I'm lost: We look for functions such that verify f(x) + f(y) = f((x+y)/(1-xy))...

[Faizer]

Hi, idk how to solve 15.(a), if you can help me, thanks a lot.
Statement : In this part, we look for functions f defined and derivable on I (with I an open interval of R containing 0) that verify: f(x) + f(y) = f((x+y)/(1-xy))
Let f be a function that satisfies these hypotheses.
14.calculate f(0) (I found f(0)=0)
15. let, a∈I with a≠0
determine λ>0 as for all h∈[-λ;λ], the equation
a+h = (a+y)/(1-ay), (of unknown y) admits exactly one solution. This solution is denoted y(h). Then show that h->y(h) is derivable on [-λ;λ]. Calculate y(0) and y'(0)

I have several leads but nothing comes to fruition.
I first thought about solving this equation :
(a+h)(1-ay)=(a+y) which brought me back to a 2nd-degree equation, but I don't think that was very helpful (or idk how to deal with it)
Then, as the lim of (a+y)/(1-ay) is -1/a
So I thought it was sufficient to choose lambda small enough so that \[a - lambda, a + lambda] does not contain -1/a. But what lambda?
Or I thought as [-λ;λ] is '' symmetrical '' it contains 0 so we can choose h=0 but with this method I can't conclude about the possible value of lambda...
I'm stuck here, so if someone could help me out
(the goal of the problem is to find the arctan function so lambda should be equal to pi/2??)
Thanks

 

[stapel]

Statement : In this part, we look for functions f defined and derivable on I (with I an open interval of R containing 0) that verify: f(x) + f(y) = f((x+y)/(1-xy))
Let f be a function that satisfies these hypotheses.
14.calculate f(0) (I found f(0)=0)
15. let, a∈I with a≠0
determine λ>0 as for all h∈[-λ;λ], the equation
a+h = (a+y)/(1-ay), (of unknown y) admits exactly one solution. This solution is denoted y(h). Then show that h->y(h) is derivable on [-λ;λ]. Calculate y(0) and y'(0)

What do you mean by "derivable"? What do you mean by a function "verifying" the given form? Were you able to do anything with what you were given at another posting of this exercise?

 

[Dr.Peterson]

Statement : In this part, we look for functions f defined and derivable on I (with I an open interval of R containing 0) that verify: f(x) + f(y) = f((x+y)/(1-xy))

I would guess that "derivable" is your word for "differentiable". I have often wished that in English the verb corresponding to "derivative" were not "differentiate". "Derive" has a very different meaning, and the two words make things confusing.

I also suppose that "verify" here means something like "satisfy" (make the equation true).

I first thought about solving this equation :
(a+h)(1-ay)=(a+y) which brought me back to a 2nd-degree equation

It would be second degree in a, but aren't you supposed to solve for y? (If it were second degree, my first thought would be to use the discriminant.)

Just solve for y as a function of h, and find its domain.

It will help if you tell us your context: What have you learned, and where does the problem come from.

It may also help if you show us the original problem exactly as given to you (an image, even if it's not in English), so we can be sure you aren't missing something.

 

[Faizer]

Ah you're right, I'm French so I never expected it was differentiate for what I call '' to derivate '' the same for to satisfy, thanks.

I would guess that "derivable" is your word for "differentiable". I have often wished that in English the verb corresponding to "derivative" were not "differentiate". "Derive" has a very different meaning, and the two words make things confusing.

I also suppose that "verify" here means something like "satisfy" (make the equation true).


It would be second degree in a, but aren't you supposed to solve for y? (If it were second degree, my first thought would be to use the discriminant.)

Just solve for y as a function of h, and find its domain.

It will help if you tell us your context: What have you learned, and where does the problem come from.

It may also help if you show us the original problem exactly as given to you (an image, even if it's not in English), so we can be sure you aren't missing something.

Ah you're right, I'm French so I never expected it was differentiate for what I call '' to derivate '' the same for to satisfy, thanks !

So I found y(h) = h/(a²+ah+1). So y is defined for every h≠(-a²-1)/a.

For the problem it's the part D '' problème inverse ''

 

[Faizer]

Ah you're right, I'm French so I never expected it was differentiate for what I call '' to derivate '' the same for to satisfy, thanks.

Ah you're right, I'm French so I never expected it was differentiate for what I call '' to derivate '' the same for to satisfy, thanks !

So I found y(h) = h/(a²+ah+1). So y is defined for every h≠(-a²-1)/a.

For the problem it's the part D '' problème inverse ''

I can also show you parts A, B and C, but I'm not sure that's very useful.

 

[Dr.Peterson]

Thanks.

I can also show you parts A, B and C, but I'm not sure that's very useful.

It's possible that seeing more context might make it clearer where the author is going here. I can see how the answer to (b) relates to the final answer (which you hadn't shown), but I'm not quite seeing the next step.

 

[blamocur]

So y is defined for every h≠(-a²-1)/a.

I agree with that, but that's not an answer to the question 15, which asks to define the whole interval [imath][-\lambda,\lambda][/imath] in which every [imath]h[/imath] satisfies [imath]h\neq -\frac{a^2+1}{a}[/imath]

 

[Faizer]

Thanks.
View attachment 36712


It's possible that seeing more context might make it clearer where the author is going here. I can see how the answer to (b) relates to the final answer (which you hadn't shown), but I'm not quite seeing the next step.

This is the whole subject

 

[Faizer]

I agree with that, but that's not an answer to the question 15, which asks to define the whole interval [imath][-\lambda,\lambda][/imath] in which every [imath]h[/imath] satisfies [imath]h\neq -\frac{a^2+1}{a}[/imath]

Yes, that's where my problem lies, I can't quite get to grips with this result/to see how to find lambda

 

[Faizer]

Yes, that's where my problem lies, I can't quite get to grips with this result/to see how to find lambda

Hi, thanks to everyone of you. That was very helpful, I was able to make progress bu my only problem now is that I'm stuck again lol.
I've shown that f(0)=0 and f'(0)=1/(a^2+1)
Now I have to express f'(a) in terms of a and f'(0)
Knowing that f(x) + f(y) =f((x+y) /(1-xy)), I've thought about something like this :
f(a) + f(y) = f((a+y) /(1-ay))
Let g(a) = f((a+y) /(1-ay))
So g'(a) =f((a+y) /(1-ay))x(a(1-ay)-y(a+y)/(1-ay)^2)
g'(a) =((-a^2y-2ay+y^2)/(1-ay)^2) x f((a+y) /(1-ay))
And g'(a) = f'(a) (because g is also equal to f(a) + f(h))
Now idk whqt to do.
Maybe there's a simplification I don't see or maybe it's not the right way.
Sorry if I'm being redundant with this problem but if someone can help thanks!