- Thread starter
- #1
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)=Xx 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![]()
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)=Xx 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?
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 $x=0.$
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 $x=0.$
I apologize for the fact that I badly reported a formula from my old notes... the correct formula is...First things first, your calculation of (2) seems false, as at x = 0 in (1) is log(4) - 2 which is definitely not 0. Check your calculation of the Taylor series.
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 $|x-1|=1$...Okay, nothing to apologize about, we all do mistakes.
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
.
Hi i'm having some problems with an task in my class.
It's as follows
Given the real function f(x)=Xx 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![]()
Although I am ever been very suspicious of a left-handed person [expecially if woman!Thank you everyone for great help understanding the math in thisended up using something called L'Hôpitals rule.
Here it's explained, (in norwegian, but math is an international language)
L'H