Taylor expansion of ln(1-x) Announcing the arrival of Valued Associate #679: Cesar Manara ...
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Taylor expansion of ln(1-x)
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 23, 2019 at 00:00UTC (8:00pm US/Eastern)Basic Taylor expansion questionStuck on Taylor expansion questionTaylor expansion of the Error functionUsing substitution while using taylor expansionTaylor expansion of a matrix to scalar functionTaylor expansion of $log(x - x^2)$ at 0?Taylor expansion of $(1-x)(1-y)$.Taylor Expansion of Eigenvector PerturbationTaylor expansion of $ln(1 + frac{2^x}{n})$How to see the following Taylor expansion?
$begingroup$
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
$$
ln(1-x) approx ln(1) + frac{d(ln(1-x))}{dx}biggvert_{x=0}[(1-x)-1] approx 0 + frac{1}{1-x}biggvert_{x=0}(-1)(-x) = x.
$$
Using the definition of Taylor expansion $f(z) approx f(a) + frac{df(z)}{dz}biggvert_{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
asked 3 hours ago
LepnakLepnak
182
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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
$$
ln(1-x) approx ln(1) + frac{d(ln(1-x))}{dx}biggvert_{x=0}[(1-x)-1] approx 0 + frac{1}{1-x}biggvert_{x=0}(-1)(-x) = x.
$$
Using the definition of Taylor expansion $f(z) approx f(a) + frac{df(z)}{dz}biggvert_{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
asked 3 hours ago
LepnakLepnak
182
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
$$
ln(1-x) approx ln(1) + frac{d(ln(1-x))}{dx}biggvert_{x=0}[(1-x)-1] approx 0 + frac{1}{1-x}biggvert_{x=0}(-1)(-x) = x.
$$
Using the definition of Taylor expansion $f(z) approx f(a) + frac{df(z)}{dz}biggvert_{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
asked 3 hours ago
LepnakLepnak
182
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.
$endgroup$
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
$$
ln(1-x) approx ln(1) + frac{d(ln(1-x))}{dx}biggvert_{x=0}[(1-x)-1] approx 0 + frac{1}{1-x}biggvert_{x=0}(-1)(-x) = x.
$$
Using the definition of Taylor expansion $f(z) approx f(a) + frac{df(z)}{dz}biggvert_{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
calculus
asked 3 hours ago
LepnakLepnak
182
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 3 hours ago
LepnakLepnak
182
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 3 hours ago
LepnakLepnak
182
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 3 hours ago
LepnakLepnak
182
asked 3 hours ago
LepnakLepnak
182
182
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.
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.
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Check out our Code of Conduct.
add a comment |
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2 Answers
2
active
oldest
votes
$begingroup$
If one considers
$$
f(x)=ln (1-x),qquad |x|<1,
$$one has
$$
f(0)=0,quad f'(x)=-frac{1}{1-x},quad f'(0)=-1,quad f''(x)=-frac{1}{(1-x)^2},quad f''(0)=-1,
$$ giving, by the Taylor expansion,
$$
f(x)=0-x-frac{x^2}2+O(x^3)
$$as $x to 0$.
answered 2 hours ago
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
2 hours ago
$begingroup$
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
2 hours ago
$begingroup$
Hmm I think I see what I did wrong. Thanks for all your answers.
$endgroup$
– Lepnak
2 hours ago
add a comment |
$begingroup$
$$y=ln(1-x)$$
$$y'=-frac{1}{1-x}=-sum_{n=0}^{infty}x^n$$
so
$$ln(1-x)=-sum_{n=0}^{infty}frac{x^{n+1}}{n+1}=-sum_{n=1}^{infty}frac{x^{n}}{n}$$
answered 2 hours ago
E.H.EE.H.E
16.8k11969
$endgroup$
add a comment |
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2 Answers
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active
oldest
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2 Answers
2
active
oldest
votes
active
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active
oldest
votes
$begingroup$
If one considers
$$
f(x)=ln (1-x),qquad |x|<1,
$$one has
$$
f(0)=0,quad f'(x)=-frac{1}{1-x},quad f'(0)=-1,quad f''(x)=-frac{1}{(1-x)^2},quad f''(0)=-1,
$$ giving, by the Taylor expansion,
$$
f(x)=0-x-frac{x^2}2+O(x^3)
$$as $x to 0$.
answered 2 hours ago
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
2 hours ago
$begingroup$
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
2 hours ago
$begingroup$
Hmm I think I see what I did wrong. Thanks for all your answers.
$endgroup$
– Lepnak
2 hours ago
add a comment |
$begingroup$
If one considers
$$
f(x)=ln (1-x),qquad |x|<1,
$$one has
$$
f(0)=0,quad f'(x)=-frac{1}{1-x},quad f'(0)=-1,quad f''(x)=-frac{1}{(1-x)^2},quad f''(0)=-1,
$$ giving, by the Taylor expansion,
$$
f(x)=0-x-frac{x^2}2+O(x^3)
$$as $x to 0$.
answered 2 hours ago
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
2 hours ago
$begingroup$
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
2 hours ago
$begingroup$
Hmm I think I see what I did wrong. Thanks for all your answers.
$endgroup$
– Lepnak
2 hours ago
add a comment |
$begingroup$
If one considers
$$
f(x)=ln (1-x),qquad |x|<1,
$$one has
$$
f(0)=0,quad f'(x)=-frac{1}{1-x},quad f'(0)=-1,quad f''(x)=-frac{1}{(1-x)^2},quad f''(0)=-1,
$$ giving, by the Taylor expansion,
$$
f(x)=0-x-frac{x^2}2+O(x^3)
$$as $x to 0$.
answered 2 hours ago
Olivier OloaOlivier Oloa
109k17178294
$endgroup$
If one considers
$$
f(x)=ln (1-x),qquad |x|<1,
$$one has
$$
f(0)=0,quad f'(x)=-frac{1}{1-x},quad f'(0)=-1,quad f''(x)=-frac{1}{(1-x)^2},quad f''(0)=-1,
$$ giving, by the Taylor expansion,
$$
f(x)=0-x-frac{x^2}2+O(x^3)
$$as $x to 0$.
answered 2 hours ago
Olivier OloaOlivier Oloa
109k17178294
edited 2 hours ago
answered 2 hours ago
Olivier OloaOlivier Oloa
109k17178294
answered 2 hours ago
Olivier OloaOlivier Oloa
109k17178294
answered 2 hours ago
Olivier OloaOlivier Oloa
109k17178294
109k17178294
$begingroup$
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
2 hours ago
$begingroup$
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
2 hours ago
$begingroup$
Hmm I think I see what I did wrong. Thanks for all your answers.
$endgroup$
– Lepnak
2 hours ago
add a comment |
$begingroup$
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
2 hours ago
$begingroup$
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
2 hours ago
$begingroup$
Hmm I think I see what I did wrong. Thanks for all your answers.
$endgroup$
– Lepnak
2 hours ago
$begingroup$
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
2 hours ago
$begingroup$
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
2 hours ago
$begingroup$
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
2 hours ago
$begingroup$
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
2 hours ago
$begingroup$
Hmm I think I see what I did wrong. Thanks for all your answers.
$endgroup$
– Lepnak
2 hours ago
$begingroup$
Hmm I think I see what I did wrong. Thanks for all your answers.
$endgroup$
– Lepnak
2 hours ago
add a comment |
$begingroup$
$$y=ln(1-x)$$
$$y'=-frac{1}{1-x}=-sum_{n=0}^{infty}x^n$$
so
$$ln(1-x)=-sum_{n=0}^{infty}frac{x^{n+1}}{n+1}=-sum_{n=1}^{infty}frac{x^{n}}{n}$$
answered 2 hours ago
E.H.EE.H.E
16.8k11969
$endgroup$
add a comment |
$begingroup$
$$y=ln(1-x)$$
$$y'=-frac{1}{1-x}=-sum_{n=0}^{infty}x^n$$
so
$$ln(1-x)=-sum_{n=0}^{infty}frac{x^{n+1}}{n+1}=-sum_{n=1}^{infty}frac{x^{n}}{n}$$
answered 2 hours ago
E.H.EE.H.E
16.8k11969
$endgroup$
add a comment |
$begingroup$
$$y=ln(1-x)$$
$$y'=-frac{1}{1-x}=-sum_{n=0}^{infty}x^n$$
so
$$ln(1-x)=-sum_{n=0}^{infty}frac{x^{n+1}}{n+1}=-sum_{n=1}^{infty}frac{x^{n}}{n}$$
answered 2 hours ago
E.H.EE.H.E
16.8k11969
$endgroup$
$$y=ln(1-x)$$
$$y'=-frac{1}{1-x}=-sum_{n=0}^{infty}x^n$$
so
$$ln(1-x)=-sum_{n=0}^{infty}frac{x^{n+1}}{n+1}=-sum_{n=1}^{infty}frac{x^{n}}{n}$$
answered 2 hours ago
E.H.EE.H.E
16.8k11969
edited 2 hours ago
answered 2 hours ago
E.H.EE.H.E
16.8k11969
answered 2 hours ago
E.H.EE.H.E
16.8k11969
answered 2 hours ago
E.H.EE.H.E
16.8k11969
16.8k11969
add a comment |
add a comment |
Lepnak is a new contributor. Be nice, and check out our Code of Conduct.
Lepnak is a new contributor. Be nice, and check out our Code of Conduct.
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