Usbrth

How to set letter above or below the symbol?

Can a zero nonce be safely used with AES-GCM if the key is random and never used again?

What is the order of Mitzvot in Rambam's Sefer Hamitzvot?

What items from the Roman-age tech-level could be used to deter all creatures from entering a small area?

Is above average number of years spent on PhD considered a red flag in future academia or industry positions?

What computer would be fastest for Mathematica Home Edition?

Using "nakedly" instead of "with nothing on"

Statistical model of ligand substitution

How does modal jazz use chord progressions?

Writing Thesis: Copying from published papers

Passing functions in C++

How should I respond to a player wanting to catch a sword between their hands?

Determine whether f is a function, an injection, a surjection

Replacing HDD with SSD; what about non-APFS/APFS?

Antler Helmet: Can it work?

Is drag coefficient lowest at zero angle of attack?

Is it possible to ask for a hotel room without minibar/extra services?

How many things? AとBがふたつ

No baking right

What can I do if my MacBook isn’t charging but already ran out?

If I can make up priors, why can't I make up posteriors?

How to say that you spent the night with someone, you were only sleeping and nothing else?

Unable to start mainnet node docker container

Unexpected result with right shift after bitwise negation

Norm of the residual

0

$begingroup$

Let A be the tridiagonal finite-element matrix for the one-dimensional diffusion equation, let $Ax = 0, A in R^nxn$ Implement the weighted-Jacobi and Gauss-Seidel iterations to solve $Ax = 0$. Consider relative reduction in residual norm $frac|}, l$ is the number of iteration. What can you say about the “asymptotic” convergence as $l$ gets large? Does this depend on $n$?

My intuition says, that since we are getting closer to the solution with each iteration $|0−Ax^l|$ and $|0−Ax^l−1|$ would be equal for large $l$, so the fraction goes to $1$, and this doesn't seem to depend on $n$. But I'm not sure, can someone give me a formal proof?

numerical-methods numerical-linear-algebra

share|cite|improve this question

edited Apr 9 at 13:08

dxdydz

asked Apr 8 at 20:35

dxdydzdxdydz

49110

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  How does $A$ depend on $n$?
  $endgroup$
  – Carl Christian
  Apr 8 at 21:11
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @CarlChristian, $n$ is the size of $A$...
  $endgroup$
  – dxdydz
  Apr 9 at 4:59
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  If we are free to choose the entries of the matrices independent of the dimension, then we can get any result we want. If on the other hand, there is an underlying pattern present, then definite conclusions are possible.
  $endgroup$
  – Carl Christian
  Apr 9 at 5:54
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  There is more than one matrix which fits the description which you have added. Are you thinking of the discrete Laplacian corresponding to a uniform grid with $n+1$ points?
  $endgroup$
  – Carl Christian
  Apr 9 at 15:37
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @CarlChristian, thats right
  $endgroup$
  – dxdydz
  Apr 9 at 16:04
  

  
  
  
  

add a comment | 

0

$begingroup$

Let A be the tridiagonal finite-element matrix for the one-dimensional diffusion equation, let $Ax = 0, A in R^nxn$ Implement the weighted-Jacobi and Gauss-Seidel iterations to solve $Ax = 0$. Consider relative reduction in residual norm $frac|}, l$ is the number of iteration. What can you say about the “asymptotic” convergence as $l$ gets large? Does this depend on $n$?

My intuition says, that since we are getting closer to the solution with each iteration $|0−Ax^l|$ and $|0−Ax^l−1|$ would be equal for large $l$, so the fraction goes to $1$, and this doesn't seem to depend on $n$. But I'm not sure, can someone give me a formal proof?

numerical-methods numerical-linear-algebra

share|cite|improve this question

edited Apr 9 at 13:08

dxdydz

asked Apr 8 at 20:35

dxdydzdxdydz

49110

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  How does $A$ depend on $n$?
  $endgroup$
  – Carl Christian
  Apr 8 at 21:11
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @CarlChristian, $n$ is the size of $A$...
  $endgroup$
  – dxdydz
  Apr 9 at 4:59
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  If we are free to choose the entries of the matrices independent of the dimension, then we can get any result we want. If on the other hand, there is an underlying pattern present, then definite conclusions are possible.
  $endgroup$
  – Carl Christian
  Apr 9 at 5:54
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  There is more than one matrix which fits the description which you have added. Are you thinking of the discrete Laplacian corresponding to a uniform grid with $n+1$ points?
  $endgroup$
  – Carl Christian
  Apr 9 at 15:37
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @CarlChristian, thats right
  $endgroup$
  – dxdydz
  Apr 9 at 16:04
  

  
  
  
  

add a comment | 

$begingroup$

Let A be the tridiagonal finite-element matrix for the one-dimensional diffusion equation, let $Ax = 0, A in R^nxn$ Implement the weighted-Jacobi and Gauss-Seidel iterations to solve $Ax = 0$. Consider relative reduction in residual norm $frac|}, l$ is the number of iteration. What can you say about the “asymptotic” convergence as $l$ gets large? Does this depend on $n$?

My intuition says, that since we are getting closer to the solution with each iteration $|0−Ax^l|$ and $|0−Ax^l−1|$ would be equal for large $l$, so the fraction goes to $1$, and this doesn't seem to depend on $n$. But I'm not sure, can someone give me a formal proof?

numerical-methods numerical-linear-algebra

share|cite|improve this question

edited Apr 9 at 13:08

dxdydz

asked Apr 8 at 20:35

dxdydzdxdydz

49110

$endgroup$

share|cite|improve this question

edited Apr 9 at 13:08

dxdydz

asked Apr 8 at 20:35

dxdydzdxdydz

49110

share|cite|improve this question

edited Apr 9 at 13:08

dxdydz

asked Apr 8 at 20:35

dxdydzdxdydz

49110

asked Apr 8 at 20:35

dxdydzdxdydz

49110

asked Apr 8 at 20:35

dxdydzdxdydz

49110

1 Answer
1

active

oldest

votes

0

$begingroup$

Without further information on $A$ there is no way to answer your questions. Regarding your intuition, depending on $A$, the methods may not even converge for every initial approximation, so the "getting closer to the solution with each iteration..." may not occur.

share|cite|improve this answer

answered Apr 9 at 8:12

PierreCarrePierreCarre

2,158215

$endgroup$

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
);



);

draft saved

draft discarded

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

Name

Email

Required, but never shown

StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3180170%2fnorm-of-the-residual%23new-answer', 'question_page');

);

Post as a guest

Name

Email

Required, but never shown

0

$begingroup$

Without further information on $A$ there is no way to answer your questions. Regarding your intuition, depending on $A$, the methods may not even converge for every initial approximation, so the "getting closer to the solution with each iteration..." may not occur.

share|cite|improve this answer

answered Apr 9 at 8:12

PierreCarrePierreCarre

2,158215

$endgroup$

add a comment | 

0

$begingroup$

Without further information on $A$ there is no way to answer your questions. Regarding your intuition, depending on $A$, the methods may not even converge for every initial approximation, so the "getting closer to the solution with each iteration..." may not occur.

share|cite|improve this answer

answered Apr 9 at 8:12

PierreCarrePierreCarre

2,158215

$endgroup$

add a comment | 

$begingroup$

Without further information on $A$ there is no way to answer your questions. Regarding your intuition, depending on $A$, the methods may not even converge for every initial approximation, so the "getting closer to the solution with each iteration..." may not occur.

share|cite|improve this answer

answered Apr 9 at 8:12

PierreCarrePierreCarre

2,158215

$endgroup$

share|cite|improve this answer

answered Apr 9 at 8:12

PierreCarrePierreCarre

2,158215

answered Apr 9 at 8:12

PierreCarrePierreCarre

2,158215

answered Apr 9 at 8:12

PierreCarrePierreCarre

2,158215