If $N(x) = x-[f^prime(x)]^-1f(x)$ and $f(x^ast)=0$, how can I prove that $N^prime(x^ast) = 0$? The Next CEO of Stack OverflowHow can I prove that $f(x):=frac1$ is not bounded above?Non Uniformly Elliptic Equations page 117 [G-T]Divergence in Definition of Laplace-Beltrami OperatorAn open $OmegasubseteqBbb R$ with $V=EcapOmega$ and $UsubsetOmega^c$, where $U,V$ are relatively open and separate $EsubseteqBbb R^n$Differential Form and Pullback of Composition of Smooth Functions (Proof Verification)Legendre-Fenchel transformation for $varphi, Big(,x, ,, f,big(,Phi,(,x,),big),Big)$.How to disprove existence of (total) derivativeProb. 17, Chap. 5 in Baby Rudin: If $f(-1)=0=f(0)=f^prime(0)$ and $f(1)=1$, then $f^(3)(x)geq 0$ for some $xin(-1,1)$There exist disjoint open sets $U_a ni a$ and $U_b ni b$ such that $overlineU_asubseteq U, overlineU_bsubseteq U$How can we prove the scaling property of the Dirac delta function rigorously?
How to solve a differential equation with a term to a power?
What does "Its cash flow is deeply negative" mean?
How do I go from 300 unfinished/half written blog posts, to published posts?
Is it ever safe to open a suspicious html file (e.g. email attachment)?
What was the first Unix version to run on a microcomputer?
How to start emacs in "nothing" mode (`fundamental-mode`)
Which tube will fit a -(700 x 25c) wheel?
Anatomically Correct Strange Women In Ponds Distributing Swords
Sending manuscript to multiple publishers
Can I equip Skullclamp on a creature I am sacrificing?
Elegant way to replace substring in a regex with optional groups in Python?
Different harmonic changes implied by a simple descending scale
How should I support this large drywall patch?
What happened in Rome, when the western empire "fell"?
Is it professional to write unrelated content in an almost-empty email?
If a black hole is created from light, can this black hole then move at speed of light?
Between two walls
Is micro rebar a better way to reinforce concrete than rebar?
I believe this to be a fraud - hired, then asked to cash check and send cash as Bitcoin
Should I tutor a student who I know has cheated on their homework?
What does convergence in distribution "in the Gromov–Hausdorff" sense mean?
How to transpose the 1st and -1th levels of arbitrarily nested array?
"and that skill is always a class skill for you" - does "always" have any meaning in Pathfinder?
Indicator light circuit
If $N(x) = x-[f^prime(x)]^-1f(x)$ and $f(x^ast)=0$, how can I prove that $N^prime(x^ast) = 0$?
The Next CEO of Stack OverflowHow can I prove that $f(x):=frac1$ is not bounded above?Non Uniformly Elliptic Equations page 117 [G-T]Divergence in Definition of Laplace-Beltrami OperatorAn open $OmegasubseteqBbb R$ with $V=EcapOmega$ and $UsubsetOmega^c$, where $U,V$ are relatively open and separate $EsubseteqBbb R^n$Differential Form and Pullback of Composition of Smooth Functions (Proof Verification)Legendre-Fenchel transformation for $varphi, Big(,x, ,, f,big(,Phi,(,x,),big),Big)$.How to disprove existence of (total) derivativeProb. 17, Chap. 5 in Baby Rudin: If $f(-1)=0=f(0)=f^prime(0)$ and $f(1)=1$, then $f^(3)(x)geq 0$ for some $xin(-1,1)$There exist disjoint open sets $U_a ni a$ and $U_b ni b$ such that $overlineU_asubseteq U, overlineU_bsubseteq U$How can we prove the scaling property of the Dirac delta function rigorously?
$begingroup$
Let $Omegasubseteq mathbbR^n$ open in $mathbbR^n$ and $fin C^2(Omega,mathbbR^n)$. Let $N(x) = x-[f^prime(x)]^-1f(x)$, supose that:
- there exist $x^astin Omega$ such that $f(x^ast)=0$,
- for all $xinOmega$, we have $N(x)inOmega$.
$f^prime(x^ast)$ is not singular,
How can I prove that $N^prime(x^ast) = 0$?
My attempt. We have $N(x^ast)=x^ast$ if, only if, $f(x^ast)=0$. My strategy is to make Taylor's development of
$$
N(x^ast+v)=N(x^ast)+N^prime(x^ast)v+O(v^2)
$$
and show that the term $N^prime(x^ast)=0$ equals zero. Note that
beginalign
N(x^ast+v)
=&
(x^ast+v)-[f^prime(x^ast+v)]^-1f(x^ast+v)
\
=&
N(x^ast)+v-[f^prime(x^ast+v)]^-1f(x^ast+v)
\
=&
N(x^ast)+v-[f^prime(x^ast)]^-1f(x^ast+v)
\
&qquad+
Big([f^prime(x^ast)]^-1-[f^prime(x^ast+v)]^-1Big)f(x^ast+v)
\
&=
N(x^ast)+v-[f^prime(x^ast)]^-1Big(f(x^ast)+f^prime(x^ast)cdot v+O(v^2)Big)
\
&qquad+
Big([f^prime(x^ast)]^-1-[f^prime(x^ast+v)]^-1Big)Big(f(x^ast)+f^prime(x^ast)cdot v+O(v^2)Big)
\
&=
N(x^ast)+v-[f^prime(x^ast)]^-1Big(f^prime(x^ast)cdot v+O(v^2)Big)
\
&qquad+
Big([f^prime(x^ast)]^-1-[f^prime(x^ast+v)]^-1Big)Big(f^prime(x^ast)cdot v+O(v^2)Big)
\
&=
N(x^ast)+O(v^2)
\
&qquad+
Big([f^prime(x^ast)]^-1-[f^prime(x^ast+v)]^-1Big)Big(f^prime(x^ast)cdot v+O(v^2)Big)
endalign
But I can not eliminate the term $ v $ from the above expression.
real-analysis derivatives taylor-expansion newton-raphson
asked 2 days ago
MathOverviewMathOverview
8,96743164
$endgroup$
add a comment |
$begingroup$
Let $Omegasubseteq mathbbR^n$ open in $mathbbR^n$ and $fin C^2(Omega,mathbbR^n)$. Let $N(x) = x-[f^prime(x)]^-1f(x)$, supose that:
- there exist $x^astin Omega$ such that $f(x^ast)=0$,
- for all $xinOmega$, we have $N(x)inOmega$.
$f^prime(x^ast)$ is not singular,
How can I prove that $N^prime(x^ast) = 0$?
My attempt. We have $N(x^ast)=x^ast$ if, only if, $f(x^ast)=0$. My strategy is to make Taylor's development of
$$
N(x^ast+v)=N(x^ast)+N^prime(x^ast)v+O(v^2)
$$
and show that the term $N^prime(x^ast)=0$ equals zero. Note that
beginalign
N(x^ast+v)
=&
(x^ast+v)-[f^prime(x^ast+v)]^-1f(x^ast+v)
\
=&
N(x^ast)+v-[f^prime(x^ast+v)]^-1f(x^ast+v)
\
=&
N(x^ast)+v-[f^prime(x^ast)]^-1f(x^ast+v)
\
&qquad+
Big([f^prime(x^ast)]^-1-[f^prime(x^ast+v)]^-1Big)f(x^ast+v)
\
&=
N(x^ast)+v-[f^prime(x^ast)]^-1Big(f(x^ast)+f^prime(x^ast)cdot v+O(v^2)Big)
\
&qquad+
Big([f^prime(x^ast)]^-1-[f^prime(x^ast+v)]^-1Big)Big(f(x^ast)+f^prime(x^ast)cdot v+O(v^2)Big)
\
&=
N(x^ast)+v-[f^prime(x^ast)]^-1Big(f^prime(x^ast)cdot v+O(v^2)Big)
\
&qquad+
Big([f^prime(x^ast)]^-1-[f^prime(x^ast+v)]^-1Big)Big(f^prime(x^ast)cdot v+O(v^2)Big)
\
&=
N(x^ast)+O(v^2)
\
&qquad+
Big([f^prime(x^ast)]^-1-[f^prime(x^ast+v)]^-1Big)Big(f^prime(x^ast)cdot v+O(v^2)Big)
endalign
But I can not eliminate the term $ v $ from the above expression.
real-analysis derivatives taylor-expansion newton-raphson
asked 2 days ago
MathOverviewMathOverview
8,96743164
$endgroup$
add a comment |
$begingroup$
Let $Omegasubseteq mathbbR^n$ open in $mathbbR^n$ and $fin C^2(Omega,mathbbR^n)$. Let $N(x) = x-[f^prime(x)]^-1f(x)$, supose that:
- there exist $x^astin Omega$ such that $f(x^ast)=0$,
- for all $xinOmega$, we have $N(x)inOmega$.
$f^prime(x^ast)$ is not singular,
How can I prove that $N^prime(x^ast) = 0$?
My attempt. We have $N(x^ast)=x^ast$ if, only if, $f(x^ast)=0$. My strategy is to make Taylor's development of
$$
N(x^ast+v)=N(x^ast)+N^prime(x^ast)v+O(v^2)
$$
and show that the term $N^prime(x^ast)=0$ equals zero. Note that
beginalign
N(x^ast+v)
=&
(x^ast+v)-[f^prime(x^ast+v)]^-1f(x^ast+v)
\
=&
N(x^ast)+v-[f^prime(x^ast+v)]^-1f(x^ast+v)
\
=&
N(x^ast)+v-[f^prime(x^ast)]^-1f(x^ast+v)
\
&qquad+
Big([f^prime(x^ast)]^-1-[f^prime(x^ast+v)]^-1Big)f(x^ast+v)
\
&=
N(x^ast)+v-[f^prime(x^ast)]^-1Big(f(x^ast)+f^prime(x^ast)cdot v+O(v^2)Big)
\
&qquad+
Big([f^prime(x^ast)]^-1-[f^prime(x^ast+v)]^-1Big)Big(f(x^ast)+f^prime(x^ast)cdot v+O(v^2)Big)
\
&=
N(x^ast)+v-[f^prime(x^ast)]^-1Big(f^prime(x^ast)cdot v+O(v^2)Big)
\
&qquad+
Big([f^prime(x^ast)]^-1-[f^prime(x^ast+v)]^-1Big)Big(f^prime(x^ast)cdot v+O(v^2)Big)
\
&=
N(x^ast)+O(v^2)
\
&qquad+
Big([f^prime(x^ast)]^-1-[f^prime(x^ast+v)]^-1Big)Big(f^prime(x^ast)cdot v+O(v^2)Big)
endalign
But I can not eliminate the term $ v $ from the above expression.
real-analysis derivatives taylor-expansion newton-raphson
asked 2 days ago
MathOverviewMathOverview
8,96743164
$endgroup$
Let $Omegasubseteq mathbbR^n$ open in $mathbbR^n$ and $fin C^2(Omega,mathbbR^n)$. Let $N(x) = x-[f^prime(x)]^-1f(x)$, supose that:
- there exist $x^astin Omega$ such that $f(x^ast)=0$,
- for all $xinOmega$, we have $N(x)inOmega$.
$f^prime(x^ast)$ is not singular,
How can I prove that $N^prime(x^ast) = 0$?
My attempt. We have $N(x^ast)=x^ast$ if, only if, $f(x^ast)=0$. My strategy is to make Taylor's development of
$$
N(x^ast+v)=N(x^ast)+N^prime(x^ast)v+O(v^2)
$$
and show that the term $N^prime(x^ast)=0$ equals zero. Note that
beginalign
N(x^ast+v)
=&
(x^ast+v)-[f^prime(x^ast+v)]^-1f(x^ast+v)
\
=&
N(x^ast)+v-[f^prime(x^ast+v)]^-1f(x^ast+v)
\
=&
N(x^ast)+v-[f^prime(x^ast)]^-1f(x^ast+v)
\
&qquad+
Big([f^prime(x^ast)]^-1-[f^prime(x^ast+v)]^-1Big)f(x^ast+v)
\
&=
N(x^ast)+v-[f^prime(x^ast)]^-1Big(f(x^ast)+f^prime(x^ast)cdot v+O(v^2)Big)
\
&qquad+
Big([f^prime(x^ast)]^-1-[f^prime(x^ast+v)]^-1Big)Big(f(x^ast)+f^prime(x^ast)cdot v+O(v^2)Big)
\
&=
N(x^ast)+v-[f^prime(x^ast)]^-1Big(f^prime(x^ast)cdot v+O(v^2)Big)
\
&qquad+
Big([f^prime(x^ast)]^-1-[f^prime(x^ast+v)]^-1Big)Big(f^prime(x^ast)cdot v+O(v^2)Big)
\
&=
N(x^ast)+O(v^2)
\
&qquad+
Big([f^prime(x^ast)]^-1-[f^prime(x^ast+v)]^-1Big)Big(f^prime(x^ast)cdot v+O(v^2)Big)
endalign
But I can not eliminate the term $ v $ from the above expression.
real-analysis derivatives taylor-expansion newton-raphson
real-analysis derivatives taylor-expansion newton-raphson
asked 2 days ago
MathOverviewMathOverview
8,96743164
asked 2 days ago
MathOverviewMathOverview
8,96743164
edited yesterday
MathOverview
asked 2 days ago
MathOverviewMathOverview
8,96743164
asked 2 days ago
MathOverviewMathOverview
8,96743164
asked 2 days ago
MathOverviewMathOverview
8,96743164
8,96743164
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Why not simply evaluating $N'(x)$ from its definition?
Assuming there exists $g(x) = f^-1(x)$ in a neighborhood around $x^*$, as your problem satisfies the conditions of the Inverse Function Theorem.
Then let $J[N](x)$ denote the Jacobian matrix of $N$ (which is preferable as notation when not working on a single dimension).
Rewriting:
$$
N(x) = x - (J[f](x))^-1f(x)
$$
By inverse function theorem:
$$
N(x) = x - J[g](x)f(x)
$$
Now, taking the Jacobian of the above expression:
$$
N(x) = x - J[g](x)f(x)
$$
Deriving:
$$
J[N](x) = J[x](x) -left(Jleft[J[g](x)right](x)right)f(x)-J[g](x)J[f](x)
$$
The first term is trivial:
$$
J[N](x) = I -left(Jleft[J[g](x)right](x)right)f(x)-J[g](x)J[f](x)
$$
The last term is the identity due to the aforementioned theorem:
$$
J[N](x) = I -left(Jleft[J[g](x)right](x)right)f(x)-I
$$
$$
J[N](x) = -left(Jleft[J[g](x)right](x)right)f(x)
$$
Now, it might be difficult to compute $Jleft[J[g](x)right](x)$, and it is also a 3 dimensional tensor, and this notation may be crappy. But we have that $f(x^*)=0$, hence that term does not matter. So evaluating at $x^*$, we get a null matrix.
answered 2 days ago
MefiticoMefitico
1,184218
$endgroup$
add a comment |
$begingroup$
Note
begineqnarray*
&&lim_tto0fracN(x^*+tv)-N(x^*)t \
&=&lim_tto0bigg v-frac[f^prime(x^*+tv)]^-1f(x^*+tv)-[f^prime(x^*)]^-1f(x^*)tbigg\
&=&v-lim_tto0frac[f^prime(x^*+tv)]^-1[f(x^*+tv)-f(x^*)]t-lim_tto0frac[f^prime(x^*+tv)]^-1-[f^prime(x^*)]^-1f(x^*)t\
&=&v-v=0.
endeqnarray*
answered yesterday
xpaulxpaul
23.4k24655
$endgroup$
add a comment |
Your Answer
StackExchange.ifUsing("editor", function ()
return StackExchange.using("mathjaxEditing", function ()
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix)
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
);
);
, "mathjax-editing");
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
);
);
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%2f3164657%2fif-nx-x-f-primex-1fx-and-fx-ast-0-how-can-i-prove-that%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
$begingroup$
Why not simply evaluating $N'(x)$ from its definition?
Assuming there exists $g(x) = f^-1(x)$ in a neighborhood around $x^*$, as your problem satisfies the conditions of the Inverse Function Theorem.
Then let $J[N](x)$ denote the Jacobian matrix of $N$ (which is preferable as notation when not working on a single dimension).
Rewriting:
$$
N(x) = x - (J[f](x))^-1f(x)
$$
By inverse function theorem:
$$
N(x) = x - J[g](x)f(x)
$$
Now, taking the Jacobian of the above expression:
$$
N(x) = x - J[g](x)f(x)
$$
Deriving:
$$
J[N](x) = J[x](x) -left(Jleft[J[g](x)right](x)right)f(x)-J[g](x)J[f](x)
$$
The first term is trivial:
$$
J[N](x) = I -left(Jleft[J[g](x)right](x)right)f(x)-J[g](x)J[f](x)
$$
The last term is the identity due to the aforementioned theorem:
$$
J[N](x) = I -left(Jleft[J[g](x)right](x)right)f(x)-I
$$
$$
J[N](x) = -left(Jleft[J[g](x)right](x)right)f(x)
$$
Now, it might be difficult to compute $Jleft[J[g](x)right](x)$, and it is also a 3 dimensional tensor, and this notation may be crappy. But we have that $f(x^*)=0$, hence that term does not matter. So evaluating at $x^*$, we get a null matrix.
answered 2 days ago
MefiticoMefitico
1,184218
$endgroup$
add a comment |
$begingroup$
Why not simply evaluating $N'(x)$ from its definition?
Assuming there exists $g(x) = f^-1(x)$ in a neighborhood around $x^*$, as your problem satisfies the conditions of the Inverse Function Theorem.
Then let $J[N](x)$ denote the Jacobian matrix of $N$ (which is preferable as notation when not working on a single dimension).
Rewriting:
$$
N(x) = x - (J[f](x))^-1f(x)
$$
By inverse function theorem:
$$
N(x) = x - J[g](x)f(x)
$$
Now, taking the Jacobian of the above expression:
$$
N(x) = x - J[g](x)f(x)
$$
Deriving:
$$
J[N](x) = J[x](x) -left(Jleft[J[g](x)right](x)right)f(x)-J[g](x)J[f](x)
$$
The first term is trivial:
$$
J[N](x) = I -left(Jleft[J[g](x)right](x)right)f(x)-J[g](x)J[f](x)
$$
The last term is the identity due to the aforementioned theorem:
$$
J[N](x) = I -left(Jleft[J[g](x)right](x)right)f(x)-I
$$
$$
J[N](x) = -left(Jleft[J[g](x)right](x)right)f(x)
$$
Now, it might be difficult to compute $Jleft[J[g](x)right](x)$, and it is also a 3 dimensional tensor, and this notation may be crappy. But we have that $f(x^*)=0$, hence that term does not matter. So evaluating at $x^*$, we get a null matrix.
answered 2 days ago
MefiticoMefitico
1,184218
$endgroup$
add a comment |
$begingroup$
Why not simply evaluating $N'(x)$ from its definition?
Assuming there exists $g(x) = f^-1(x)$ in a neighborhood around $x^*$, as your problem satisfies the conditions of the Inverse Function Theorem.
Then let $J[N](x)$ denote the Jacobian matrix of $N$ (which is preferable as notation when not working on a single dimension).
Rewriting:
$$
N(x) = x - (J[f](x))^-1f(x)
$$
By inverse function theorem:
$$
N(x) = x - J[g](x)f(x)
$$
Now, taking the Jacobian of the above expression:
$$
N(x) = x - J[g](x)f(x)
$$
Deriving:
$$
J[N](x) = J[x](x) -left(Jleft[J[g](x)right](x)right)f(x)-J[g](x)J[f](x)
$$
The first term is trivial:
$$
J[N](x) = I -left(Jleft[J[g](x)right](x)right)f(x)-J[g](x)J[f](x)
$$
The last term is the identity due to the aforementioned theorem:
$$
J[N](x) = I -left(Jleft[J[g](x)right](x)right)f(x)-I
$$
$$
J[N](x) = -left(Jleft[J[g](x)right](x)right)f(x)
$$
Now, it might be difficult to compute $Jleft[J[g](x)right](x)$, and it is also a 3 dimensional tensor, and this notation may be crappy. But we have that $f(x^*)=0$, hence that term does not matter. So evaluating at $x^*$, we get a null matrix.
answered 2 days ago
MefiticoMefitico
1,184218
$endgroup$
Why not simply evaluating $N'(x)$ from its definition?
Assuming there exists $g(x) = f^-1(x)$ in a neighborhood around $x^*$, as your problem satisfies the conditions of the Inverse Function Theorem.
Then let $J[N](x)$ denote the Jacobian matrix of $N$ (which is preferable as notation when not working on a single dimension).
Rewriting:
$$
N(x) = x - (J[f](x))^-1f(x)
$$
By inverse function theorem:
$$
N(x) = x - J[g](x)f(x)
$$
Now, taking the Jacobian of the above expression:
$$
N(x) = x - J[g](x)f(x)
$$
Deriving:
$$
J[N](x) = J[x](x) -left(Jleft[J[g](x)right](x)right)f(x)-J[g](x)J[f](x)
$$
The first term is trivial:
$$
J[N](x) = I -left(Jleft[J[g](x)right](x)right)f(x)-J[g](x)J[f](x)
$$
The last term is the identity due to the aforementioned theorem:
$$
J[N](x) = I -left(Jleft[J[g](x)right](x)right)f(x)-I
$$
$$
J[N](x) = -left(Jleft[J[g](x)right](x)right)f(x)
$$
Now, it might be difficult to compute $Jleft[J[g](x)right](x)$, and it is also a 3 dimensional tensor, and this notation may be crappy. But we have that $f(x^*)=0$, hence that term does not matter. So evaluating at $x^*$, we get a null matrix.
answered 2 days ago
MefiticoMefitico
1,184218
edited yesterday
answered 2 days ago
MefiticoMefitico
1,184218
answered 2 days ago
MefiticoMefitico
1,184218
answered 2 days ago
MefiticoMefitico
1,184218
1,184218
add a comment |
add a comment |
$begingroup$
Note
begineqnarray*
&&lim_tto0fracN(x^*+tv)-N(x^*)t \
&=&lim_tto0bigg v-frac[f^prime(x^*+tv)]^-1f(x^*+tv)-[f^prime(x^*)]^-1f(x^*)tbigg\
&=&v-lim_tto0frac[f^prime(x^*+tv)]^-1[f(x^*+tv)-f(x^*)]t-lim_tto0frac[f^prime(x^*+tv)]^-1-[f^prime(x^*)]^-1f(x^*)t\
&=&v-v=0.
endeqnarray*
answered yesterday
xpaulxpaul
23.4k24655
$endgroup$
add a comment |
$begingroup$
Note
begineqnarray*
&&lim_tto0fracN(x^*+tv)-N(x^*)t \
&=&lim_tto0bigg v-frac[f^prime(x^*+tv)]^-1f(x^*+tv)-[f^prime(x^*)]^-1f(x^*)tbigg\
&=&v-lim_tto0frac[f^prime(x^*+tv)]^-1[f(x^*+tv)-f(x^*)]t-lim_tto0frac[f^prime(x^*+tv)]^-1-[f^prime(x^*)]^-1f(x^*)t\
&=&v-v=0.
endeqnarray*
answered yesterday
xpaulxpaul
23.4k24655
$endgroup$
add a comment |
$begingroup$
Note
begineqnarray*
&&lim_tto0fracN(x^*+tv)-N(x^*)t \
&=&lim_tto0bigg v-frac[f^prime(x^*+tv)]^-1f(x^*+tv)-[f^prime(x^*)]^-1f(x^*)tbigg\
&=&v-lim_tto0frac[f^prime(x^*+tv)]^-1[f(x^*+tv)-f(x^*)]t-lim_tto0frac[f^prime(x^*+tv)]^-1-[f^prime(x^*)]^-1f(x^*)t\
&=&v-v=0.
endeqnarray*
answered yesterday
xpaulxpaul
23.4k24655
$endgroup$
Note
begineqnarray*
&&lim_tto0fracN(x^*+tv)-N(x^*)t \
&=&lim_tto0bigg v-frac[f^prime(x^*+tv)]^-1f(x^*+tv)-[f^prime(x^*)]^-1f(x^*)tbigg\
&=&v-lim_tto0frac[f^prime(x^*+tv)]^-1[f(x^*+tv)-f(x^*)]t-lim_tto0frac[f^prime(x^*+tv)]^-1-[f^prime(x^*)]^-1f(x^*)t\
&=&v-v=0.
endeqnarray*
answered yesterday
xpaulxpaul
23.4k24655
answered yesterday
xpaulxpaul
23.4k24655
answered yesterday
xpaulxpaul
23.4k24655
answered yesterday
xpaulxpaul
23.4k24655
23.4k24655
add a comment |
add a comment |
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%2f3164657%2fif-nx-x-f-primex-1fx-and-fx-ast-0-how-can-i-prove-that%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
