SteveL27
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matphysik said:matphysik said:Hi. If the function is analytic, then it`s continuous.
CORRECTION:
f (as given) takes ℝ into ℝ so that, y=f(x)=Re{f(x)}. Let z↦log₁z take ℂ\(-∞,0] onto ℂ. Write, f(x)=(ax²+6x-8)/(a+6x-8x²)≡(x-A)(x-B)/(x-C)(x-D). For `a` to be determined such that f takes ℝ onto ℝ. ONE possibility for f(x) being onto occurs if ALL of {A,B,C,D} are complex numbers.Then,
Re[log₁f(x)]=Re[log₁(x-A)+...-log₁(x-D)] takes ℝ onto ℝ\(-∞,0]. Next, let z↦log₂z take ℂ\[0,∞) onto ℂ. Then, Re[log₂f(x)]=Re[log₂(x-A)+...-log₂(x-D)] takes ℝ onto ℝ\[0,∞). So that, given such `a` f will take ℝ onto ℝ\{0} (since, (-∞,0]∩[0,∞)={0}). Last, since Re{f(0)}=Re{(A)(B)/(C)(D)} we have proved (for this particular case) that f takes ℝ onto ℝ.
I'm sorry, I had a hard time following that. You used the phrase "for this particular case," but I don't see any particular case.
Consider f(z) = z^2. It's onto the complex numbers as a function of complex numbers; but its restriction to the reals is not onto the reals. Why not? Well, f(x) = x^2 does not hit -1, for example. If you take z = i, then f(z) = z^2 does hit -1. But Re(i) = 0 so you haven't really gotten any benefit out of taking the real part of a complex function.
Earlier you said that any a < -9/8 gives an onto function (with f as in the original problem); but I showed that with a = -2, the resulting rational function has two complex roots and no real ones, so as a function from the reals to the reals, it fails to hit 0 and is therefore not onto.
Do you have a specific 'a' that works?
(edit)
I think you're getting in trouble with y=f(x)=Re{f(x)}. It's true that you can take the real part of f(z) to get a function from C to R. But it doesn't follow that you now have a function from R to R that preserves the attributes you care about. You only have a function from C to R. When you restrict the domain to R, you lose surjectivity, as the example of f(z) = z^2 shows.
Another example is f(z) = e^z. We have f(i*pi) = -1, but the real part of i*pi is 0; and the the real part of -1 is -1. You can't freely interchange Re and f.
By the way, what is log₁? What does the subscript '1' do? Do you just mean complex natural log?
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