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 ~.##