# Is this possible?

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• Dec 18th 2009, 08:12 AM
rainer
Is this possible?
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
• Dec 18th 2009, 08:22 AM
Plato
Quote:

Originally Posted by rainer
Is it possible to have a function $\displaystyle f$ such that: $\displaystyle \frac{f}{2}=\int_{a}^{b}f'$ ?

As written the question is meaningless.
Here is why: $\displaystyle \int_{a}^{b}f'$ is a number if $\displaystyle f'$ is intergable.
On the other hand, what is $\displaystyle \frac{f}{2}?$. Is it a function? a number? What?
• Dec 18th 2009, 09:45 AM
rainer
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
Plato
Try $\displaystyle f(x) = \exp \left( {\frac{x}{{2(b - a)}}} \right)$.
• Dec 18th 2009, 09:59 AM
rainer
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
Drexel28
Quote:

Originally Posted by rainer
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?

This is just completely convoluted, no offense. $\displaystyle x=0$ works, for clearly $\displaystyle f(x)=0\implies f'(x)=0\implies \int_a^{b}f'(x)\text{ }dx=0=\frac{x}{2}=\frac{0}{2}$, but I doubt that is what you're after.
• Dec 19th 2009, 07:24 AM
rainer
Yeah, sorry. After sleeping on it, I think I know what I'm trying to ask now:

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

such that: $\displaystyle \frac{f(x)}{2}=\int_{a}^{b}g'(x)dt$
• Dec 19th 2009, 08:16 AM
HallsofIvy
Quote:

Originally Posted by rainer
Yeah, sorry. After sleeping on it, I think I know what I'm trying to ask now:

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

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

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 $\displaystyle G= \int_a^b g(x)dx$ 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
rainer
Quote:

Originally Posted by HallsofIvy
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 $\displaystyle Ce^{2x}$.

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
Defunkt
rainer, by definition, what is $\displaystyle \int_a^b g(x)$dt?
• Dec 19th 2009, 12:22 PM
drop10
Interrupting just for fun(Rock):

$\displaystyle \int_a^b{g(x)} dt = b*g(x) - a*g(x)$
Since we are integrating with respect to t, $\displaystyle g(x)$ is treated like a constant.
• Dec 19th 2009, 12:39 PM
skeeter
Quote:

Originally Posted by drop10
Interrupting just for fun(Rock):

$\displaystyle \int_a^b{g(x)} dt = b*g(x) - a*g(x)$
Since we are integrating with respect to t, $\displaystyle g(x)$ is treated like a constant.

some more fun ... what if x is a function of t ?
• Dec 19th 2009, 12:53 PM
drop10
Quote:

Originally Posted by skeeter
some more fun ... what if x is a function of t ?

Do you mean something like: $\displaystyle \int_a^b{g(x)}dt, x=h(t)$

Wouldn't that still imply that the notation is wrong since you would ultimately ask $\displaystyle \int_a^b{g(t)}dt$ for some function g. I am actually curious now as to how to do this(Doh)
• Dec 19th 2009, 12:56 PM
rainer
Quote:

Originally Posted by Defunkt
rainer, by definition, what is $\displaystyle \int_a^b g(x)$dt?

I wouldn't have known how to answer that. Thanks.

But

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

What am I missing?
• Dec 19th 2009, 01:09 PM
Plato
Rainer, you have succeeded is confusing this whole question. First you ask the following
Quote:

Originally Posted by rainer
Does there exist a function $\displaystyle f(x)$ such that:
$\displaystyle \int_{a}^{b}f'(x)dt = \frac{f(x)}{2}$

I replied with a function that does exactly that.
Quote:

Originally Posted by Plato
Try $\displaystyle f(x) = \exp \left( {\frac{x}{{2(b - a)}}} \right)$.

Now you have changed the whole question. Do you know what you want?
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