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Math Help - Logical equivalence

  1. #1
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    Logical equivalence

    Suppose y is a known function. Is it then correct to write

    f(x) = y \ \Leftrightarrow \ f'(x) = y'

    I mean, f(x) = y is true. And if and only if f(x) = y is true, then so is f'(x) = y', right?
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  2. #2
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    Re: Logical equivalence

    Quote Originally Posted by SweatingBear View Post
    Suppose y is a known function. Is it then correct to write f(x) = y \ \Leftrightarrow \ f'(x) = y'
    I mean, f(x) = y is true. And if and only if f(x) = y is true, then so is f'(x) = y', right?
    1) There are functions that do not have derivatives.

    2) x^3+2~\&~x^3-2 have the same derivative but are different functions.
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  3. #3
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    Re: Logical equivalence

    to put it in a different way: equivalent is not equal. we "lump stuff together" when we make equivalences.

    you can define f to be equivalent to g if f'(x) = g'(x), that is a perfectly good equivalence. but it is NOT equality.

    y = f(x) = 2 and y = g(x) = 0 have the same derivative, and in some sense "act the same". but clearly 2 isn't 0.

    this is why we put the +C after integrals, because we don't get a unique function, we get an equivalence class of functions.
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  4. #4
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    Re: Logical equivalence

    What if I write

    f(x) = x^2 + x + 1 \ \Leftrightarrow \ f'(x) = 2x + 1

    Still not valid?
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  5. #5
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    Re: Logical equivalence

    My lecturer advised his students to do mental workout in the morning: to take a constant and differentiate it. Next morning, he would say, take a bigger constant.
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  6. #6
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    Re: Logical equivalence

    Hello, SweatingBear!

    Suppose y is a known function.
    Is it then correct to write: . f(x) = y \;\; \Leftrightarrow \;\; f'(x) = y'

    I mean, f(x) = y is true.
    And if and only if f(x) = y is true, then so is f'(x) = y', right?

    I would drop the "if and only if".

    The converse is not always true.
    . . If f'(x) = y', then f(x) may be equal to y.


    Example:

    \begin{Bmatrix}f(x) &=& x^2 - 8 \\ y &=& x^2 + 7 \end{Bmatrix}\quad\Rightarrow\quad \begin{Bmatrix} f'(x) &=& 2x \\ y' &=& 2x\end{Bmatrix}

    The derivatives are equal, but the functions are not equal.

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  7. #7
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    Re: Logical equivalence

    no. if we let g(x) = x^2 + x + 2 then g'(x) = 2x + 1, so we cannot conclude from f'(x) = 2x + 1, that f(x) = x^2 + 2x + 1.

    we have "=>" but not "<=" therefore not "<=>".
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