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 $ln(1 + frac2^xn)$How to see the following Taylor expansion?
Should I use a zero-interest credit card for a large one-time purchase?
If Windows 7 doesn't support WSL, then what does Linux subsystem option mean?
An adverb for when you're not exaggerating
How do I find out the mythology and history of my Fortress?
Take 2! Is this homebrew Lady of Pain warlock patron balanced?
SF book about people trapped in a series of worlds they imagine
AppleTVs create a chatty alternate WiFi network
Drawing without replacement: why is the order of draw irrelevant?
Why should I vote and accept answers?
Why wasn't DOSKEY integrated with COMMAND.COM?
ArcGIS Pro Python arcpy.CreatePersonalGDB_management
Did Deadpool rescue all of the X-Force?
How would a mousetrap for use in space work?
Do wooden building fires get hotter than 600°C?
How to install press fit bottom bracket into new frame
Is CEO the "profession" with the most psychopaths?
How could we fake a moon landing now?
Chinese Seal on silk painting - what does it mean?
Why does it sometimes sound good to play a grace note as a lead in to a note in a melody?
Why do we need to use the builder design pattern when we can do the same thing with setters?
How do living politicians protect their readily obtainable signatures from misuse?
Find 108 by using 3,4,6
How to react to hostile behavior from a senior developer?
Illegal assignment from sObject to Id
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 $ln(1 + frac2^xn)$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
$$
ln(1-x) approx ln(1) + fracd(ln(1-x))dxbiggvert_x=0[(1-x)-1] approx 0 + frac11-xbiggvert_x=0(-1)(-x) = x.
$$
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
asked 2 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
$$
ln(1-x) approx ln(1) + fracd(ln(1-x))dxbiggvert_x=0[(1-x)-1] approx 0 + frac11-xbiggvert_x=0(-1)(-x) = x.
$$
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
asked 2 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
$$
ln(1-x) approx ln(1) + fracd(ln(1-x))dxbiggvert_x=0[(1-x)-1] approx 0 + frac11-xbiggvert_x=0(-1)(-x) = x.
$$
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
asked 2 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
$$
ln(1-x) approx ln(1) + fracd(ln(1-x))dxbiggvert_x=0[(1-x)-1] approx 0 + frac11-xbiggvert_x=0(-1)(-x) = x.
$$
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
calculus
asked 2 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 2 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 2 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 2 hours ago
LepnakLepnak
182
asked 2 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.
Lepnak is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
If one considers
$$
f(x)=ln (1-x),qquad |x|<1,
$$one has
$$
f(0)=0,quad f'(x)=-frac11-x,quad f'(0)=-1,quad f''(x)=-frac1(1-x)^2,quad f''(0)=-1,
$$ giving, by the Taylor expansion,
$$
f(x)=0-x-fracx^22+O(x^3)
$$as $x to 0$.
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
1 hour ago
$begingroup$
Hmm I think I see what I did wrong. Thanks for all your answers.
$endgroup$
– Lepnak
1 hour ago
add a comment |
$begingroup$
$$y=ln(1-x)$$
$$y'=-frac11-x=-sum_n=0^inftyx^n$$
so
$$ln(1-x)=-sum_n=0^inftyfracx^n+1n+1=-sum_n=1^inftyfracx^nn$$
answered 2 hours ago
E.H.EE.H.E
16.8k11969
$endgroup$
add a comment |
Your Answer
StackExchange.ready(function()
var channelOptions =
tags: "".split(" "),
id: "69"
;
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function()
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled)
StackExchange.using("snippets", function()
createEditor();
);
else
createEditor();
);
function createEditor()
StackExchange.prepareEditor(
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader:
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
,
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
);
);
Lepnak is a new contributor. Be nice, and check out our Code of Conduct.
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3193068%2ftaylor-expansion-of-ln1-x%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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 $x to 0$.
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
1 hour ago
$begingroup$
Hmm I think I see what I did wrong. Thanks for all your answers.
$endgroup$
– Lepnak
1 hour ago
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 $x to 0$.
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
1 hour ago
$begingroup$
Hmm I think I see what I did wrong. Thanks for all your answers.
$endgroup$
– Lepnak
1 hour ago
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 $x to 0$.
answered 2 hours ago
Olivier OloaOlivier Oloa
109k17178294
$endgroup$
If one considers
$$
f(x)=ln (1-x),qquad |x|<1,
$$one has
$$
f(0)=0,quad f'(x)=-frac11-x,quad f'(0)=-1,quad f''(x)=-frac1(1-x)^2,quad f''(0)=-1,
$$ giving, by the Taylor expansion,
$$
f(x)=0-x-fracx^22+O(x^3)
$$as $x to 0$.
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
1 hour ago
$begingroup$
Hmm I think I see what I did wrong. Thanks for all your answers.
$endgroup$
– Lepnak
1 hour 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
1 hour ago
$begingroup$
Hmm I think I see what I did wrong. Thanks for all your answers.
$endgroup$
– Lepnak
1 hour 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
1 hour 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
1 hour ago
$begingroup$
Hmm I think I see what I did wrong. Thanks for all your answers.
$endgroup$
– Lepnak
1 hour ago
$begingroup$
Hmm I think I see what I did wrong. Thanks for all your answers.
$endgroup$
– Lepnak
1 hour ago
add a comment |
$begingroup$
$$y=ln(1-x)$$
$$y'=-frac11-x=-sum_n=0^inftyx^n$$
so
$$ln(1-x)=-sum_n=0^inftyfracx^n+1n+1=-sum_n=1^inftyfracx^nn$$
answered 2 hours ago
E.H.EE.H.E
16.8k11969
$endgroup$
add a comment |
$begingroup$
$$y=ln(1-x)$$
$$y'=-frac11-x=-sum_n=0^inftyx^n$$
so
$$ln(1-x)=-sum_n=0^inftyfracx^n+1n+1=-sum_n=1^inftyfracx^nn$$
answered 2 hours ago
E.H.EE.H.E
16.8k11969
$endgroup$
add a comment |
$begingroup$
$$y=ln(1-x)$$
$$y'=-frac11-x=-sum_n=0^inftyx^n$$
so
$$ln(1-x)=-sum_n=0^inftyfracx^n+1n+1=-sum_n=1^inftyfracx^nn$$
answered 2 hours ago
E.H.EE.H.E
16.8k11969
$endgroup$
$$y=ln(1-x)$$
$$y'=-frac11-x=-sum_n=0^inftyx^n$$
so
$$ln(1-x)=-sum_n=0^inftyfracx^n+1n+1=-sum_n=1^inftyfracx^nn$$
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.
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.
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3193068%2ftaylor-expansion-of-ln1-x%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
