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    does f_1 \circ f_2 =f_1 (x) \circ f_2 (x) ?
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    (Original post by will'o'wisp2)
    does f_1 \circ f_2 =f_1 (x) \circ f_2 (x) ?
    Haven't seen it written like that on the RHS, but yes I don't see why it would be different.

    Technically would make more sense to write it as (f_1 \circ f_2)(x) = f_1(x) \circ f_2(x)
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    (Original post by RDKGames)
    Haven't seen it written like that on the RHS, but yes I don't see why it would be different.

    Technically would make more sense to write it as (f_1 \circ f_2)(x) = f_1(x) \circ f_2(x)
    ok thanks man
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    (Original post by will'o'wisp2)
    does f_1 \circ f_2 =f_1 (x) \circ f_2 (x) ?
    (Original post by RDKGames)
    Haven't seen it written like that on the RHS, but yes I don't see why it would be different.

    Technically would make more sense to write it as (f_1 \circ f_2)(x) = f_1(x) \circ f_2(x)
    Some context would be nice, but by default I'd interpret (f_1 \circ f_2) (x) to be f_1(f_2(x)). (And without context, I have no idea what f_1(x) \circ f_2(x) would mean).
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    (Original post by DFranklin)
    Some context would be nice, but by default I'd interpret (f_1 \circ f_2) (x) to be f_1(f_2(x)). (And without context, I have no idea what f_1(x) \circ f_2(x) would mean).
    Agreed, hence my comment about not seeing it before, but f_1(x) \circ f_2(x)=f_1(f_2(x)) would be my go-to 'assumption' without any context - hopefully it is not a notation that is used in his textbooks/course.
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    (Original post by RDKGames)
    Agreed, hence my comment about not seeing it before, but f_1(x) \circ f_2(x)=f_1(f_2(x)) would be my go-to 'assumption' without any context - hopefully it is not a notation that is used in his textbooks/course.
    The fact that the LHS is explictly saying x is the argument to f_1 seems counter to interpreting it to equal the RHS.

    I would basically hate to put money on any interpretation of this (including what the OP actually thinks either side means) without more context.
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    (Original post by DFranklin)
    The fact that the LHS is explictly saying x is the argument to f_1 seems counter to interpreting it to equal the RHS.

    I would basically hate to put money on any interpretation of this (including what the OP actually thinks either side means) without more context.
    the rhs is me making my own stuff up xD

    but all it is really was just if they're like the same thing and mean the same thing

    it's a composite function basically so i thought it might not be clear if i just use f_2 \circ f_1 =3 for example so i didn't know if i needed to stick the x there or not
 
 
 
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