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Taylor expansion of $ln(1-x)$

2

$begingroup$

I was just wondering where the minus sign in the first term of the Taylor expansion of $ ln(1-x) $ comes from? In wikipedia page and everywhere else $ln(1-x)$ is given by
$$
ln(1-x) = -x-dots
$$
But assuming $x$ is small and expand around $1$, I got
$$
ln(1-x) approx ln(1) + fracd(ln(1-x))dxbiggvert_x=0[(1-x)-1] approx 0 + frac11-xbiggvert_x=0(-1)(-x) = x.
$$
Using the definition of Taylor expansion $f(z) approx f(a) + fracdf(z)dzbiggvert_z=a(z-a) $, where here $z=1-x$, $f(z) = ln(1-z)$ and $a=1$.

I know you can get $ln(1-x) approx -x$ by e.g. substitute $xrightarrow -x$ into the expansion of $ln(1+x)$ and through other methods etc. But I still don't quite get how you can get the minus sign from Taylor expansion alone. Thanks.

calculus

share|cite|improve this question

edited Apr 19 at 6:16

YuiTo Cheng

2,75641038

asked Apr 19 at 1:32

LepnakLepnak

162

New contributor

Lepnak is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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add a comment | 

2

$begingroup$

I was just wondering where the minus sign in the first term of the Taylor expansion of $ ln(1-x) $ comes from? In wikipedia page and everywhere else $ln(1-x)$ is given by
$$
ln(1-x) = -x-dots
$$
But assuming $x$ is small and expand around $1$, I got
$$
ln(1-x) approx ln(1) + fracd(ln(1-x))dxbiggvert_x=0[(1-x)-1] approx 0 + frac11-xbiggvert_x=0(-1)(-x) = x.
$$
Using the definition of Taylor expansion $f(z) approx f(a) + fracdf(z)dzbiggvert_z=a(z-a) $, where here $z=1-x$, $f(z) = ln(1-z)$ and $a=1$.

I know you can get $ln(1-x) approx -x$ by e.g. substitute $xrightarrow -x$ into the expansion of $ln(1+x)$ and through other methods etc. But I still don't quite get how you can get the minus sign from Taylor expansion alone. Thanks.

calculus

share|cite|improve this question

edited Apr 19 at 6:16

YuiTo Cheng

2,75641038

asked Apr 19 at 1:32

LepnakLepnak

162

New contributor

Lepnak is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

add a comment | 

$begingroup$

I was just wondering where the minus sign in the first term of the Taylor expansion of $ ln(1-x) $ comes from? In wikipedia page and everywhere else $ln(1-x)$ is given by
$$
ln(1-x) = -x-dots
$$
But assuming $x$ is small and expand around $1$, I got
$$
ln(1-x) approx ln(1) + fracd(ln(1-x))dxbiggvert_x=0[(1-x)-1] approx 0 + frac11-xbiggvert_x=0(-1)(-x) = x.
$$
Using the definition of Taylor expansion $f(z) approx f(a) + fracdf(z)dzbiggvert_z=a(z-a) $, where here $z=1-x$, $f(z) = ln(1-z)$ and $a=1$.

I know you can get $ln(1-x) approx -x$ by e.g. substitute $xrightarrow -x$ into the expansion of $ln(1+x)$ and through other methods etc. But I still don't quite get how you can get the minus sign from Taylor expansion alone. Thanks.

calculus

share|cite|improve this question

edited Apr 19 at 6:16

YuiTo Cheng

2,75641038

asked Apr 19 at 1:32

LepnakLepnak

162

New contributor

Lepnak is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

share|cite|improve this question

edited Apr 19 at 6:16

YuiTo Cheng

2,75641038

asked Apr 19 at 1:32

LepnakLepnak

162

New contributor

Lepnak is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

share|cite|improve this question

edited Apr 19 at 6:16

YuiTo Cheng

2,75641038

asked Apr 19 at 1:32

LepnakLepnak

162

New contributor

Lepnak is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

edited Apr 19 at 6:16

YuiTo Cheng

2,75641038

edited Apr 19 at 6:16

YuiTo Cheng

2,75641038

asked Apr 19 at 1:32

LepnakLepnak

162

New contributor

Lepnak is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

asked Apr 19 at 1:32

LepnakLepnak

162

2 Answers
2

active

oldest

votes

2

$begingroup$

If one considers
$$
f(x)=ln (1-x),qquad |x|<1,
$$one has
$$
f(0)=0,quad f'(x)=-frac11-x,quad f'(0)=-1,quad f''(x)=-frac1(1-x)^2,quad f''(0)=-1,
$$ giving, by the Taylor expansion,
$$
f(x)=0-x-fracx^22+O(x^3)
$$as $x to 0$.

share|cite|improve this answer

edited Apr 19 at 1:46

answered Apr 19 at 1:42

Olivier OloaOlivier Oloa

109k17178294

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks for the answer but what about the $(z-a)$ part in the Taylor expansion $f(z) = f(a)+f^prime(a)(z-a)$? Substitute $z=1-x$ and $a=1$ gives a $-x$ though?
  $endgroup$
  – Lepnak
  Apr 19 at 2:04
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The Taylor series centred at $0$ is $$f(x)=f(0)+f'(0)x +cdots.$$ Use $f(0)$ and $f'(0)$ from Olivier Oloa's answer and you should get the right answer. In your OP, you are actually expanding $f(x)$ around $0$, not around $1$ (where $f(x)=ln (1-x)$). So $a=0$. By the way, if you substitute $z=1-x$ where $f(z)=ln (1-z)$, you would get $ln(1-(1-x))=ln x$, rather than $ln(1-x)$ (which is what you want). So no need to do this substitution.
  $endgroup$
  – Minus One-Twelfth
  Apr 19 at 2:28
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Hmm I think I see what I did wrong. Thanks for all your answers.
  $endgroup$
  – Lepnak
  Apr 19 at 2:37
  

  
  
  
  

add a comment | 

3

$begingroup$

$$y=ln(1-x)$$
$$y'=-frac11-x=-sum_n=0^inftyx^n$$
so
$$ln(1-x)=-sum_n=0^inftyfracx^n+1n+1=-sum_n=1^inftyfracx^nn$$

share|cite|improve this answer

edited Apr 19 at 1:59

answered Apr 19 at 1:52

E.H.EE.H.E

17.1k11969

$endgroup$

add a comment | 

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2

$begingroup$

If one considers
$$
f(x)=ln (1-x),qquad |x|<1,
$$one has
$$
f(0)=0,quad f'(x)=-frac11-x,quad f'(0)=-1,quad f''(x)=-frac1(1-x)^2,quad f''(0)=-1,
$$ giving, by the Taylor expansion,
$$
f(x)=0-x-fracx^22+O(x^3)
$$as $x to 0$.

share|cite|improve this answer

edited Apr 19 at 1:46

answered Apr 19 at 1:42

Olivier OloaOlivier Oloa

109k17178294

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks for the answer but what about the $(z-a)$ part in the Taylor expansion $f(z) = f(a)+f^prime(a)(z-a)$? Substitute $z=1-x$ and $a=1$ gives a $-x$ though?
  $endgroup$
  – Lepnak
  Apr 19 at 2:04
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The Taylor series centred at $0$ is $$f(x)=f(0)+f'(0)x +cdots.$$ Use $f(0)$ and $f'(0)$ from Olivier Oloa's answer and you should get the right answer. In your OP, you are actually expanding $f(x)$ around $0$, not around $1$ (where $f(x)=ln (1-x)$). So $a=0$. By the way, if you substitute $z=1-x$ where $f(z)=ln (1-z)$, you would get $ln(1-(1-x))=ln x$, rather than $ln(1-x)$ (which is what you want). So no need to do this substitution.
  $endgroup$
  – Minus One-Twelfth
  Apr 19 at 2:28
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Hmm I think I see what I did wrong. Thanks for all your answers.
  $endgroup$
  – Lepnak
  Apr 19 at 2:37
  

  
  
  
  

add a comment | 

2

$begingroup$

If one considers
$$
f(x)=ln (1-x),qquad |x|<1,
$$one has
$$
f(0)=0,quad f'(x)=-frac11-x,quad f'(0)=-1,quad f''(x)=-frac1(1-x)^2,quad f''(0)=-1,
$$ giving, by the Taylor expansion,
$$
f(x)=0-x-fracx^22+O(x^3)
$$as $x to 0$.

share|cite|improve this answer

edited Apr 19 at 1:46

answered Apr 19 at 1:42

Olivier OloaOlivier Oloa

109k17178294

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks for the answer but what about the $(z-a)$ part in the Taylor expansion $f(z) = f(a)+f^prime(a)(z-a)$? Substitute $z=1-x$ and $a=1$ gives a $-x$ though?
  $endgroup$
  – Lepnak
  Apr 19 at 2:04
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The Taylor series centred at $0$ is $$f(x)=f(0)+f'(0)x +cdots.$$ Use $f(0)$ and $f'(0)$ from Olivier Oloa's answer and you should get the right answer. In your OP, you are actually expanding $f(x)$ around $0$, not around $1$ (where $f(x)=ln (1-x)$). So $a=0$. By the way, if you substitute $z=1-x$ where $f(z)=ln (1-z)$, you would get $ln(1-(1-x))=ln x$, rather than $ln(1-x)$ (which is what you want). So no need to do this substitution.
  $endgroup$
  – Minus One-Twelfth
  Apr 19 at 2:28
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Hmm I think I see what I did wrong. Thanks for all your answers.
  $endgroup$
  – Lepnak
  Apr 19 at 2:37
  

  
  
  
  

add a comment | 

$begingroup$

If one considers
$$
f(x)=ln (1-x),qquad |x|<1,
$$one has
$$
f(0)=0,quad f'(x)=-frac11-x,quad f'(0)=-1,quad f''(x)=-frac1(1-x)^2,quad f''(0)=-1,
$$ giving, by the Taylor expansion,
$$
f(x)=0-x-fracx^22+O(x^3)
$$as $x to 0$.

share|cite|improve this answer

edited Apr 19 at 1:46

answered Apr 19 at 1:42

Olivier OloaOlivier Oloa

109k17178294

$endgroup$

share|cite|improve this answer

edited Apr 19 at 1:46

answered Apr 19 at 1:42

Olivier OloaOlivier Oloa

109k17178294

answered Apr 19 at 1:42

Olivier OloaOlivier Oloa

109k17178294

answered Apr 19 at 1:42

Olivier OloaOlivier Oloa

109k17178294

3

$begingroup$

$$y=ln(1-x)$$
$$y'=-frac11-x=-sum_n=0^inftyx^n$$
so
$$ln(1-x)=-sum_n=0^inftyfracx^n+1n+1=-sum_n=1^inftyfracx^nn$$

share|cite|improve this answer

edited Apr 19 at 1:59

answered Apr 19 at 1:52

E.H.EE.H.E

17.1k11969

$endgroup$

add a comment | 

3

$begingroup$

$$y=ln(1-x)$$
$$y'=-frac11-x=-sum_n=0^inftyx^n$$
so
$$ln(1-x)=-sum_n=0^inftyfracx^n+1n+1=-sum_n=1^inftyfracx^nn$$

share|cite|improve this answer

edited Apr 19 at 1:59

answered Apr 19 at 1:52

E.H.EE.H.E

17.1k11969

$endgroup$

add a comment | 

$begingroup$

$$y=ln(1-x)$$
$$y'=-frac11-x=-sum_n=0^inftyx^n$$
so
$$ln(1-x)=-sum_n=0^inftyfracx^n+1n+1=-sum_n=1^inftyfracx^nn$$

share|cite|improve this answer

edited Apr 19 at 1:59

answered Apr 19 at 1:52

E.H.EE.H.E

17.1k11969

$endgroup$

share|cite|improve this answer

edited Apr 19 at 1:59

answered Apr 19 at 1:52

E.H.EE.H.E

17.1k11969

answered Apr 19 at 1:52

E.H.EE.H.E

17.1k11969

answered Apr 19 at 1:52

E.H.EE.H.E

17.1k11969