Functional equation problem

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  • #1
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Homework Statement


Let ##f:R^+ \rightarrow R## be a strictly increasing function such that ##f(x) > -\frac{1}{x} \, \forall \,x>0## and ##\displaystyle f(x)f\left(f(x)+\frac{1}{x}\right)=1 \, \forall \, x>0##. Find

a. f(1)
b. Maximum value of f(x) in [1,2]
c. Minimum value of f(x) in [1,2]

Homework Equations





The Attempt at a Solution


Honestly, I don't know where to begin with. I think I need to somehow find f(x) but I don't see how. I am completely clueless. :(

Part b and c are easy once I have f(x).
 

Answers and Replies

  • #2
Curious3141
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Homework Statement


Let ##f:R^+ \rightarrow R## be a strictly increasing function such that ##f(x) > -\frac{1}{x} \, \forall \,x>0## and ##\displaystyle f(x)f\left(f(x)+\frac{1}{x}\right)=1 \, \forall \, x>0##. Find

a. f(1)
b. Maximum value of f(x) in [1,2]
c. Minimum value of f(x) in [1,2]

Homework Equations





The Attempt at a Solution


Honestly, I don't know where to begin with. I think I need to somehow find f(x) but I don't see how. I am completely clueless. :(

Part b and c are easy once I have f(x).
Try putting ##f(x) = \frac{k}{x}##. You'll get two solutions for k, one positive, the other negative.

Only one of these fulfills the parameters of the question (strictly increasing over the given domain).

By the way, note that the values you get for k are rather special, in fact you might even say they're worth their weight in gold. :wink:
 
  • #3
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Try putting ##f(x) = \frac{k}{x}##. You'll get two solutions for k, one positive, the other negative.

Only one of these fulfills the parameters of the question (strictly increasing over the given domain).

By the way, note that the values you get for k are rather special, in fact you might even say they're worth their weight in gold. :wink:
Since f(x)=k/x,
$$\frac{k}{x}f\left(\frac{k+1}{x}\right)=1 \Rightarrow \frac{k^2}{k+1}=1$$
Solving for k,
$$k=\frac{1±\sqrt{5}}{2}$$
Since f(x) is strictly increasing, therefore
$$f(x)=\frac{1-\sqrt{5}}{2x}$$
For b part, max is at x=2 and for c, minimum is at x=1.

But I still don't see what's special about these values of k. And how did you even think of f(x)=k/x? :confused:
 
  • #4
Curious3141
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Since f(x)=k/x,
$$\frac{k}{x}f\left(\frac{k+1}{x}\right)=1 \Rightarrow \frac{k^2}{k+1}=1$$
Solving for k,
$$k=\frac{1±\sqrt{5}}{2}$$
Since f(x) is strictly increasing, therefore
$$f(x)=\frac{1-\sqrt{5}}{2x}$$
For b part, max is at x=2 and for c, minimum is at x=1.

But I still don't see what's special about these values of k. And how did you even think of f(x)=k/x? :confused:
https://en.wikipedia.org/wiki/Golden_ratio

Mainly, it was an inspired guess, although there are some strong indicators of the form of the function.

If you manipulate the functional equation setting ##f(x) = y##, and ##x = f^{-1}(y) = g(y)##, you'll ultimately be able to come up with:

$$g(y)[g(\frac{1}{y}) - y] = 1$$

(call this equation 1)

This strongly suggests that ##g## has to be an algebraic function (as opposed to a transcendental one). This means that its inverse, ##f##, also has to be algebraic. Of course, we cannot restrict ourselves to polynomials alone, so the field is still wide, but at least we know the ballpark.

We try putting ##g(y) = ky^n## into equation 1 as a first-go. ##n## can be any real number. I'm working with equation 1 as opposed to the original functional definition because the algebra is a LOT easier. From this, we immediately get:

$$k^2 - ky^{n+1} = 1$$

Now, we're given that this holds for all values of ##y## in the range (since the original functional definition applies to all values of ##x## in the domain). Since the RHS is a constant, the LHS has to be independent of ##y##, which means the only admissible value of ##n## is ##-1##. From that, we can quickly solve for ##k## and get the definition for ##g##. It's trivial to get ##f## from this (the function is a self-inverse).

Our simple "first-go" trial worked out, so we need to seek no further. But if it hadn't, we'd have to try more complicated forms, of course.
 
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  • #5
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Nice post Curious, thanks a lot! :)

I did try to use the inverse in the examination and I guess I even reached that equation 1 but I couldn't analyse the way you did. Nicely done, thank you. :smile:
 
  • #6
Curious3141
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Nice post Curious, thanks a lot! :)

I did try to use the inverse in the examination and I guess I even reached that equation 1 but I couldn't analyse the way you did. Nicely done, thank you. :smile:
You're most welcome, as always.

I hope the exam went fine in general?
 
  • #7
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I hope the exam went fine in general?
It was a small test, nothing too much to worry about. The test paper is to be discussed tomorrow (Monday). There were 2-3 problems I couldn't do and I found this problem to be interesting. I couldn't resist myself for tomorrow so I posted it here. :)
 
  • #8
Curious3141
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It was a small test, nothing too much to worry about. The test paper is to be discussed tomorrow (Monday). There were 2-3 problems I couldn't do and I found this problem to be interesting. I couldn't resist myself for tomorrow so I posted it here. :)
Ah, that's good. BTW, I'd be interested to know if your teacher's approach is different (more systematic).
 
  • #9
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Ah, that's good. BTW, I'd be interested to know if your teacher's approach is different (more systematic).
Sure, I will let you know once its discussed. :)
 
  • #10
haruspex
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I would not have thought it was acceptable to assume f is continuous, let alone differentiable or algebraic. And we cannot assume there is a unique function satisfying the conditions, so finding an algebraic solution is not in itself sufficient. It may even be the case that f is not uniquely determined, yet the specific questions have unique answers.
Maybe f can be proved to be differentiable. It's not hard to show that 0 < f(x+dx) - f(x) < dx/(x(x+dx)).
 
  • #11
Curious3141
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Haruspex is right. I wasn't thinking straight, and there are unjustifiable assumptions in my method.

There's a more elegant method to find f(1) without finding f(x).

Start with setting f(1) = F (a constant).

Then F.f(F+1) = 1

f(F+1) = 1/F

Using g as the inverse of f, as per my previous post.

g(1/F) = F + 1

Use the relationship I derived before, setting y = F:

g(1/F) = 1/[g(F) - (1/F)] = 1/[1 - (1/F)

Equate the two, simplify,

F^2 - F - 1 = 0

There are two roots here (again related to the golden ratio). To justify discarding the positive root, we need to establish that f(x) is negative throughout the domain. I remember I had a proof for this in my working, but I needed to assume that f(x) is differentiable (so I can apply product and chain rule to the functional definition). So some assumptions are definitely needed, IMO.

I need to get to work, so I'll follow up on this later.
 
  • #12
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Haruspex is right. I wasn't thinking straight, and there are unjustifiable assumptions in my method.

There's a more elegant method to find f(1) without finding f(x).

Start with setting f(1) = F (a constant).

Then F.f(F+1) = 1

f(F+1) = 1/F

Using g as the inverse of f, as per my previous post.

g(1/F) = F + 1

Use the relationship I derived before, setting y = F:

g(1/F) = 1/[g(F) - (1/F)] = 1/[1 - (1/F)

Equate the two, simplify,

F^2 - F - 1 = 0

There are two roots here (again related to the golden ratio). To justify discarding the positive root, we need to establish that f(x) is negative throughout the domain. I remember I had a proof for this in my working, but I needed to assume that f(x) is differentiable (so I can apply product and chain rule to the functional definition). So some assumptions are definitely needed, IMO.

I need to get to work, so I'll follow up on this later.
Sorry, I am late, I couldn't access internet on Monday.

I don't exactly remember what my teacher did but it was along the following lines:
Substitute f(x)+1/x=y
Then f(y)=1/f(x).
Also,
f(y)f(f(y)+1/y)=1
$$\Rightarrow f\left(\frac{1}{f(x)}+\cfrac{1}{f(x)+\cfrac{1}{x}}\right)=f(x)$$
My teacher said if f(a)=f(b) and the function is strictly increasing, then a=b.
$$\frac{1}{f(x)}+\cfrac{1}{f(x)+\cfrac{1}{x}}=x$$
Solving for f(x) and using the fact that f(x) is strictly increasing,
$$f(x)=\frac{1-\sqrt{5}}{2x}$$
Is this method correct?
 
  • #13
haruspex
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Sorry, I am late, I couldn't access internet on Monday.

I don't exactly remember what my teacher did but it was along the following lines:
Substitute f(x)+1/x=y
Then f(y)=1/f(x).
Also,
f(y)f(f(y)+1/y)=1
$$\Rightarrow f\left(\frac{1}{f(x)}+\cfrac{1}{f(x)+\cfrac{1}{x}}\right)=f(x)$$
My teacher said if f(a)=f(b) and the function is strictly increasing, then a=b.
$$\frac{1}{f(x)}+\cfrac{1}{f(x)+\cfrac{1}{x}}=x$$
Solving for f(x) and using the fact that f(x) is strictly increasing,
$$f(x)=\frac{1-\sqrt{5}}{2x}$$
Is this method correct?
It is essentially the same as Curious3141's method. It just looks more cumbersome.
 
  • #14
Curious3141
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It is essentially the same as Curious3141's method. It just looks more cumbersome.
No, this is better because the functional equation f(x) is actually derived, rather than just getting f(1) (as per my second attempt) or assuming and then verifying an algebraic form k/x (as per my first attempt). I'd say this derivation is sound because it assumed no more than what's given in the question.

When I saw the question, I almost immediately "knew" intuitively that the function had to involve the golden ratio in some way. However, I couldn't quite get to a rigorous proof. In Math, inspiration and intuition are nice, but rigour is critical.
 
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  • #15
haruspex
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No, this is better because the functional equation f(x) is actually derived,
Sorry, you're right. For some reason I misread it as solving for a specific x.
 

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