A question that give wrong answer?

  • Thread starter Thread starter vkash
  • Start date Start date
  • #51
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?
 
Last edited:
Physics news on Phys.org
  • #52
"By the way, what is log₁? What does the subscript '1' do? Do you just mean complex natural log?"

They are different branches of the complex logarithm.

"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."

Find B²-4AC<0 for BOTH quadratics in order to find an interval for `a` for each case, then find the intersection of the two intervals for `a` obtained.
 
  • #53
:redface:
 
  • #54
matphysik said:
"By the way, what is log₁? What does the subscript '1' do? Do you just mean complex natural log?"

They are different branches of the complex logarithm.

Oh, I'd never seen that notation before. Thanks.

matphysik said:
Find B²-4AC<0 for BOTH quadratics in order to find an interval for `a` for each case, then find the intersection of the two intervals for `a` obtained.

I think you have a clever idea to use complex analysis to attack this problem. It's clear that the discriminants of both the top and bottom polynomials are the same, namely 32a + 36; and that this is negative exactly whenever a < -9/8.

What you've found, assuming the rest of your proof holds up, are functions from the complex numbers that are onto the reals. Unfortunately the original problems specifies f:R -> R.

Now you are making the leap to say that restricting the domain to R preserves surjectivity. But this is demonstrably false, since if you take a = -2 you get a rational function that has two complex roots and no real ones. That means it doesn't hit 0. So you've lost surjectivity when you restrict the domain to the reals.

I already presented this example earlier and you have not commented on it.

I also gave the example of f(z) = z^2, which is surjective when considered as a function from C to C; but is not surjective onto the reals when the domain is restricted to R.

I've just said the exact things as I did earlier. One, there is a clear counterexample to your claim that {x : discriminant < 0} gives a solution to the problem. And two, that your ambitious attempt to use complex analysis is a good idea, and DOES produce a surjection C -> R. But it does not produce a surjection R -> R.
matphysik said:
:redface:

Back atcha :smile:
 
Last edited:
  • #55
SteveL27 said:
Oh, I'd never seen that notation before. Thanks.



I think you have a clever idea to use complex analysis to attack this problem. It's clear that the discriminants of both the top and bottom polynomials are the same, namely 32a + 36; and that this is negative exactly whenever a < -9/8.

What you've found, assuming the rest of your proof holds up, are functions from the complex numbers that are onto the reals. Unfortunately the original problems specifies f:R -> R.

Now you are making the leap to say that restricting the domain to R preserves surjectivity. But this is demonstrably false, since if you take a = -2 you get a rational function that has two complex roots and no real ones. That means it doesn't hit 0. So you've lost surjectivity when you restrict the domain to the reals.

I already presented this example earlier and you have not commented on it.

I also gave the example of f(z) = z^2, which is surjective when considered as a function from C to C; but is not surjective onto the reals when the domain is restricted to R.

I've just said the exact things as I did earlier. One, there is a clear counterexample to your claim that {x : discriminant < 0} gives a solution to the problem. And two, that your ambitious attempt to use complex analysis is a good idea, and DOES produce a surjection C -> R. But it does not produce a surjection R -> R.




Back atcha :smile:

Lol, Yeah. I`ve never really thought deeply about a question like this. Even though at first glance it seems simple. It`s not.

The plan put forth by `HallsofIvy` is NOT good either. For you end up with a quadratic in `a` without any meaningful conditions imposed upon it.

I tried (unsuccessfully) to use the well known properties of the complex log to make some progress indirectly. But what i wrote are just attempts, and not a polished end result. Oh well.
 
  • #56
vkash said:
I also think so.
Actually my main motto for this post was to know why method written by HallsOfIvy is wrong and now i got that.
I does not post it(real answer and my question) directly in first post because i want to know is there any mathematical dude here who think so. unfortunately no such dude come here to answer.

NO. What you`re doing is giving up when the going gets tough.:-p
 
  • #57
matphysik said:
NO. What you`re doing is giving up when the going gets tough.:-p
Do you not read micromass 13th post. He explain it.
micromass said:
HoI's method is perfectly correct. His method will give that the function f is onto if a≥2. However, since the function must be entire, it must also hold that a+6x−8x2 has no roots. But no a≥2 satisfies this. So there are no solutions.
You are little bit correct that i am giving up. Mainly because of around nil knowledge of complex numbers.
 
  • #58
RandomMystery said:
Oh, I think I know what it is know. Isn't it weird how there is an -8 on top where there is an a in the bottom and vise versa? You simply put in a=-8 and the answer is always 1!

((-8)*x^2+6x-8) / ((-8)+6x-8*x^2)=1

-8 is smaller than -9/8 so the calculus was true.
I don't quite get what the question is though.

NO. Then f will take ℝ onto {1}. We want f to take ℝ onto ℝ. As a matter of fact, for a=-8 we have f(x)≡1.
 
  • #59
Calculus proof of that such an a does not exist:
First of all, the denominator will have to be non-zero for all x. Hence the discriminant Delta = 6^2-4(-8)a = 36+32a will have to negative, i.e. a < -36/32 = -9/8.

For such a the function is well-defined. Now consider \lim_{x \to \infty} f(x) = \lim_{x \to \infty} \frac{ax^2+6x-8}{a+6x-8x^2} = \lim_{x \to \infty} \frac{ax^2+6x-8}{a+6x-8x^2} = \lim_{x \to \infty} \frac{a+6\frac{1}{x}-8\frac{1}{x^2}}{a\frac{1}{x^2}+6\frac{1}{x}-8} = \frac{a}{-8}.

Similarly \lim_{x \to -\infty} f(x) = \frac{a}{-8}.

So there are real numbers A and B such that |f(x)-\frac{a}{-8}| &lt; 1 for all x > A, and |f(x)-\frac{a}{-8}| &lt; 1 for all x < B, and we can assume that A > B.
i
f(x) on [B,A] is continuous and thus bounded, so there is an M such that |f(x)|<M for all x in [A,B]. Hence |f(x)| &lt; \max(1+|\frac{a}{-8}|,M) for all x in (-\infty,B) \cup [B,A] \cup (A,\infty) = \mathbb{R}, and is thus not surjective.
 
  • #60
disregardthat said:
Calculus proof of that such an a does not exist:
First of all, the denominator will have to be non-zero for all x. Hence the discriminant Delta = 6^2-4(-8)a = 36+32a will have to negative, i.e. a < -36/32 = -9/8.

For such a the function is well-defined. Now consider \lim_{x \to \infty} f(x) = \lim_{x \to \infty} \frac{ax^2+6x-8}{a+6x-8x^2} = \lim_{x \to \infty} \frac{ax^2+6x-8}{a+6x-8x^2} = \lim_{x \to \infty} \frac{a+6\frac{1}{x}-8\frac{1}{x^2}}{a\frac{1}{x^2}+6\frac{1}{x}-8} = \frac{a}{-8}.

Similarly \lim_{x \to -\infty} f(x) = \frac{a}{-8}.

So there are real numbers A and B such that |f(x)-\frac{a}{-8}| &lt; 1 for all x > A, and |f(x)-\frac{a}{-8}| &lt; 1 for all x < B, and we can assume that A > B.
i
f(x) on [B,A] is continuous and thus bounded, so there is an M such that |f(x)|<M for all x in [A,B]. Hence |f(x)| &lt; \max(1+|\frac{a}{-8}|,M) for all x in (-\infty,B) \cup [B,A] \cup (A,\infty) = \mathbb{R}, and is thus not surjective.

Your proof looks good!:approve:
 
  • #61
disregardthat said:
Calculus proof of that such an a does not exist:
First of all, the denominator will have to be non-zero for all x. Hence the discriminant Delta = 6^2-4(-8)a = 36+32a will have to negative, i.e. a < -36/32 = -9/8.

For such a the function is well-defined. Now consider \lim_{x \to \infty} f(x) = \lim_{x \to \infty} \frac{ax^2+6x-8}{a+6x-8x^2} = \lim_{x \to \infty} \frac{ax^2+6x-8}{a+6x-8x^2} = \lim_{x \to \infty} \frac{a+6\frac{1}{x}-8\frac{1}{x^2}}{a\frac{1}{x^2}+6\frac{1}{x}-8} = \frac{a}{-8}.

Similarly \lim_{x \to -\infty} f(x) = \frac{a}{-8}.

So there are real numbers A and B such that |f(x)-\frac{a}{-8}| &lt; 1 for all x > A, and |f(x)-\frac{a}{-8}| &lt; 1 for all x < B, and we can assume that A > B.
i
f(x) on [B,A] is continuous and thus bounded, so there is an M such that |f(x)|<M for all x in [A,B]. Hence |f(x)| &lt; \max(1+|\frac{a}{-8}|,M) for all x in (-\infty,B) \cup [B,A] \cup (A,\infty) = \mathbb{R}, and is thus not surjective.


Given, f(x)= (ax² + 6x - 8)/(a + 6x - 8x²). We are asked to find `a` s.t. f is onto (surjective). Set, (Ax² + Bx + C)(a + 6x - 8x²) = ax² + 6x - 8, for {A,B,C} to be determined. Expanding and collecting terms gives:

Aa + 6B - 8C = a
-6A - 8B = 0
A=0 ⇒ B=0
Ba + 6C = 6 ⇒ C=1
Ca = -8 ⇒ a=-8.

There is, in fact, only one possible value of `a`, and the given f cannot take ℝ onto ℝ.

Thus, f(x) ≡ 1. Viz., f: ℝ→{1}⊂ℝ. The lesson to be learned is that the given problem statement is ambiguous, and misleading.
 
  • #62
mathphysics, what you have proved is that in order for f(x) to be a polynomial a must be -8. But the question is whether f(x) is surjective.
 
  • #63
NO. What i did was divide the numerator quadratic by the denominator quadratic directly, in order to reveal the true identity of f(x).

ADDENDUM:

If we should assume the simplest non-polynomial in this case, say:

(Ax² + Bx + C + D/x)(a + 6x - 8x²) = ax² + 6x - 8.

Then expanding, and collecting terms gives:

A=0
-8B + 6A=0 ⇒ B=0
-8C + 6B + Aa=a ⇒ C=a/-8
-8D + 6C + Ba=6
6D + Ca=-8
6Da + Ca²=-8a ⇒ C= -8/a
Da=0 ⇒ D=0
-8Da + 6Ca=6a ⇒ C=1 ⇒ a= -8.
 
Last edited:
  • #64
disregardthat said:
Calculus proof of that such an a does not exist:
First of all, the denominator will have to be non-zero for all x. Hence the discriminant Delta = 6^2-4(-8)a = 36+32a will have to negative, i.e. a < -36/32 = -9/8.

For such a the function is well-defined. Now consider \lim_{x \to \infty} f(x) = \lim_{x \to \infty} \frac{ax^2+6x-8}{a+6x-8x^2} = \lim_{x \to \infty} \frac{ax^2+6x-8}{a+6x-8x^2} = \lim_{x \to \infty} \frac{a+6\frac{1}{x}-8\frac{1}{x^2}}{a\frac{1}{x^2}+6\frac{1}{x}-8} = \frac{a}{-8}.

Similarly \lim_{x \to -\infty} f(x) = \frac{a}{-8}.

So there are real numbers A and B such that |f(x)-\frac{a}{-8}| &lt; 1 for all x > A, and |f(x)-\frac{a}{-8}| &lt; 1 for all x < B, and we can assume that A > B.
i
f(x) on [B,A] is continuous and thus bounded, so there is an M such that |f(x)|<M for all x in [A,B]. Hence |f(x)| &lt; \max(1+|\frac{a}{-8}|,M) for all x in (-\infty,B) \cup [B,A] \cup (A,\infty) = \mathbb{R}, and is thus not surjective.
I love this proof.
 
  • #65
matphysik said:
NO. What i did was divide the numerator quadratic by the denominator quadratic directly, in order to reveal the true identity of f(x).

And it turns out to be ... wealthy playboy Bruce Wayne!
 
  • #66
disregardthat said:
Calculus proof of that such an a does not exist:
First of all, the denominator will have to be non-zero for all x. Hence the discriminant Delta = 6^2-4(-8)a = 36+32a will have to negative, i.e. a < -36/32 = -9/8.

For such a the function is well-defined. Now consider \lim_{x \to \infty} f(x) = \lim_{x \to \infty} \frac{ax^2+6x-8}{a+6x-8x^2} = \lim_{x \to \infty} \frac{ax^2+6x-8}{a+6x-8x^2} = \lim_{x \to \infty} \frac{a+6\frac{1}{x}-8\frac{1}{x^2}}{a\frac{1}{x^2}+6\frac{1}{x}-8} = \frac{a}{-8}.

Similarly \lim_{x \to -\infty} f(x) = \frac{a}{-8}.

So there are real numbers A and B such that |f(x)-\frac{a}{-8}| &lt; 1 for all x > A, and |f(x)-\frac{a}{-8}| &lt; 1 for all x < B, and we can assume that A > B.
i
f(x) on [B,A] is continuous and thus bounded, so there is an M such that |f(x)|<M for all x in [A,B]. Hence |f(x)| &lt; \max(1+|\frac{a}{-8}|,M) for all x in (-\infty,B) \cup [B,A] \cup (A,\infty) = \mathbb{R}, and is thus not surjective.

Why can't we make a equal some function?
a = -8x
\frac{a}{-8}=\frac{-8x}{-8}
Here the function is globally onto.

I don't understand why you choose y=0 in the discriminant:
Delta = 6^2-4(-8)a
instead of
Delta = (6 - 6y)^2 - 4(a + 8y)(-8 - ay)





RandomMystery said:
I think you can still use the quadratic equation:

gif.latex?x=-\frac{(6-6y)\pm&space;\sqrt{(6-6y)^2-4(a&plus;8y)(-8-ay)}}{2(a+8y)}.gif

We can either replace a for one function that solves the dilemma, or we could make a into a piece wise value (or function).
The question says find the value of a, well the value of a could be different for each x
 
  • #67
Why can't we make a equal some function?
a = -8x
\frac{a}{-8}=\frac{-8x}{-8}
Here the function is globally onto.
I don't think there is anything stopping someone from making a into a function. However, the OP seems to have made it clear later on that a does not depend on x.

I don't understand why you choose y=0 in the discriminant:
Delta = 6^2-4(-8)a
instead of
Delta = (6 - 6y)^2 - 4(a + 8y)(-8 - ay)
The OP states that the domain of f is R. For this to be true, f must not contain points of discontinuity. We can see that f will have such points of discontinuity whenever the denominator is equal to zero. The only way to prevent the denominator from equaling zero is to choose a value of a such that quadratic has no real roots. Such an a must make the discriminant less than zero.

I think your confusion is coming from the fact that DisregardThat is not using the method suggested by HallsOfIvy.
 

Similar threads

Replies
2
Views
2K
Replies
47
Views
8K
Replies
1
Views
422
Replies
7
Views
2K
Replies
1
Views
594
Replies
1
Views
2K
Back
Top