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Thread: Functions

  1. #11
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    Default Re: Functions

    Quote Originally Posted by BuleriaChk View Post
    To the reader: There are members on this forum who insist on spamming my threads with ignorant and/irrelevant comments. Please read my ignore list before continuing..
    -----------------------------------------------
    Since some on this forum seem unfamiliar with the concept of function (especially as regards the derivative) I thought I'd provide a direct reference link for those new to the subject.

    FUNCTION

    (It is with a view to helping John Gabriel understand the role of curvature, who at present seems to be only vaguely familiar with the concept of a straight line, but can (almost) calculate the slope correctly ...

    However, the reader may need to refer first to the equation of a straight line....

    Attachment 1138

    The correct slope is given by

    where m = 0.

    (John Gabriel's red line is too long and in the wrong vertical position because of m)
    One then takes the limit on n->0 for the linear derivative:

    ,

    which eliminates (i.e. makes insignificant) higher orders of f(x) (i.e., curvature) when n gets small enough for an arbitrary function f(x) at an arbitrary position x ; i.e., (x, f(x)) = (x, y(x)) ..
    There are only two points labeled on your illustration that are on the function f, and your line y=Ax+b does not go through them both. It only goes through one of them.
    Last edited by grapes; 09-20-2016 at 07:24 AM.

  2. #12
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    Default Re: Functions

    Quote Originally Posted by grapes View Post
    There are only two points labeled on your illustration that are on the function f, and your line y=Ax+b does not go through them both. It only goes through one of them.
    The line y = Ax + b goes through four points.

    1. (0,b) (b is negative; it is the y intercept)

    2. (x-m,0)

    3.(x', f(x-m))

    4. (x+n, f(x+n))

    Which point do you think it does not go through?

    Point 3 is on the correct "run" for the definition of the slope. If m = 0, it would be the point (x,f(x))
    Last edited by BuleriaChk; 09-20-2016 at 11:10 AM.
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  3. #13
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    Default Re: Functions

    Quote Originally Posted by emperorzelos View Post
    (snip)
    The Naked Emperor wants to go back to the very beginning (sets) under the illusion that this is the way to resolve issues involving functions related to derivatives. Of course, his first set doesn't have a metric, and is irrelevant to a discussion of derivatives. Just imaginary points just floating around in a Venn diagram somewhere.

    Those that understand calculus already knew about those and have moved on to ranges and domains. And independent variables...

    If all you have is a hammer, then everything looks like a nail....

    Like most fundamentalist religious fanatics -

    ".. and he's not even wrong!" .. W. Pauli
    Last edited by BuleriaChk; 09-20-2016 at 11:37 AM.
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  4. #14
    Moderator grapes's Avatar
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    Default Re: Functions

    Quote Originally Posted by BuleriaChk View Post
    Quote Originally Posted by grapes
    There are only two points labeled on your illustration that are on the function f, and your line y=Ax+b does not go through them both. It only goes through one of them.
    The line y = Ax + b goes through four points.

    1. (0,b) (b is negative; it is the y intercept)

    2. (x-m,0)

    3.(x', f(x-m))

    4. (x+n, f(x+n))

    Which point do you think it does not go through?
    There are two points labeled that are on the function: (x-m, f(x-m)) and (x+n, f(x+n))

  5. #15
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    Default Re: Functions

    Quote Originally Posted by grapes View Post
    There are two points labeled that are on the function: (x-m, f(x-m)) and (x+n, f(x+n))
    It doesn't go through (x-m, f(x-m)) because John Gabriel's slope is wrong, and so therefore is not on the line y = Ax + b.

    If m = 0, the point would be (x',f(x)) (Point 3 in Post #13) thus defining the correct slope.
    (x' then becomes identified with x (x'=x) , so the point is (x,f(x))
    Last edited by BuleriaChk; 09-20-2016 at 12:13 PM.
    _______________________________________
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  6. #16
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    Default Re: Functions

    Quote Originally Posted by BuleriaChk View Post
    The Naked Emperor wants to go back to the very beginning (sets) under the illusion that this is the way to resolve issues involving functions related to derivatives. Of course, his first set doesn't have a metric, and is irrelevant to a discussion of derivatives. Just imaginary points just floating around in a Venn diagram somewhere.

    Those that understand calculus already knew about those and have moved on to ranges and domains. And independent variables...

    If all you have is a hammer, then everything looks like a nail....

    Like most fundamentalist religious fanatics -

    ".. and he's not even wrong!" .. W. Pauli
    It is cute how you cite Pauli when you don't even understand the basic structures of mathematics that you try to use and when I try to tell you about them, all you do is cry "buzzwords", sorry but you not understanding things does not make them buzzwords.

  7. #17
    Moderator grapes's Avatar
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    Default Re: Functions

    Quote Originally Posted by BuleriaChk View Post
    It doesn't go through (x-m, f(x-m)) because John Gabriel's slope is wrong, and so therefore is not on the line y = Ax + b.

    If m = 0, the point would be (x',f(x)) (Point 3 in Post #13) thus defining the correct slope.
    (x' then becomes identified with x (x'=x) , so the point is (x,f(x))
    Where is (x', f(x')), on your graph?

  8. #18
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    Default Re: Functions

    Quote Originally Posted by grapes View Post
    Where is (x', f(x')), on your graph?
    At (x,f(x)) when m = 0
    (Try to use your imagination to see how m = 0 changes things)
    (Draw some graphs yourself and move things around...)
    _______________________________________
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    The Relativistic Unit Circle 03/28/2017 07:40 AM PST
    Proof of Fermat's Last Theorem Updates 03/19/2017 8:23 PM PST
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  9. #19
    Moderator grapes's Avatar
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    Default Re: Functions

    Quote Originally Posted by BuleriaChk View Post
    Quote Originally Posted by grapes View Post
    Where is (x', f(x')), on your graph?
    At (x,f(x)) when m = 0
    I'm asking about your posted graph

    Clearly, m is not zero on that graph
    (Try to use your imagination to see how m = 0 changes things)
    (Draw some graphs yourself and move things around...)
    I just want to know where you think (x', f(x') would be on your graph?

  10. #20
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    Default Re: Functions

    Quote Originally Posted by grapes View Post
    I'm asking about your posted graph

    Clearly, m is not zero on that graph

    I just want to know where you think (x', f(x') would be on your graph?
    I am showing the ERROR in John's definition of the slope. The ERROR comes about because m 0.
    To CORRECT the error, one sets m = 0. The point in ERROR moves to (x',f(x)) where it becomes (x,f(x)), and thus is on the line y=Ax + b which is the correct equation of the line. At that point, one replaces x' with x (at the left end of the green line).

    The correct diagram for the equation of a line can be found in many sources on the internet. (Try "Math is Fun")

    When m = 0, John's equation then becomes the standard prototype for the derivative (a line with slope = A):



    Note that at this point, the function has not been specified; only that it runs through the two points (x,f(x)) and ((x+n),f(x+n))

    Notice that the "run" (n) is parallel to the x axis, so can be represented as n = x for the LINE y = Ax + b

    To allow for the slope of an arbitrary function f(x) (i.e., without specifying the function explicitly) one replaces n = x with the catchall term h, because now the slope rotates along the curve of the function (whatever it is) as the limit is reached. (There is a .gif of this in the "toast" link below below from "Math Is Fun")

    See the first post in the thread "Derivatives" (and ignore the rest)

    The equation then becomes the standard definition of the derivative:




    ------------------------------------------------------------------------------------


    For physics, a potential energy is represented as a function V(x); if h is the quantum of action, the derivative represents the case where h -> 0 (at 0, A(x) = 0 if A(x) = V(x) = , t = 1 (one period of the DeBroglie wave function, and h is the action of a single photo-electron).

    (When the quantum potential energy goes to 0 (the electron doesn't exist), classical physics (Newton's equations of force and , and Maxwell's equations apply (only masses and EM fields) - and time-space Relativity where x=ct is a "ruler creation process").

    (This last result is a topic for another thread I'm working on. This is just to show its relevance to physics - Newton used his to calculate orbits.)
    Last edited by BuleriaChk; 09-21-2016 at 02:24 PM.
    _______________________________________
    "Flamenco Chuck" Keyser
    The Relativistic Unit Circle 03/28/2017 07:40 AM PST
    Proof of Fermat's Last Theorem Updates 03/19/2017 8:23 PM PST
    Ignore List -The Peanut Gallery.

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