Let's Solve A Functional Equation | 2nd Method?

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[titan]

My friend (I still don't know your real name, Sybermath 😢), I absolutely LOVE your videos; and you're such a talented mathematician. Please, my friend, keep it up. THANK YOU! ❤

[saishashank]

Put "-x" in place of "x"
Now, add both equations..
You get,
f(x+√{x²-1}) = - f(-x+√{x²-1})
=> x+√{x²-1} = x-√{x²-1}
=> x²-1=0
=> x=±1

[NadiehFan]

For anyone familiar with hyperbolic functions and their inverses the result was quite expected because we have

cosh(u) = (eᵘ + e⁻ᵘ)/2 = (e²ᵘ + 1)/(2eᵘ)

If we set cosh(u) = x and we want to find an expression for the inverse function arcosh(x) = u in terms of x we can set eᵘ = t which then gives (t² + 1)/(2t) = x or

t² − 2xt + 1 = 0

So t = x ± √(x² − 1). For the domain ℝ the range of the cosh function is [1, ∞) so the domain of arcosh is [1, ∞), that is, x ≥ 1. Both values of t = x ± √(x² − 1) are real and positive for x ≥ 1, so, since eᵘ = t implies u = ln t we get two values u = ln(x ± √(x² − 1)) which are each others opposite since the product of the roots of the quadratic in t is 1. This is because cosh is an even function and therefore not invertible if we take ℝ as its domain.

If we restrict the domain to [0, ∞) then cosh(u) = x with u ≥ 0 and then the function is invertible and we have u = ln(x + √(x² − 1)) and therefore

arcosh(x) = ln(x + √(x² − 1))

for x ≥ 1.

We could use this to solve the functional equation

f(x + √(x² − 1)) = x

in a single line, because if we set x + √(x² − 1) = t and t = eᵘ then x = cosh(u) = (e²ᵘ + 1)/(2eᵘ) = (t² + 1)/(2t) so f(t) = (t² + 1)/(2t). This, however, supposes that x ≥ 1.

Note that since t = x + √(x² − 1) and t = x − √(x² − 1) both imply x = (t² + 1)/(2t) the functional equation

f(x − √(x² − 1)) = x

has the same solution f(t) = (t² + 1)/(2t) but that for |x| ≥ 1, x + √(x² − 1) can only take values on [−1, 0) ∪ [1, ∞) and x − √(x² − 1) can only take values on (−∞, −1] ∪ (0, 1]. This is in agreement with the fact that, considered as a real function, f(t) = (t² + 1)/(2t) has the domain ℝ\{0} and the range (−∞, −1] ∪ [1, ∞).

If we have x ≤ −1 and we set x + √(x² − 1) = τ then τ is negative but then we have (−x) − √((−x)² − 1) = −τ = t where −x ≥ 1 and t is positive, so with t = eᵘ we then again have −x = cosh(u) = (e²ᵘ + 1)/(2eᵘ) = (t² + 1)/(2t) and therefore x = (t² + 1)/(−2t) = (τ² + 1)/(2τ) so f(τ) = (τ² + 1)/(2τ). This is a simple consequence of the fact that x + √(x² − 1) = t and x − √(x² − 1) = t both imply x = (t² + 1)/(2t).

[jimschneider]

One interesting point is that although the function is defined on the reals for all nonzero x, the argument of the function in the functional equation is complex for |x| < 1. In fact, digging a little deeper leads to the observation that if x is a complex root of unity, then f(x) will be a real number on the interval [ -1, 1 ].

Actually, a minor correction: if x is a complex root of unity with a positive imaginary part, f(x) will be a real number on the interval [ -1, 1 ]. I have not characterized the function for the other half of the group complex roots of unity.

[yoav]

Let x+sqrt(x^2-1)=u and x-sqrt(x^2-1)=z, and notice uz=1 so z=1/u. Now x=0.5(u+z)=0.5(u+1/u) so f(x)=0.5(x+1/x)😊

[cav]

Yu are excellent. I suppose that you work with a team because it is imposible for me to think that only one person can solve and produce a lot of problems, that are sometimes very, very complex.

[matniet]

You could also let x = cosθ and use Euler's formula

[robertcotton]

Domain of original was x could not be between -1 and 1

[FisicTrapella]

WA -> f(t+Sqrt[Power[t, 2]-1])=t solve f(t) for t=x+Sqrt[Power[x, 2]-1]
it works 🤗🤗

[yurenchu]

f(x + √(x² - 1)) = x

Let
y = x + √(x² - 1)
y - x = √(x² - 1)
(y - x)² = x² - 1
y² - 2xy + x² = x² - 1
y² - 2xy + 1 = 0
y² + 1 = 2xy
x = (y² + 1)/(2y)

==>
f(y) = (y²+1)/(2y) = ½(y + 1/y)

[sngmn]

Let x = ch(y)
f(x + sqrt(...) = f( ch(y) + sh(y) ) = f(e^y)
Let e^y = t => y = ln(t) => x = ch(ln(t)) = [e^ln(t) + e^-ln(t)] / 2 = t/2 + 1/2t
f(t) = t/2 + 1/2t

[honestadministrator]

Writing z = x + √ ( x^2 - 1)
z^2 - 2 z x + 1 = 0
x = z /2 + 1 /( 2 z)
f ( z) = (z + 1/z) /2