Question

Solve for f(x)

Answer 1

\[\Large \int\limits_a^b f(x)dx= \int\limits_{1/b}^{1/a}f(x)dx\]

Answer 2

Nontrivial solution, right?

Answer 3

Haha yeah, I suppose so. I mean where's the fun in f(x)=0 haha

Answer 4

f(x) = 0
is same as y= 0
X axis :P

Answer 5

F(b) - F(a) = F(1/a)- F(1/b)
F(b) - F(1/a) = F(a)- F(1/b)
\(\Large \int\limits_{1/a}^b f(x)dx= \int\limits_{1/b}^{a}f(x)dx\)

not sure how that would help :P

Answer 6

Here's my reasoning so far:

Flip the limits \[\Large \int\limits_a^b f(x)dx= -\int\limits_{1/a}^{1/b}f(x)dx\]

Do a substitution \[\Large \int\limits_a^b f(x)dx= -\int\limits_{a}^{b}f( \frac{1}{x} ) \frac{1}{x^2} dx\]

Combine them \[\Large 0= \int\limits_a^b[ f(x) + f( \frac{1}{x} ) \frac{1}{x^2} ]dx\]

The only way this integral can be zero for any arbitrary values of a and b is for it to be 0 everywhere, so:

\[\Large 0= f(x) + f( \frac{1}{x} ) \frac{1}{x^2} \]

This leaves us with \[\Large \frac{1}{x} f( \frac{1}{x}) = x f(x)\] but I am not sure how to go from here.

Answer 7

a not =b
right ?

Answer 8

and u forgot the negative sign

Answer 9

I don't think it matters. I think the main limitation is a and b can't be equal to 0 since that would give us division by zero in the limits of the integral to begin with.

@hartnn what step did I forget a negative sign?

Answer 10

1/x f(1/x) = -x f(x)

Answer 11

if a=b
then both integral = 0

Answer 12

Oh whoops I did the substitution wrong. It will change the sign on the integral above, so I ended up with the correct result because I forgot to change it there: \[\Large - \int\limits_{u=1/a}^{u=1/b}f(u)du\] Choose: \[\Large u= x^{-1} \\ \Large du = -x^{-2}dx\] Then we get: \[\Large \int\limits_{x=a}^{x=b}f(\frac{1}{x}) \frac{1}{x^2}dx\]

@hartnn Sure, if a=b then the integral = 0, however that will still be consistent with whatever f(x) can be.

Answer 13

@hartnn if I say \[\Large \int\limits_0^a e^xdx = e^a-1\] There's nothing wrong with saying a=0, as it will be true here as well. It's not a limitation, just a property of all integrals.

Answer 14

i am with you :)
if a=b,
f(x) =infinite solutions

Answer 15

@hartnn I suppose, but a and b are variables and a=b is just one set of possible values they can take on in their domain, but it's unrestricted. Besides that's too easy!

Answer 16

FIXED VERSION:

Initial problem:\[\Large \int\limits_a^b f(x)dx= \int\limits_{1/b}^{1/a}f(x)dx\]

Flip the limits \[\Large \int\limits_a^b f(x)dx= -\int\limits_{1/a}^{1/b}f(x)dx\]

Do a substitution \[\Large \int\limits_a^b f(x)dx= \int\limits_{a}^{b}f( \frac{1}{x} ) \frac{1}{x^2} dx\]

Combine them \[\Large 0= \int\limits_a^b[ f(x) -f( \frac{1}{x} ) \frac{1}{x^2} ]dx\]

The only way this integral can be zero for any arbitrary values of a and b is for it to be 0 everywhere, so:

\[\Large 0= f(x) - f( \frac{1}{x} ) \frac{1}{x^2} \]

This leaves us with \[\Large \frac{1}{x} f( \frac{1}{x}) = x f(x)\] but I am not sure how to go from here.

Answer 17

Power series?

Answer 18

I am trying to avoid that but maybe that's what we have to play around with lol. It just looks so nice and symmetrical though in this form. If you replace x with 1/x it becomes itself again. The variable is outside itself on both sides... There's gotta be a way...

Answer 19

Actually there's an easier way to get to this result, just let b=1 and take the derivative with respect to a: \[\Large \int\limits_a^b f(x)dx= \int\limits_{1/b}^{1/a}f(x)dx\]\[\Large -f(a)=f(\frac{1}{a}) \frac{-1}{a^2}\]

Answer 20

I hate to give away the answer, but by just playing around I "guessed" it. \[\Large f(x)=\frac{1}{x}\] Which does indeed satisfy the equation and integral. But I can't figure out a good reasoning to get there. For instance, what if I wanted to figure out this related question? \[\Large xf(\frac{1}{x})=\frac{1}{x}f(x)\]

Answer 21

Actually by just guessing, this equation I just wrote has the answer f(x)=x lol. Yeah why did you delete that it looked like you were right.

Answer 22

\(
\Large \frac{1}{x} f( \frac{1}{x}) = x f(x)\)
f(x) =1/x
you are right

i think you made the typo in your last comment

Answer 23

I'm wondering if there's any merit to this idea... We take this equation
\[\frac{1}{x} f\left( \frac{1}{x}\right) = x f(x)\]
and take some partial derivatives, namely w.r.t \(\dfrac{1}{x}\) and \(x\). Then you have
\[\text{Partials wrt }\dfrac{1}{x}:\\
\begin{align*}
\frac{1}{x} f\left(\frac{1}{x}\right) &= x f(x)\\\\
&= \frac{1}{\frac{1}{x}} f\left(\frac{1}{\frac{1}{x}}\right)\\\\
\frac{\partial}{\partial \frac{1}{x}}[\cdots]&=\frac{\partial}{\partial \frac{1}{x}}[\cdots]\\\\
f\left(\frac{1}{x}\right)+ \frac{1}{x}f'\left(\frac{1}{x}\right)&=-\frac{1}{\frac{1}{x^2}}f\left(\frac{1}{\frac{1}{x}}\right)-\frac{1}{\frac{1}{x^3}}f'\left(\frac{1}{\frac{1}{x}}\right)\\\\
f\left(\frac{1}{x}\right)+ \frac{1}{x}f'\left(\frac{1}{x}\right)&=-x^2f(x)-x^3f'(x)
\end{align*}\]

\[\text{Partials wrt }x:\\
\begin{align*}
\frac{1}{x} f\left(\frac{1}{x}\right) &= x f(x)\\\\
\frac{\partial}{\partial x}[\cdots]&=\frac{\partial}{\partial x}[\cdots]\\\\
-\frac{1}{x^2}f\left(\frac{1}{x}\right)-\frac{1}{x^3}f'\left(\frac{1}{x}\right)&=f(x)+xf'(x)
\end{align*}\]
Denoting \(\alpha=f(x)\) and \(\beta=f\left(\dfrac{1}{x}\right)\), we have the system
\[\begin{cases}\begin{align*}
-x^2\alpha-x^3\alpha'&=\beta+\dfrac{1}{x}\beta'\\\\
\alpha+x\alpha'&=-\dfrac{1}{x^2}\beta-\dfrac{1}{x^3}\beta'\end{align*}
\end{cases}\]

Answer 24

Hmm maybe not, the derivatives aren't the same in both equations...

Answer 25

There is an interesting level of symmetry going on here, but I can't seem to figure out how to reason out the correct answer.

@hartnn The reason it seems like that is because I'm just talking about a similar problem to what we want to solve would look like. I'm not really sure but it is interesting to play with.

Answer 26

Actually I just noticed. Suppose that instead f(x)=1/x is only part of the answer, then we can say: \[\Large f(x)=\frac{1}{x}+g(x)\] plug it into the equation \[\Large xf(x)=\frac{1}{x}f(\frac{1}{x})\]and we get: \[\Large x[ \frac{1}{x}+g(x)]=\frac{1}{x}[x+g(\frac{1}{x})]\] which simplifies down to \[\Large xg(x)=\frac{1}{x}g(\frac{1}{x})\] which has the same solution, so we can do this an arbitrary number of times. \[\Large f(x)=C \frac{1}{x}\] So it appears as though this is a more general solution. But is it the most general solution?

Answer 27

I think the thing to notice here is that \(f\) has to satisfy symmetry across the line \(y=x\). I don't think anything but \(\dfrac{C}{x}\) satisfies this. You could use some crazy non-function though, but then you'd have to redefine the definite integral.