So, ##f(1) = -1~.##How can I find the value of f(1)

[YoungPhysicist]

Homework Statement

$$f(x+\frac{1}{x}) = x^2+\frac{1}{x^2}\\ f(1) = ?$$

Homework Equations

The Attempt at a Solution

I thought maybe I can find a $x$ that $x+\frac{1}{x} =1$ that I can substitute, but that $x$ is a imaginary number so I am not sure what should I do next.

 

[member 587159]

Just some quick random thoughts. This may get you started.

Did you find a function that satisfies the given condition?

$x^2 + \frac{1}{x^2} = (x+\frac{1}{x})^2 -2$

We have $f(x+\frac{1}{x}) = (x+\frac{1}{x})^2 -2$

and we can take $f(y) = y^2 - 2$ as an example and from this you can conjecture that $f(1) = -1$.

Now, this is one specific example. Maybe there are other functions left so there is still work left for you to do.

In fact, the condition $f(x+\frac{1}{x}) = x^2 + \frac{1}{x^2}$ determines the function $f$ uniquely on the set $\{\frac{1}{x}+x \}$ where $x$ runs over an appropriate domain, so you might want to determine this set.

 

[YoungPhysicist]

Just some quick random thoughts. This may get you started.

Did you find a function that satisfies the given condition?

$x^2 + \frac{1}{x^2} = (x+\frac{1}{x})^2 -2$

We have $f(x+\frac{1}{x}) = (x+\frac{1}{x})^2 -2$

and we can take $f(y) = y^2 - 2$ as an example and from this you can conjecture that $f(1) = -1$.

Thanks! That is the insight I need!

In fact, the condition f(x+1x)=x2+1x2f(x+1x)=x2+1x2f(x+\frac{1}{x}) = x^2 + \frac{1}{x^2} determines the function fff uniquely on the set {1x+1}{1x+1}\{\frac{1}{x}+1 \} so you might want to determine this set.

Do you mean making it the form of $f(x)$?

 

[member 587159]

Do you mean making it the form of $f(x)$?

Yes, from the condition $f(x+1/x) = x^2 + 1/x^2$, we want to go to something of the form $f(x)$. Fix an $y$ in the domain of $f$. If there exists $x$ such that $y = x+1/x$, then you know that $f(y) = x^2 + 1/x^2$ so this is what I mean with that the given condition implies that you know how $f$ behaves on the set $\{x+1/x\}$. I'm not sure if this is the right way to proceed as I didn't solve the exercice myself, but it might give you some clue to come to a solution.

 

[WWGD]

Homework Statement

$$f(x+\frac{1}{x}) = x^2+\frac{1}{x^2}\\ f(1) = ?$$

Homework Equations

The Attempt at a Solution

I thought maybe I can find a $x$ that $x+\frac{1}{x} =1$ that I can substitute, but that $x$ is a imaginary number so I am not sure what should I do next.

Can we assume f is continuous? Is it defined for x=0?EDIT: Is the domain the Reals? The range/codomain also Real?

 

[YoungPhysicist]

Can we assume f is continuous? Is it defined for x=0?

The problem didn’t specify that, but I think it is not defined at 0 since there are no exceptions written with the function.

 

[Mark44]

Is it defined for x=0?EDIT: Is the domain the Reals? The range/codomain also Real?

Pretty obviously, the function is not defined at x = 0.

The problem didn’t specify that, but I think it is not defined at 0 since there are no exceptions written with the function.

With no other information given, it's reasonable to assume that the domain is $\{x \in \mathbb R : x \ne 0 \}$.

 

[WWGD]

Pretty obviously, the function is not defined at x = 0.

With no other information given, it's reasonable to assume that the domain is $\{x \in \mathbb R : x \ne 0 \}$.

Not quite; there are often piece-wise definitions, e.g., sin(1/x) for $x\neq 0$ ; 0 when $x=0$. Specially when the discontinuity is removable, but not necessarily.

 

[Mark44]

Not quite; there are often piece-wise definitions, e.g., sin(1/x) for $x\neq 0$ ; 0 when $x=0$. Specially when the discontinuity is removable, but not necessarily.

Sure, but since no piecewise definition was given, it's not reasonable to assume that the function is defined at x = 0.

 

[Hiero]

I thought maybe I can find a $x$ that $x+\frac{1}{x} =1$ that I can substitute, but that $x$ is a imaginary number so I am not sure what should I do next.

That is a logical approach. Sure x will be imaginary, but you will end up squaring x everywhere anyway, so the answer will still come out real and correct.
[edit: not to imply that any F(x²)∈ℝ]

The approach of @Math_QED is far more elegant and should be learned from. But there is nothing wrong with your approach, except that you stopped.

 

[YoungPhysicist]

That is a logical approach. Sure x will be imaginary, but you will end up squaring x everywhere anyway, so the answer will still come out real and correct.

Oh right!

 

[SammyS]

The issue of Domain, etc. is pretty interesting for the formulation of the function being discussed in this thread.

The function, $f ~ ,$ is only given indirectly, and that's only by its composition with another function, let's call it $g ~ ,$ where $\ g(x) = x + \dfrac{1}{x} \, .$

Thanks to @Math_QED we see that $\ f(g(x)) = \left( g(x) \right)^2 -2 \, . \$ Furthermore, to use this idea to determine the value $f(1)\,, \$ all that seems to be required is to set $g(x)$ to be 1. However, there is no real number, $x$, for which $g(x)=1 ~ .$

If we consider $g$ to be a real function, then the domain of $g$ cannot include zero. (By the way: This has nothing to with the issue of 0 being in the domain of $f ~.$ ) The natural domain (a.k.a. implicit domain) of $g$ as a real function is ℝ\{0} . The image of $g$ for this domain is $( -\infty,\, -2] ~\cup ~ [ 2, \, \infty)$, (often referred to as the Range of $g$).

The Problem Statement asked for the value of $f(1)$, so I assume we need to extend the domain of $g$. Let's see if that's possible while keeping $f$ as being a real function, and Ideally have $f$ continue to be defined by $f(u) = u^2 - 2 ~ .$

Let's use $G$ to denote the extension of $g$. Furthermore, consider $x$ to now be complex with real part, $a ~ ,$ and imaginary part, $b~.$ I.e. let $x = a + bi ~.$ Finally, find the domain, D_G ⊂ ℂ, such that the Range of $G$ is real. $( G(D_G) \subset \mathbb{R} )\$

$G(a + bi) = a + bi +\dfrac{1}{a + bi}$

$= a + bi +\dfrac{a - bi}{a^2+b^2}$

$= \dfrac{a(a^2+b^2+1)}{a^2+b^2}+\dfrac{b(a^2+b^2-1)}{a^2+b^2}i$​

.
Thus, if $a^2 + b^2 =1 ~$, then $G(a + bi)$ is real and in fact $G(a + bi) = 2a ~.$

Of course, if $b = 0~ ,$ then $x$ is purely real and $G(x) = x + \dfrac{1}{x} \, ,\$ as desired.

With this extension of $g$, the function, $f$ is given by $f(u) = u^2 - 2 ~, \$ with a domain of $\mathbb{R} \,.$

Oh, by the way: If $x = \pm i,$ $G(x) = 0~,$ giving $f(0) = -2 ~.$

 

[WWGD]

The issue of Domain, etc. is pretty interesting for the formulation of the function being discussed in this thread.

The function, $f ~ ,$ is only given indirectly, and that's only by its composition with another function, let's call it $g ~ ,$ where $\ g(x) = x + \dfrac{1}{x} \, .$

Thanks to @Math_QED we see that $\ f(g(x)) = \left( g(x) \right)^2 -2 \, . \$ Furthermore, to use this idea to determine the value $f(1)\,, \$ all that seems to be required is to set $g(x)$ to be 1. However, there is no real number, $x$, for which $g(x)=1 ~ .$

If we consider $g$ to be a real function, then the domain of $g$ cannot include zero. (By the way: This has nothing to with the issue of 0 being in the domain of $f ~.$ ) The natural domain (a.k.a. implicit domain) of $g$ as a real function is ℝ\{0} . The image of $g$ for this domain is $( -\infty,\, -2] ~\cup ~ [ 2, \, \infty)$, (often referred to as the Range of $g$).

The Problem Statement asked for the value of $f(1)$, so I assume we need to extend the domain of $g$. Let's see if that's possible while keeping $f$ as being a real function, and Ideally have $f$ continue to be defined by $f(u) = u^2 - 2 ~ .$

Let's use $G$ to denote the extension of $g$. Furthermore, consider $x$ to now be complex with real part, $a ~ ,$ and imaginary part, $b~.$ I.e. let $x = a + bi ~.$ Finally, find the domain, D_G ⊂ ℂ, such that the Range of $G$ is real. $( G(D_G) \subset \mathbb{R} )\$

I agree with you, but I would believe functions would at least be assumed continuous ( of course, where defined, which is all one can do). Otherwise, I think it is difficult if not impossible to narrow things down enough.
$G(a + bi) = a + bi +\dfrac{1}{a + bi}$

$= a + bi +\dfrac{a - bi}{a^2+b^2}$

$= \dfrac{a(a^2+b^2+1)}{a^2+b^2}+\dfrac{b(a^2+b^2-1)}{a^2+b^2}i$​

.
Thus, if $a^2 + b^2 =1 ~$, then $G(a + bi)$ is real and in fact $G(a + bi) = 2a ~.$

Of course, if $b = 0~ ,$ then $x$ is purely real and $G(x) = x + \dfrac{1}{x} \, ,\$ as desired.

With this extension of $g$, the function, $f$ is given by $f(u) = u^2 - 2 ~, \$ with a domain of $\mathbb{R} \,.$

Oh, by the way: If $x = \pm i,$ $G(x) = 0~,$ giving $f(0) = -2 ~.$