- #36

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even for x neighbouring to ##\xi## , ##2^np_n(x)## is totally different than ##2^n(\xi \pm \epsilon)^n##.

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- Thread starter jaus tail
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- #36

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- #37

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I'm not saying it won't work, only that it requires more care. It's not obvious, at least to me, why it will work. I assume that some topological argument might be needed to make the step from iterations to the continuum. I would prefer a theorem before cases, but that's a personal view.even for x neighbouring to ##\xi## , ##2^np_n(x)## is totally different than ##2^n(\xi+-\epsilon)^n##. I don't think the limit ##\lim_{x->\xi}p_n(x)## for x any but not ##\xi## is equal to ##2^n(\xi+-\epsilon)^n##.

- #38

WWGD

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If you are using that to approximate ##f ## in ##[0,1]## then the limit should be finite, as continuous function in compact is bounded. But then you have a limit of the type ##\infty . \epsilon ## which should be dominated by ##\epsilon ## in order for the limit to be finite.

- #39

WWGD

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- #40

StoneTemplePython

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Another approach would be to show that for any any ##x \in (-1,0)##, after a large enough finite number of iterations, the resulting x becomes 'trapped' in ##\{(-1, 0.99], [0.01, 0)\}## and with respect to the magnitude of the function the effect of the x scaling over two iterations is a multiple less than one fourth, which overwhelms the scaling of ##2^2 = 4##, and hence the infinite product decays toward zero. This can be shown for 3 iterations at a time as well. It's sufficient to show this for 10 iterations for starting x = 0.5.

With respect to the ##2x## factor. Another way to think about it is how it relates to ##GM \leq AM## where the inequality is strict unless all values are the same. If we wanted to have some fun, look at the magnitude of the scalars ##\big \vert \approx 0\big \vert, \big \vert \approx -1\big \vert##, and see that the arithmetic mean is ##\frac{1}{2}## which is strictly larger than the geometric mean. Hence the effect of scaling by ##x## in our infinite product must overwhelm the scaling by 2. The infinite product decays toward zero.

- #41

jaus tail

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I tried using derivatives.

f(x) = 2x(f(x

Let f(x^2 - 1) = g(x)

So

f(x) = 2x g(x)

f'(x) = 2xg'(x) + 2g(x)

f'(x) = 2xf'(x)(2x) + 2f(x^2 - 1)

f'(x) (1-4x^2) = 2f(x^2 - 1)

So we get f'(x) = 2f(x^2 - 1)/(1-4x^2)

Now we got that f(-1) = f(1) = 0.

So if f(x) is non zero at any point between 1 and -1, then f'(x) must be positive some place and negative some place.

But above we see that f'(x) is an even function.

So f'(x) is not satisfying condition of positive and negative. It's either positive or negative and if it's one of them then it can have only 1 zero.

And so we can say that f(x) is constant at zero.

Aren't all continuous functions derivative? I'm not sure what's the difference. Sine and consine are continous and differentiable. So i guess it should be true.

- #42

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Aren't all continuous functions derivative? I'm not sure what's the difference. Sine and consine are continous and differentiable. So i guess it should be true.

Not all continuous functions are differentiable. One example is the modulus ##|x|##, which is not differentiable at ##x = 0##.

There is, in fact, the Weierstrass function, which is continuous but nowhere differentiable!

https://en.wikipedia.org/wiki/Weierstrass_function

- #43

jaus tail

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1) f(x) is differentiable then we can go with my solution.

and 2) f(x) is not differentiable. Any idea how to prove in this case?

- #44

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We've only been taught of limits and derivatives and integration.

I tried using derivatives.

f(x) = 2x(f(x^{2}-a))

Let f(x^2 - 1) = g(x)

So

f(x) = 2x g(x)

f'(x) = 2xg'(x) + 2g(x)

f'(x) = 2xf'(x)(2x) + 2f(x^2 - 1)

f'(x) (1-4x^2) = 2f(x^2 - 1)

So we get f'(x) = 2f(x^2 - 1)/(1-4x^2)

Now we got that f(-1) = f(1) = 0.

So if f(x) is non zero at any point between 1 and -1, then f'(x) must be positive some place and negative some place.

But above we see that f'(x) is an even function.

So f'(x) is not satisfying condition of positive and negative. It's either positive or negative and if it's one of them then it can have only 1 zero.

And so we can say that f(x) is constant at zero.

First:

##g'(x) = 2xf'(x^2-1) \ne 2xf'(x)##

But also, I don't understand how you get that an even function can't take positive and negative values.

I wonder where you got this question, because it seems to be a little beyond your level at the moment.

- #45

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1) f(x) is differentiable then we can go with my solution.

and 2) f(x) is not differentiable. Any idea how to prove in this case?

2) f(x) is not differentiable. Any idea how to prove in this case?

See @Ray Vickson's solution in post #14.

- #46

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If you are not allowed to use integral and derivatives then you can follow my approach explained in post #15, #20 and #28 which I believe is also correct.

- #47

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If you are using that to approximate ##f ## in ##[0,1]## then the limit should be finite, as continuous function in compact is bounded. But then you have a limit of the type ##\infty . \epsilon ## which should be dominated by ##\epsilon ## in order for the limit to be finite.

##2^np_n(x)## is dominated by ##p_n(x)## for any ##x\neq\xi## its ##\lim_{n->\infty}{p_n(x)}=0## and the convergence of ##p_n(x)## to 0 is such that it overcomes ##2^n##. I believe I prove this at post #28.

- #48

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##2^np_n(x)## is dominated by ##p_n(x)## for any ##x\neq\xi## its ##\lim_{n->\infty}{p_n(x)}=0## and the convergence of ##p_n(x)## to 0 is such that it overcomes ##2^n##. I believe I prove this at post #28.

If you take an interval close enough to -1 then you can generate a sequence of function values of increasing magnitude without bound in that interval. Which contradicts continuity, unless the function is zero on the interval. After that your argument simplifies for other points.

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