Is it possible to have a function $\displaystyle f$ such that:

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

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

Thanks

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- Dec 18th 2009, 08:12 AM #1

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- Dec 18th 2009, 08:22 AM #2

- Dec 18th 2009, 09:45 AM #3

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Let me try that again.

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

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

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

- Dec 18th 2009, 09:57 AM #4

- Dec 18th 2009, 09:59 AM #5

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I'll try just one more time:

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

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

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?

- Dec 19th 2009, 12:03 AM #6

- Dec 19th 2009, 07:24 AM #7

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- Dec 19th 2009, 08:16 AM #8

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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 $\displaystyle G= \int_a^b g(x)dx$ and $\displaystyle G/2= \int_a^b g'(x) dx$.

Of course, $\displaystyle \int_a^b g'(x)dx= g(b)- g(a)$ by the "Fundamental Theorem of Calculus" so that is asking if there exist a function g such that$\displaystyle \int_a^b g(x)dx= \frac{g(b)-g(a)}{2}$. That, in turn, is asking whether there is a function, g, whose anti-derivative is g/2- $\displaystyle \int g(x)dx= g(x)/2$. 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 $\displaystyle g(x)= Ce^{x/2}$ for any constant C. In that case, the two equations become $\displaystyle f= \int_a^b Ce^{x/2}dx= \frac{C}{2}(e^{b/2}- e^{a/2})$ and $\displaystyle \frac{f}{2}= \frac{C}{2}\int_a^b e^{x/2}dx= \frac{C}{2}(2(e^{b/2}- e^{a/2})$ 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 $\displaystyle f(x)= \int_a^x g(t)dt$ and $\displaystyle f(x)/2= \int_a^x g'(t) dt$. Differentiating those equatkons gives $\displaystyle f'(x)= g(x)$ 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 $\displaystyle g(x)= Ce^{2x}$ and, since the integral of $\displaystyle Ce^{2x}$ is $\displaystyle \frac{C}{2}e^{2x}$, which is still a constant time $\displaystyle e^{2x}$, f(x) is also $\displaystyle Ce^{2x}$.

- Dec 19th 2009, 09:53 AM #9

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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 $\displaystyle f(x)=\int_a^b g(x)dt$ and $\displaystyle \frac{f(x)}{2}=\int_a^b g'(x)dt$ :

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

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

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

I might just be chasing my tail around in circles--does this amount to a pointless statement?

- Dec 19th 2009, 11:59 AM #10

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- Dec 19th 2009, 12:22 PM #11

- Dec 19th 2009, 12:39 PM #12

- Dec 19th 2009, 12:53 PM #13

- Dec 19th 2009, 12:56 PM #14

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- Dec 19th 2009, 01:09 PM #15