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- Thread starter fridtern
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- Thread starter
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- Feb 13, 2012

- 1,704

http://mathhelpboards.com/analysis-50/never-ending-dispute-2060.htmlHi i'm having some problems with an task in my class.

It's as follows

Given the real function f(x)=X^{x}for X ∈ (0,∞).

Then i'm to expand the definition area to [0,∞), f shall still be continuous. What does f(0) have to be?

Thanks for any help

Kind regards

$\chi$ $\sigma$

- Mar 22, 2013

- 573

There is no analytic, or even meromorphic or even a well-defined extension of x^x at x = 0. You can calculate the neighborhood of that particular function around zero, however.

A well-defined AC on C\{0} can be obtained by choosing appropriate branches of complex logs for the function exp(x log(x)).

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- #4

- Feb 7, 2012

- 2,740

Hi fridtern, and welcome to MHB! This graph may help you :Hi i'm having some problems with an task in my class.

It's as follows

Given the real function f(x)=X^{x}for X ∈ (0,∞).

Then i'm to expand the definition area to [0,∞), f shall still be continuous. What does f(0) have to be?

[graph]x9fjteythg[/graph] (Click on the graph for an enlargement.)

Nobody is saying that the extension should be analytic or meromorphic. But the limit \(\displaystyle \lim_{x\searrow0}x^x\) clearly exists, and provides a continuous extension for $f(x)$ at $x=0.$

- Mar 22, 2013

- 573

As I have already indicated, the neighborhood can be used instead of that particular point instead, so indeed that stands.But the limit \(\displaystyle \lim_{x\searrow0}x^x\) clearly exists, and provides a continuous extension for $f(x)$ at $x=0.$

- Feb 13, 2012

- 1,704

I think that this graph, kindly supplied by 'Monster Wolfram' may help fridtern even more!...Hi fridtern, and welcome to MHB! This graph may help you :

[graph]x9fjteythg[/graph] (Click on the graph for an enlargement.)

Nobody is saying that the extension should be analytic or meromorphic. But the limit \(\displaystyle \lim_{x\searrow0}x^x\) clearly exists, and provides a continuous extension for $f(x)$ at $x=0.$

The basis to arrive to this result lies on the identity $\displaystyle x^{x} = e^{x\ \ln x}$ and, because the function $\displaystyle f(x) = x\ \ln x$ is defined for any value real or complex of x, the same is for $g(x)= e^{f(x)}$. The 'objection' regarding the value of f(x) in x=0 has been 'disproved' in my post 'A neved ending dispute...' with the following steps...

a) the Taylor expansion of $f(x)= x\ \ln x$ around x=1 is...

$\displaystyle f(x) = x\ \ln x = (x-1) + \sum_{n=2}^{\infty} (-1)^{n}\ \frac{(x-1)^{n}}{n\ (n-1)}\ (1)$

b) setting in (1) x=0 the series becomes...

$\displaystyle f(0) = -1 + 1 - \frac{1}{2} + \frac{1}{2} - \frac{1}{3} + ... \frac{1}{n-1} - \frac{1}{n} + ... = 0\ (2)$

An important point that has to remarked is that the (2) doesn't mean $\displaystyle \lim_{x \rightarrow 0} f(x) = 0$ because x remain 'fixed' at 0 and is n which tends to infinity... the (20) means simply $f(0) = 0$...

Kind regards

$\chi$ $\sigma$

Last edited:

- Mar 22, 2013

- 573

- Feb 13, 2012

- 1,704

I apologize for the fact that I badly reported a formula from my old notes... the correct formula is...

$\displaystyle f(x)= x\ \ln x = (x-1) + \sum_{n=2}^{\infty} (-1)^{n}\ \frac{(x-1)^{n}}{n\ (n-1)}\ (1)$

Sorry again! ...

Kind regards

$\chi$ $\sigma$

- Mar 22, 2013

- 573

Now, about the continuation, you cannot use the Taylor series to evaluate at x = 0, or equivalently x - 1 = 1, as the theory says it's radius of convergence is |x - 1| < 1.

Balarka

.

- Feb 13, 2012

- 1,704

As far as I remember the 'theory' says something different... and precisely that a Taylor expansion around $x_{0}$ of an f(x) that is analytic in $x=x_{0}$ converges for all $|x-x_{0}| < r$ and diverges for all $|x-x_{0}| > r$, where r is called 'radious of convergence'. An obvious question is: what does it happen on the radious of convergence, i.e. where is $|x - x_{0}| = r$?... the response may be different for different f(x).... sometimes it coverges everywhere, sometime it diverges everywhere and sometime it converges in some points and diverges in some other points... in the case of 'my' Taylor expansion of $f(x) = x\ \ln x$ it converges everywhere on $|x-1|=1$...

Now, about the continuation, you cannot use the Taylor series to evaluate at x = 0, or equivalently x - 1 = 1, as the theory says it's radius of convergence is |x - 1| < 1.

Balarka

.

Kind regards

$\chi$ $\sigma$

Last edited:

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- #11

Hi i'm having some problems with an task in my class.

It's as follows

Given the real function f(x)=X^{x}for X ∈ (0,∞).

Then i'm to expand the definition area to [0,∞), f shall still be continuous. What does f(0) have to be?

Thanks for any help

Thank you everyone for great help understanding the math in this ended up using something called L'Hôpitals rule.

Here it's explained, (in norwegian, but math is an internasional language )

L'H

- Feb 13, 2012

- 1,704

Although I am ever been very suspicious of a left-handed person [expecially if woman! ...], I have to say that Inge Christin did a very good lesson!... only one aspect has to be pointed out, i.e. that l'Hopital rule is a very performant method for finding the $\lim_{ x \rightarrow a} f(x)$ and in general it nothing say about f(a) because f(x) can be not continous in x = a...Thank you everyone for great help understanding the math in this ended up using something called L'Hôpitals rule.

Here it's explained, (in norwegian, but math is an international language )

L'H

Kind regards

$\chi$ $\sigma$