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Math Help - Is this possible?

  1. #1
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    Is this possible?

    Is it possible to have a function f such that:

    \frac{f}{2}=\int_{a}^{b}f' ?

    Examples would be welcome. But I am mostly interested in whether it is possible or not.

    Thanks
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  2. #2
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    Quote Originally Posted by rainer View Post
    Is it possible to have a function f such that: \frac{f}{2}=\int_{a}^{b}f' ?
    As written the question is meaningless.
    Here is why: \int_{a}^{b}f' is a number if f' is intergable.
    On the other hand, what is \frac{f}{2}?. Is it a function? a number? What?
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  3. #3
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    Let me try that again.

    Does there exist a function f(x) such that:

    \int_{a}^{b}f'(x)dt = \frac{f(x)}{2}

    In words-- does a function exist such that the definite integral over some interval a,b with respect to some dummy variable "t" of the derivative of the function equal the original function divided by two?

    Ah, and I'm not sure whether to call f(x) a function or a number. In the problem I am trying to solve it is the arithmetic-geometric mean of two numbers.

    Oh wait a second, this is really messed up.... let me get back to you
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  4. #4
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    Try f(x) = \exp \left( {\frac{x}{{2(b - a)}}} \right).
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  5. #5
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    I'll try just one more time:

    Does there exist some number x and some function f(x)=x such that:

    \int_{a}^{b}f'(x)dt = \frac{x}{2}

    In words-- does there exist some number x and some function f(x)=x such that the definite integral of the derivative of the function over some interval a,b with respect to some dummy variable "t" equals x divided by two?
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  6. #6
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by rainer View Post
    I'll try just one more time:

    Does there exist some number x and some function f(x)=x such that:

    \int_{a}^{b}f'(x)dt = \frac{x}{2}

    In words-- does there exist some number x and some function f(x)=x such that the definite integral of the derivative of the function over some interval a,b with respect to some dummy variable "t" equals x divided by two?
    This is just completely convoluted, no offense. x=0 works, for clearly f(x)=0\implies f'(x)=0\implies \int_a^{b}f'(x)\text{ }dx=0=\frac{x}{2}=\frac{0}{2}, but I doubt that is what you're after.
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  7. #7
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    Yeah, sorry. After sleeping on it, I think I know what I'm trying to ask now:


    I need a function f(x)=\int_{a}^{b}g(x)dt

    such that: \frac{f(x)}{2}=\int_{a}^{b}g'(x)dt
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  8. #8
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    Quote Originally Posted by rainer View Post
    Yeah, sorry. After sleeping on it, I think I know what I'm trying to ask now:


    I need a function f(x)=\int_{a}^{b}g(x)dt

    such that: \frac{f(x)}{2}=\int_{a}^{b}g'(x)dt
    You are back to the same problem you had initially. As Plato told you in his response, the way you have that written that, the left side of the equation (both equations, now) is function of x while the right side is a number!

    One way I can makes sense of it is if the "x" on the left side, of each equation, is intended to be a constant- a number. If that is the case, then f(x) is just some other number. In that case you are asking either, "given a function g(x), does there exist a function g, and a number, G, such that G= \int_a^b g(x)dx and G/2= \int_a^b g'(x) dx.

    Of course, \int_a^b g'(x)dx= g(b)- g(a) by the "Fundamental Theorem of Calculus" so that is asking if there exist a function g such that \int_a^b g(x)dx= \frac{g(b)-g(a)}{2}. That, in turn, is asking whether there is a function, g, whose anti-derivative is g/2- \int g(x)dx= g(x)/2. Differentiating both sides of that, we are asking whether there is a function g(x) such that g'(x)= g(x)/2.

    The answer to that, of course, is "Yes, there is". And the function, g, is any solution to the differential equation y'= y/2. The general solution to that is g(x)= Ce^{x/2} for any constant C. In that case, the two equations become f= \int_a^b Ce^{x/2}dx= \frac{C}{2}(e^{b/2}- e^{a/2}) and \frac{f}{2}= \frac{C}{2}\int_a^b e^{x/2}dx= \frac{C}{2}(2(e^{b/2}- e^{a/2}) which is correct.

    Another way to make sense of it is to assume you really did not mean the limits of integration but are looking at f being some anti-derivative of g.
    That is, you want f(x)= \int_a^x g(t)dt and f(x)/2= \int_a^x g'(t) dt. Differentiating those equatkons gives f'(x)= g(x) and [tex]f'(x)/2= g'(x)[tex].

    Again, it is not f that is important but g. From the second equation, f'(x)= 2g'(x) so we have f'(x)= 2g(x)= g(x) as before. Now this is true for g(x)= Ce^{2x} and, since the integral of Ce^{2x} is \frac{C}{2}e^{2x}, which is still a constant time e^{2x}, f(x) is also Ce^{2x}.
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    Quote Originally Posted by HallsofIvy View Post
    You are back to the same problem you had initially. As Plato told you in his response, the way you have that written that, the left side of the equation (both equations, now) is function of x while the right side is a number!

    One way I can makes sense of it is if the "x" on the left side, of each equation, is intended to be a constant- a number. If that is the case, then f(x) is just some other number. In that case you are asking either, "given a function g(x), does there exist a function g, and a number, G, such that G= \int_a^b g(x)dx and G/2= \int_a^b g'(x) dx.

    Of course, \int_a^b g'(x)dx= g(b)- g(a) by the "Fundamental Theorem of Calculus" so that is asking if there exist a function g such that \int_a^b g(x)dx= \frac{g(b)-g(a)}{2}. That, in turn, is asking whether there is a function, g, whose anti-derivative is g/2- \int g(x)dx= g(x)/2. Differentiating both sides of that, we are asking whether there is a function g(x) such that g'(x)= g(x)/2.

    The answer to that, of course, is "Yes, there is". And the function, g, is any solution to the differential equation y'= y/2. The general solution to that is g(x)= Ce^{x/2} for any constant C. In that case, the two equations become f= \int_a^b Ce^{x/2}dx= \frac{C}{2}(e^{b/2}- e^{a/2}) and \frac{f}{2}= \frac{C}{2}\int_a^b e^{x/2}dx= \frac{C}{2}(2(e^{b/2}- e^{a/2}) which is correct.

    Another way to make sense of it is to assume you really did not mean the limits of integration but are looking at f being some anti-derivative of g.
    That is, you want f(x)= \int_a^x g(t)dt and f(x)/2= \int_a^x g'(t) dt. Differentiating those equatkons gives f'(x)= g(x) and [tex]f'(x)/2= g'(x)[tex].

    Again, it is not f that is important but g. From the second equation, f'(x)= 2g'(x) so we have f'(x)= 2g(x)= g(x) as before. Now this is true for g(x)= Ce^{2x} and, since the integral of Ce^{2x} is \frac{C}{2}e^{2x}, which is still a constant time e^{2x}, f(x) is also Ce^{2x}.
    Ok, yes, that advice works for when the integral/derivative is dx, but I need it with respect to some dummy variable, not x.

    In the meantime I think I've come up with a solution that satisfies f(x)=\int_a^b g(x)dt and \frac{f(x)}{2}=\int_a^b g'(x)dt :

    \frac{\pi}{2x}=\int_{0}^{\frac{\pi}{2}}\frac{\pi\l  eft(1+\tanh\left(\frac{-\ln{\sin\theta}}{2} \right ) \right )}{4x} d\theta

    \frac{\pi}{4x}=\int_{0}^{\frac{\pi}{2}}\frac{\math  rm{d} }{\mathrm{d} \theta}\frac{\pi\left(1+\tanh\left(\frac{-\ln{\sin\theta}}{2} \right ) \right )}{4x} d\theta

    In this solution f(x)=\frac{\pi}{2x} and g(x)=\frac{\pi\left(1+\tanh\left(\frac{-\ln{\sin\theta}}{2} \right ) \right )}{4x}

    I might just be chasing my tail around in circles--does this amount to a pointless statement?
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  10. #10
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    rainer, by definition, what is \int_a^b g(x)dt?
    Last edited by Defunkt; December 19th 2009 at 12:13 PM.
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  11. #11
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    Interrupting just for fun:

    \int_a^b{g(x)} dt = b*g(x) - a*g(x)
    Since we are integrating with respect to t, g(x) is treated like a constant.
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  12. #12
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    Quote Originally Posted by drop10 View Post
    Interrupting just for fun:

    \int_a^b{g(x)} dt = b*g(x) - a*g(x)
    Since we are integrating with respect to t, g(x) is treated like a constant.
    some more fun ... what if x is a function of t ?
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  13. #13
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    Quote Originally Posted by skeeter View Post
    some more fun ... what if x is a function of t ?
    Do you mean something like: \int_a^b{g(x)}dt, x=h(t)

    Wouldn't that still imply that the notation is wrong since you would ultimately ask \int_a^b{g(t)}dt for some function g. I am actually curious now as to how to do this
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  14. #14
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    Quote Originally Posted by Defunkt View Post
    rainer, by definition, what is \int_a^b g(x)dt?
    I wouldn't have known how to answer that. Thanks.

    But

    \frac{\pi}{2}\frac{\pi\left(1+\tanh\left(\frac{-\ln{\sin\frac{\pi}{2}}}{2} \right ) \right )}{4x}=\frac{\pi^2}{8x}\neq\frac{\pi}{2x}

    What am I missing?
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  15. #15
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    Rainer, you have succeeded is confusing this whole question. First you ask the following
    Quote Originally Posted by rainer View Post
    Does there exist a function f(x) such that:
    \int_{a}^{b}f'(x)dt = \frac{f(x)}{2}
    I replied with a function that does exactly that.
    Quote Originally Posted by Plato View Post
    Try f(x) = \exp \left( {\frac{x}{{2(b - a)}}} \right).
    Now you have changed the whole question. Do you know what you want?
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