is a mapping from A to B also.

    But I still am not clear as to what the question is asking.
    How do you deduce that?

    I guess i was asking 'what are pointwise addition and multiplication?'.

    So far i'm thinking along the lines of...because f and g are maps from A to B and B has a certain operation defined on it, you can also perform this operation on f and g...
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  6. #6
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    Quote Originally Posted by Obstacle1 View Post
    How do you deduce that?
    If each of f & g is a mapping of A to B and * is the binary operation on B, then $\displaystyle \forall x \in A\, \Rightarrow \; f(x) \in B \wedge g(x) \in B$. Therefore, $\displaystyle f(x)*g(x) \in B$ is a well defined operation; thus defining $\displaystyle \left( {f*g} \right)(x) = f(x)*g(x)$ creates a mapping $\displaystyle f*g:A \mapsto B$.
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  7. #7
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    Quote Originally Posted by Plato View Post
    If each of f & g is a mapping of A to B and * is the binary operation on B, then $\displaystyle \forall x \in A\, \Rightarrow \; f(x) \in B \wedge g(x) \in B$. Therefore, $\displaystyle f(x)*g(x) \in B$ is a well defined operation; thus defining $\displaystyle \left( {f*g} \right)(x) = f(x)*g(x)$ creates a mapping $\displaystyle f*g:A \mapsto B$.

    ok. i see now, quite simple really.
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