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Math Help - total derivatives of increasing functions

  1. #1
    eco
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    total derivatives of increasing functions

    Suppose y(x) is strictly increasing and y'''(x)=0.

    Since it's an increasing function, for dx > 0 I know:

    dy = y'dx + .5y''dx^2 > 0.

    Dividing by dx:

    dy/dx = y' + .5y''dx > 0.

    But what if I instead take partial derivative of dy wrt dx? That is:

    \partial (dy)/\partial (dx) = y' + y''dx.

    I now have an extra .5y''dx. Do I still know this is positive? Since y(x) is increasing, it seems that dy grows with dx, so intuitively it seems so. Also, I know for small enough dx, the sign of (y' + y''dx) is the same as the sign of y', which is positive. But on the other hand y'' can be any sign and dx might be large. So it seems not.

    Which is right? Thanks in advance for your help.
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    Forum Admin topsquark's Avatar
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    Quote Originally Posted by eco View Post
    Suppose y(x) is strictly increasing and y'''(x)=0.

    Since it's an increasing function, for dx > 0 I know:

    dy = y'dx + .5y''dx^2 > 0.

    Dividing by dx:

    dy/dx = y' + .5y''dx > 0.

    But what if I instead take partial derivative of dy wrt dx? That is:

    \partial (dy)/\partial (dx) = y' + y''dx.

    I now have an extra .5y''dx. Do I still know this is positive? Since y(x) is increasing, it seems that dy grows with dx, so intuitively it seems so. Also, I know for small enough dx, the sign of (y' + y''dx) is the same as the sign of y', which is positive. But on the other hand y'' can be any sign and dx might be large. So it seems not.

    Which is right? Thanks in advance for your help.
    I need you to clarify: What does dy = y'dx + .5y''dx^2 mean? This expression makes no sense to me.

    I think what you are reaching for is
    dy = c~dx + xy''~dx

    though most would write this as
    y' = c + xy''

    You can certainly take a partial derivative of a function y(x), but it is the same as an ordinary derivative.

    Since y''' = 0 we know that y(x) = a + bx + (1/2)cx^2 where a, b, c are constants. We also know that y' = b + cx > 0 for all x. What does this say about b + cx?

    -Dan
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  3. #3
    eco
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    The expression

    dy = y'dx + .5y''dx^2

    Comes from the Taylor expansion. The first two terms are sufficient since y'''=0.
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    Forum Admin topsquark's Avatar
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    Quote Originally Posted by eco View Post
    The expression

    dy = y'dx + .5y''dx^2

    Comes from the Taylor expansion. The first two terms are sufficient since y'''=0.
    The problem is the dy and the dx^2. They cannot really be in the same expression. Consider when we put this equation into a more "ordinary" form by dividing both sides by dx:
    \frac{dy}{dx} = y' + \frac{y''}{2}dx

    dx is infinitesimal and thus can be dropped giving merely y' = y'. Another problem is that you are including functions where constants should be. (Namely y'.) Let me explain:

    To do a Taylor expansion we must first pick a point (near the evaluation point) with which to estimate. Call this point x = x_0. (In many cases x0 = 0. This is a MacLaurin expansion. It often happens that this series converges for all x which is why it is so useful.)

    The Taylor expansion of y' is then:
    y'(x) = y'(x_0) + y''(x_0)(x - x_0) + \frac{1}{2}y'''(x_0)(x - x_0^2) + ~...

    where y''(x_0), for example, is the second derivative of y evaluated at x = x0. In this case, as you say, y'''(x) = 0, so the second term is the end of the expansion. Thus
    y'(x) = y'(x_0) + y''(x_0)(x - x_0)

    y'(x0) and y''(x0) are both constants, not functions of x. (y''(x) = const anyway in this case.) So to put the equation into the form you were trying to get it to:
    dy = (a + bx) dx

    where a = y'(x0) - x0*y''(x0) and b = y''(x0).

    -Dan
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  5. #5
    eco
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    topsquark,

    thanks for being patient with me -- I'm sorry my notation is confusing you. dy and dx are just changes. y' is the derivative. So (confusingly) dy/dx is not a derivative, it's just a quotient.

    Recall I have y(x) a strictly increasing function and y'''=0.

    So I know for dx>0 that

    dy = y'dx + .5*y''*(dx)^2 > 0.

    Since dx > 0 I can divide:

    y' + .5*y''*dx > 0.

    On the other hand if I just take the derivative of dy wrt to dx I have

    y' + y''*dx.

    The question is: can I sign this? Intuitively, since y is increasing in x, dy would seem to increase as I take a bigger dx. (And for arbitrarily small dx it's just y' which is positive.) But since y'' can be any sign and dx is arbitrary, I don't see how I can prove it.

    Thanks again for sticking with me.
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    Forum Admin topsquark's Avatar
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    Quote Originally Posted by eco View Post
    topsquark,

    thanks for being patient with me -- I'm sorry my notation is confusing you. dy and dx are just changes. y' is the derivative. So (confusingly) dy/dx is not a derivative, it's just a quotient.

    Recall I have y(x) a strictly increasing function and y'''=0.

    So I know for dx>0 that

    dy = y'dx + .5*y''*(dx)^2 > 0.

    Since dx > 0 I can divide:

    y' + .5*y''*dx > 0.
    Let's do this. Your Taylor series is truncated (that is it ends) so it always converges. Thus we can simply choose x0 = 0. Then dx = x - 0 = x and we can get rid of this nasty notation.

    Recall that y' and y'' are constants based on the evaluation point x0 that is chosen. They are not functions! Now, y'' = const for all x, so let's just call that y''(x) = a or something. y' represents y'(0), so let's call that b. This gives
    y' + (1/2)y''*x > 0 Implies b + (1/2)ax > 0

    What does this tell you about a and b?

    -Dan

    Edit: As for your other method, it gives the same results, just with a different constant for a.
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