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Expanding $x^x$ to series with $o(x)$ polynomials

2

$begingroup$

I have some doubts in finding $x^x$ series.
I know that from Taylor theorem I have
$$ f(x) = f(a) + f'(a)(x-a) + frac{f''(a)}{2!}(x-a)^2 + cdots + frac{f^{(k)}(a)}{k!}(x-a)^k + o(x^k) $$

I want to have $o(x)$ polynomials so:

Let $$ f(x) = x^x $$
$$ f'(x) = (e^{xcdot ln x}) = (ln x + 1)x^x$$
In use of taylor I should
$$ f(x) = f(0) + frac{f'(0)(x)}{1!} + o(x) $$
but $f(0)$ is not defined $0^0$. Also $f'(0)$ is not defined...
But I found informations that it should be
$$ 1+ ln x cdot x + o(x)$$
I don't know why :-(

real-analysis taylor-expansion

share|cite|improve this question

asked Mar 23 at 12:42

VirtualUserVirtualUser

1,321317

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  notice that $0^0 =1$
  $endgroup$
  – Jneven
  Mar 23 at 12:46
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  $x^x$ is not defined for $x<0$ so you can't possibly expand it in a Taylor series about the origin. However, since $x^x=e^{xlog{x}}$ you can substitute $xlog{x}$ into the Taylor series for $e^x$
  $endgroup$
  – saulspatz
  Mar 23 at 12:49
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Not every function has a Taylor series (it has to be differntiable arbitrarily many times). $x^x$ is one of those that do not have Taylor expansion around 0 (because it's not differentiable at that point). It has a Taylor expansion around any point $a>0$ though.
  $endgroup$
  – Adam Latosiński
  Mar 23 at 12:55
  

  
  
  
  

add a comment | 

2

$begingroup$

I have some doubts in finding $x^x$ series.
I know that from Taylor theorem I have
$$ f(x) = f(a) + f'(a)(x-a) + frac{f''(a)}{2!}(x-a)^2 + cdots + frac{f^{(k)}(a)}{k!}(x-a)^k + o(x^k) $$

I want to have $o(x)$ polynomials so:

Let $$ f(x) = x^x $$
$$ f'(x) = (e^{xcdot ln x}) = (ln x + 1)x^x$$
In use of taylor I should
$$ f(x) = f(0) + frac{f'(0)(x)}{1!} + o(x) $$
but $f(0)$ is not defined $0^0$. Also $f'(0)$ is not defined...
But I found informations that it should be
$$ 1+ ln x cdot x + o(x)$$
I don't know why :-(

real-analysis taylor-expansion

share|cite|improve this question

asked Mar 23 at 12:42

VirtualUserVirtualUser

1,321317

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  notice that $0^0 =1$
  $endgroup$
  – Jneven
  Mar 23 at 12:46
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  $x^x$ is not defined for $x<0$ so you can't possibly expand it in a Taylor series about the origin. However, since $x^x=e^{xlog{x}}$ you can substitute $xlog{x}$ into the Taylor series for $e^x$
  $endgroup$
  – saulspatz
  Mar 23 at 12:49
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Not every function has a Taylor series (it has to be differntiable arbitrarily many times). $x^x$ is one of those that do not have Taylor expansion around 0 (because it's not differentiable at that point). It has a Taylor expansion around any point $a>0$ though.
  $endgroup$
  – Adam Latosiński
  Mar 23 at 12:55
  

  
  
  
  

add a comment | 

$begingroup$

I have some doubts in finding $x^x$ series.
I know that from Taylor theorem I have
$$ f(x) = f(a) + f'(a)(x-a) + frac{f''(a)}{2!}(x-a)^2 + cdots + frac{f^{(k)}(a)}{k!}(x-a)^k + o(x^k) $$

I want to have $o(x)$ polynomials so:

Let $$ f(x) = x^x $$
$$ f'(x) = (e^{xcdot ln x}) = (ln x + 1)x^x$$
In use of taylor I should
$$ f(x) = f(0) + frac{f'(0)(x)}{1!} + o(x) $$
but $f(0)$ is not defined $0^0$. Also $f'(0)$ is not defined...
But I found informations that it should be
$$ 1+ ln x cdot x + o(x)$$
I don't know why :-(

real-analysis taylor-expansion

share|cite|improve this question

asked Mar 23 at 12:42

VirtualUserVirtualUser

1,321317

$endgroup$

share|cite|improve this question

asked Mar 23 at 12:42

VirtualUserVirtualUser

1,321317

share|cite|improve this question

asked Mar 23 at 12:42

VirtualUserVirtualUser

1,321317

asked Mar 23 at 12:42

VirtualUserVirtualUser

1,321317

asked Mar 23 at 12:42

VirtualUserVirtualUser

1,321317

2 Answers
2

active

oldest

votes

2

$begingroup$

As you have written
$$x^x=e^{xln{x}}$$
and as the series expansion of $e^x$ converges for all values of $xinmathbb{C}$ one can write
$$e^x=frac{x^0}{0!}+frac{x^1}{1!}+frac{x^2}{2!}+...$$
$$e^{xln{x}}=frac{(xln{x})^0}{0!}+frac{(xln{x})^1}{1!}+frac{(xln{x})^2}{2!}+...$$
$$therefore x^x=1+xln{x}+frac{x^2ln^2{x}}{2}+frac{x^3ln^3{x}}{6}+...$$

share|cite|improve this answer

answered Mar 23 at 12:48

Peter ForemanPeter Foreman

7,2411319

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes, that trick really helps me, thanks!
  $endgroup$
  – VirtualUser
  Mar 23 at 15:42
  

  
  
  
  

add a comment | 

1

$begingroup$

Although $f(0)$ and $f'(0)$ are not defined, one can write$$lim_{xto 0^+} f(x)=lim_{xto 0^+} x^x=e^{lim_{xto 0^+} xln x}=e^0=1$$In that case $${f(x)=x^xsim 1\f'(x)=x^x(ln x+1)sim ln x+1}$$both when $xto0^+.$ Here is a plot of the function:

share|cite|improve this answer

edited Mar 23 at 13:02

answered Mar 23 at 12:55

Mostafa AyazMostafa Ayaz

18.1k31040

$endgroup$

add a comment | 

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2

$begingroup$

As you have written
$$x^x=e^{xln{x}}$$
and as the series expansion of $e^x$ converges for all values of $xinmathbb{C}$ one can write
$$e^x=frac{x^0}{0!}+frac{x^1}{1!}+frac{x^2}{2!}+...$$
$$e^{xln{x}}=frac{(xln{x})^0}{0!}+frac{(xln{x})^1}{1!}+frac{(xln{x})^2}{2!}+...$$
$$therefore x^x=1+xln{x}+frac{x^2ln^2{x}}{2}+frac{x^3ln^3{x}}{6}+...$$

share|cite|improve this answer

answered Mar 23 at 12:48

Peter ForemanPeter Foreman

7,2411319

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes, that trick really helps me, thanks!
  $endgroup$
  – VirtualUser
  Mar 23 at 15:42
  

  
  
  
  

add a comment | 

2

$begingroup$

As you have written
$$x^x=e^{xln{x}}$$
and as the series expansion of $e^x$ converges for all values of $xinmathbb{C}$ one can write
$$e^x=frac{x^0}{0!}+frac{x^1}{1!}+frac{x^2}{2!}+...$$
$$e^{xln{x}}=frac{(xln{x})^0}{0!}+frac{(xln{x})^1}{1!}+frac{(xln{x})^2}{2!}+...$$
$$therefore x^x=1+xln{x}+frac{x^2ln^2{x}}{2}+frac{x^3ln^3{x}}{6}+...$$

share|cite|improve this answer

answered Mar 23 at 12:48

Peter ForemanPeter Foreman

7,2411319

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes, that trick really helps me, thanks!
  $endgroup$
  – VirtualUser
  Mar 23 at 15:42
  

  
  
  
  

add a comment | 

$begingroup$

As you have written
$$x^x=e^{xln{x}}$$
and as the series expansion of $e^x$ converges for all values of $xinmathbb{C}$ one can write
$$e^x=frac{x^0}{0!}+frac{x^1}{1!}+frac{x^2}{2!}+...$$
$$e^{xln{x}}=frac{(xln{x})^0}{0!}+frac{(xln{x})^1}{1!}+frac{(xln{x})^2}{2!}+...$$
$$therefore x^x=1+xln{x}+frac{x^2ln^2{x}}{2}+frac{x^3ln^3{x}}{6}+...$$

share|cite|improve this answer

answered Mar 23 at 12:48

Peter ForemanPeter Foreman

7,2411319

$endgroup$

share|cite|improve this answer

answered Mar 23 at 12:48

Peter ForemanPeter Foreman

7,2411319

answered Mar 23 at 12:48

Peter ForemanPeter Foreman

7,2411319

answered Mar 23 at 12:48

Peter ForemanPeter Foreman

7,2411319

1

$begingroup$

Although $f(0)$ and $f'(0)$ are not defined, one can write$$lim_{xto 0^+} f(x)=lim_{xto 0^+} x^x=e^{lim_{xto 0^+} xln x}=e^0=1$$In that case $${f(x)=x^xsim 1\f'(x)=x^x(ln x+1)sim ln x+1}$$both when $xto0^+.$ Here is a plot of the function:

share|cite|improve this answer

edited Mar 23 at 13:02

answered Mar 23 at 12:55

Mostafa AyazMostafa Ayaz

18.1k31040

$endgroup$

add a comment | 

1

$begingroup$

Although $f(0)$ and $f'(0)$ are not defined, one can write$$lim_{xto 0^+} f(x)=lim_{xto 0^+} x^x=e^{lim_{xto 0^+} xln x}=e^0=1$$In that case $${f(x)=x^xsim 1\f'(x)=x^x(ln x+1)sim ln x+1}$$both when $xto0^+.$ Here is a plot of the function:

share|cite|improve this answer

edited Mar 23 at 13:02

answered Mar 23 at 12:55

Mostafa AyazMostafa Ayaz

18.1k31040

$endgroup$

add a comment | 

$begingroup$

Although $f(0)$ and $f'(0)$ are not defined, one can write$$lim_{xto 0^+} f(x)=lim_{xto 0^+} x^x=e^{lim_{xto 0^+} xln x}=e^0=1$$In that case $${f(x)=x^xsim 1\f'(x)=x^x(ln x+1)sim ln x+1}$$both when $xto0^+.$ Here is a plot of the function:

share|cite|improve this answer

edited Mar 23 at 13:02

answered Mar 23 at 12:55

Mostafa AyazMostafa Ayaz

18.1k31040

$endgroup$

share|cite|improve this answer

edited Mar 23 at 13:02

answered Mar 23 at 12:55

Mostafa AyazMostafa Ayaz

18.1k31040

answered Mar 23 at 12:55

Mostafa AyazMostafa Ayaz

18.1k31040

answered Mar 23 at 12:55

Mostafa AyazMostafa Ayaz

18.1k31040