Does calculating the limit of a given real-valued (recursive) sequence already imply its convergence?












0












$begingroup$


Assuming we are given a real-valued recursive sequence $(a_n)_{n in mathbb{N}}$ by its starting point $a_1$ and its recursive function $a_{n+1} = varphi(a_n)$



EDIT: Since this question caused some confusion or i failed to be clear i'd like to add the recursive sequence:



$$a_1 = sqrt{2}, a_{n+1} = sqrt{2a_n}$$



In order to prove the convergence of $(a_n)_{n}$ i would try to prove boundedness and monotonicity (if that's the case). Then i would try to compute its limit.



However, a few days ago i've been asked whether it would not be sufficient enough to simply compute $$lim_{n to infty} a_n$$



without even proving whether or not $(a_n)_n$ is convergent at all.



Ive been asked that if we basically could find an $a$ for which $$ lim_{n to infty} a_n = a$$ holds
we could immediately conclude that the given sequence was convergent in the first place.



I was quite sceptical but i failed to provide a reasonable answer why it would certainly not sufficient enough. My first reaction was that it might be possible that $a$ might not be the limit but only an accumulation point. But i'm not sure if that's either the case or if i'm completely mistaken.



Can you possibly help me with an answer to that question? Can i simply compute $lim_{n to infty} a_n = a$ prior to proving the existence of such an $a$ at all?



This is all restricted to $mathbb{R}$ but any general answer is highly appreciated as well.



Thank you very much for any help.






real-analysis calculus sequences-and-series convergence







share|cite|improve this question











$endgroup$












  • $begingroup$
    I do not know what the exact question is. With $varphi(x) = x^2$ your hypotetical limit will satisfy $a=a^2$ which would give $a=1$ or $a=0$. But of course the sequence diverges for $|a_1| >1$
    $endgroup$
    – gammatester
    Dec 6 '18 at 22:32












  • $begingroup$
    I don't know what you are asking. A sequence is convergent if and only if the limit exists. What distinction are you making?
    $endgroup$
    – saulspatz
    Dec 6 '18 at 22:40










  • $begingroup$
    If $limlimits_{n to infty} a_n$ exists then the sequence converges, but it difficult to see how you know it exists just by calculating successive terms which appear to approach a particular value without reaching it
    $endgroup$
    – Henry
    Dec 6 '18 at 22:41










  • $begingroup$
    Sorry for any confusion i might have caused. I will edit my question to provide a sequence
    $endgroup$
    – Zest
    Dec 6 '18 at 22:41










  • $begingroup$
    It's still not clear what you're asking here; I think explaining how you compute the limit would help matters here. Note that saying e.g. 'the limit is the unique value $L$ satisfying $L=sqrt{2L}$' isn't actually computation in any relevant sense...
    $endgroup$
    – Steven Stadnicki
    Dec 6 '18 at 22:49


















0












$begingroup$


Assuming we are given a real-valued recursive sequence $(a_n)_{n in mathbb{N}}$ by its starting point $a_1$ and its recursive function $a_{n+1} = varphi(a_n)$



EDIT: Since this question caused some confusion or i failed to be clear i'd like to add the recursive sequence:



$$a_1 = sqrt{2}, a_{n+1} = sqrt{2a_n}$$



In order to prove the convergence of $(a_n)_{n}$ i would try to prove boundedness and monotonicity (if that's the case). Then i would try to compute its limit.



However, a few days ago i've been asked whether it would not be sufficient enough to simply compute $$lim_{n to infty} a_n$$



without even proving whether or not $(a_n)_n$ is convergent at all.



Ive been asked that if we basically could find an $a$ for which $$ lim_{n to infty} a_n = a$$ holds
we could immediately conclude that the given sequence was convergent in the first place.



I was quite sceptical but i failed to provide a reasonable answer why it would certainly not sufficient enough. My first reaction was that it might be possible that $a$ might not be the limit but only an accumulation point. But i'm not sure if that's either the case or if i'm completely mistaken.



Can you possibly help me with an answer to that question? Can i simply compute $lim_{n to infty} a_n = a$ prior to proving the existence of such an $a$ at all?



This is all restricted to $mathbb{R}$ but any general answer is highly appreciated as well.



Thank you very much for any help.






real-analysis calculus sequences-and-series convergence







share|cite|improve this question











$endgroup$












  • $begingroup$
    I do not know what the exact question is. With $varphi(x) = x^2$ your hypotetical limit will satisfy $a=a^2$ which would give $a=1$ or $a=0$. But of course the sequence diverges for $|a_1| >1$
    $endgroup$
    – gammatester
    Dec 6 '18 at 22:32












  • $begingroup$
    I don't know what you are asking. A sequence is convergent if and only if the limit exists. What distinction are you making?
    $endgroup$
    – saulspatz
    Dec 6 '18 at 22:40










  • $begingroup$
    If $limlimits_{n to infty} a_n$ exists then the sequence converges, but it difficult to see how you know it exists just by calculating successive terms which appear to approach a particular value without reaching it
    $endgroup$
    – Henry
    Dec 6 '18 at 22:41










  • $begingroup$
    Sorry for any confusion i might have caused. I will edit my question to provide a sequence
    $endgroup$
    – Zest
    Dec 6 '18 at 22:41










  • $begingroup$
    It's still not clear what you're asking here; I think explaining how you compute the limit would help matters here. Note that saying e.g. 'the limit is the unique value $L$ satisfying $L=sqrt{2L}$' isn't actually computation in any relevant sense...
    $endgroup$
    – Steven Stadnicki
    Dec 6 '18 at 22:49
















0












0








0





$begingroup$


Assuming we are given a real-valued recursive sequence $(a_n)_{n in mathbb{N}}$ by its starting point $a_1$ and its recursive function $a_{n+1} = varphi(a_n)$



EDIT: Since this question caused some confusion or i failed to be clear i'd like to add the recursive sequence:



$$a_1 = sqrt{2}, a_{n+1} = sqrt{2a_n}$$



In order to prove the convergence of $(a_n)_{n}$ i would try to prove boundedness and monotonicity (if that's the case). Then i would try to compute its limit.



However, a few days ago i've been asked whether it would not be sufficient enough to simply compute $$lim_{n to infty} a_n$$



without even proving whether or not $(a_n)_n$ is convergent at all.



Ive been asked that if we basically could find an $a$ for which $$ lim_{n to infty} a_n = a$$ holds
we could immediately conclude that the given sequence was convergent in the first place.



I was quite sceptical but i failed to provide a reasonable answer why it would certainly not sufficient enough. My first reaction was that it might be possible that $a$ might not be the limit but only an accumulation point. But i'm not sure if that's either the case or if i'm completely mistaken.



Can you possibly help me with an answer to that question? Can i simply compute $lim_{n to infty} a_n = a$ prior to proving the existence of such an $a$ at all?



This is all restricted to $mathbb{R}$ but any general answer is highly appreciated as well.



Thank you very much for any help.






real-analysis calculus sequences-and-series convergence







share|cite|improve this question











$endgroup$




Assuming we are given a real-valued recursive sequence $(a_n)_{n in mathbb{N}}$ by its starting point $a_1$ and its recursive function $a_{n+1} = varphi(a_n)$



EDIT: Since this question caused some confusion or i failed to be clear i'd like to add the recursive sequence:



$$a_1 = sqrt{2}, a_{n+1} = sqrt{2a_n}$$



In order to prove the convergence of $(a_n)_{n}$ i would try to prove boundedness and monotonicity (if that's the case). Then i would try to compute its limit.



However, a few days ago i've been asked whether it would not be sufficient enough to simply compute $$lim_{n to infty} a_n$$



without even proving whether or not $(a_n)_n$ is convergent at all.



Ive been asked that if we basically could find an $a$ for which $$ lim_{n to infty} a_n = a$$ holds
we could immediately conclude that the given sequence was convergent in the first place.



I was quite sceptical but i failed to provide a reasonable answer why it would certainly not sufficient enough. My first reaction was that it might be possible that $a$ might not be the limit but only an accumulation point. But i'm not sure if that's either the case or if i'm completely mistaken.



Can you possibly help me with an answer to that question? Can i simply compute $lim_{n to infty} a_n = a$ prior to proving the existence of such an $a$ at all?



This is all restricted to $mathbb{R}$ but any general answer is highly appreciated as well.



Thank you very much for any help.






real-analysis calculus sequences-and-series convergence




real-analysis calculus sequences-and-series convergence






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 6 '18 at 22:43







Zest

















asked Dec 6 '18 at 22:18









ZestZest

32111




32111












  • $begingroup$
    I do not know what the exact question is. With $varphi(x) = x^2$ your hypotetical limit will satisfy $a=a^2$ which would give $a=1$ or $a=0$. But of course the sequence diverges for $|a_1| >1$
    $endgroup$
    – gammatester
    Dec 6 '18 at 22:32












  • $begingroup$
    I don't know what you are asking. A sequence is convergent if and only if the limit exists. What distinction are you making?
    $endgroup$
    – saulspatz
    Dec 6 '18 at 22:40










  • $begingroup$
    If $limlimits_{n to infty} a_n$ exists then the sequence converges, but it difficult to see how you know it exists just by calculating successive terms which appear to approach a particular value without reaching it
    $endgroup$
    – Henry
    Dec 6 '18 at 22:41










  • $begingroup$
    Sorry for any confusion i might have caused. I will edit my question to provide a sequence
    $endgroup$
    – Zest
    Dec 6 '18 at 22:41










  • $begingroup$
    It's still not clear what you're asking here; I think explaining how you compute the limit would help matters here. Note that saying e.g. 'the limit is the unique value $L$ satisfying $L=sqrt{2L}$' isn't actually computation in any relevant sense...
    $endgroup$
    – Steven Stadnicki
    Dec 6 '18 at 22:49




















  • $begingroup$
    I do not know what the exact question is. With $varphi(x) = x^2$ your hypotetical limit will satisfy $a=a^2$ which would give $a=1$ or $a=0$. But of course the sequence diverges for $|a_1| >1$
    $endgroup$
    – gammatester
    Dec 6 '18 at 22:32












  • $begingroup$
    I don't know what you are asking. A sequence is convergent if and only if the limit exists. What distinction are you making?
    $endgroup$
    – saulspatz
    Dec 6 '18 at 22:40










  • $begingroup$
    If $limlimits_{n to infty} a_n$ exists then the sequence converges, but it difficult to see how you know it exists just by calculating successive terms which appear to approach a particular value without reaching it
    $endgroup$
    – Henry
    Dec 6 '18 at 22:41










  • $begingroup$
    Sorry for any confusion i might have caused. I will edit my question to provide a sequence
    $endgroup$
    – Zest
    Dec 6 '18 at 22:41










  • $begingroup$
    It's still not clear what you're asking here; I think explaining how you compute the limit would help matters here. Note that saying e.g. 'the limit is the unique value $L$ satisfying $L=sqrt{2L}$' isn't actually computation in any relevant sense...
    $endgroup$
    – Steven Stadnicki
    Dec 6 '18 at 22:49


















$begingroup$
I do not know what the exact question is. With $varphi(x) = x^2$ your hypotetical limit will satisfy $a=a^2$ which would give $a=1$ or $a=0$. But of course the sequence diverges for $|a_1| >1$
$endgroup$
– gammatester
Dec 6 '18 at 22:32






$begingroup$
I do not know what the exact question is. With $varphi(x) = x^2$ your hypotetical limit will satisfy $a=a^2$ which would give $a=1$ or $a=0$. But of course the sequence diverges for $|a_1| >1$
$endgroup$
– gammatester
Dec 6 '18 at 22:32














$begingroup$
I don't know what you are asking. A sequence is convergent if and only if the limit exists. What distinction are you making?
$endgroup$
– saulspatz
Dec 6 '18 at 22:40




$begingroup$
I don't know what you are asking. A sequence is convergent if and only if the limit exists. What distinction are you making?
$endgroup$
– saulspatz
Dec 6 '18 at 22:40












$begingroup$
If $limlimits_{n to infty} a_n$ exists then the sequence converges, but it difficult to see how you know it exists just by calculating successive terms which appear to approach a particular value without reaching it
$endgroup$
– Henry
Dec 6 '18 at 22:41




$begingroup$
If $limlimits_{n to infty} a_n$ exists then the sequence converges, but it difficult to see how you know it exists just by calculating successive terms which appear to approach a particular value without reaching it
$endgroup$
– Henry
Dec 6 '18 at 22:41












$begingroup$
Sorry for any confusion i might have caused. I will edit my question to provide a sequence
$endgroup$
– Zest
Dec 6 '18 at 22:41




$begingroup$
Sorry for any confusion i might have caused. I will edit my question to provide a sequence
$endgroup$
– Zest
Dec 6 '18 at 22:41












$begingroup$
It's still not clear what you're asking here; I think explaining how you compute the limit would help matters here. Note that saying e.g. 'the limit is the unique value $L$ satisfying $L=sqrt{2L}$' isn't actually computation in any relevant sense...
$endgroup$
– Steven Stadnicki
Dec 6 '18 at 22:49






$begingroup$
It's still not clear what you're asking here; I think explaining how you compute the limit would help matters here. Note that saying e.g. 'the limit is the unique value $L$ satisfying $L=sqrt{2L}$' isn't actually computation in any relevant sense...
$endgroup$
– Steven Stadnicki
Dec 6 '18 at 22:49












1 Answer
1






active

oldest

votes


















4












$begingroup$

Take a sequence $(a_n)$ defined by $a_0=0$ and $a_n=2a_{n-1}+1$. If you assume that the limit $L=lim_{ntoinfty}(a_n)$ exists then by taking the limit of both sides you get $L=-1$. Does this answer your question?






share|cite|improve this answer











$endgroup$













  • $begingroup$
    this is exactly what i was looking for. i really appreciate every comment and your answer. thank you very much.
    $endgroup$
    – Zest
    Dec 6 '18 at 22:54






  • 1




    $begingroup$
    @Zest You are welcome! I’m glad to help.
    $endgroup$
    – William Sun
    Dec 6 '18 at 22:55










  • $begingroup$
    one last question: the given sequence isn't convergent since it is not bounded right?
    $endgroup$
    – Zest
    Dec 6 '18 at 23:06






  • 1




    $begingroup$
    @Zest The answer is Yes assuming you are working in $mathbb{R}$, in which case all convergent sequences are bounded.
    $endgroup$
    – William Sun
    Dec 6 '18 at 23:08






  • 1




    $begingroup$
    @Zest $L=-1$ is not a limit for the recursively defined sequence in my example. The reasoning is “If the limit of the sequence exists, then it must be $-1$.” But this does not provide with any clue whether the limit exists or not.
    $endgroup$
    – William Sun
    Dec 6 '18 at 23:19











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
});


}
});














draft saved

draft discarded


















StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3029129%2fdoes-calculating-the-limit-of-a-given-real-valued-recursive-sequence-already-i%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown

























1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









4












$begingroup$

Take a sequence $(a_n)$ defined by $a_0=0$ and $a_n=2a_{n-1}+1$. If you assume that the limit $L=lim_{ntoinfty}(a_n)$ exists then by taking the limit of both sides you get $L=-1$. Does this answer your question?






share|cite|improve this answer











$endgroup$













  • $begingroup$
    this is exactly what i was looking for. i really appreciate every comment and your answer. thank you very much.
    $endgroup$
    – Zest
    Dec 6 '18 at 22:54






  • 1




    $begingroup$
    @Zest You are welcome! I’m glad to help.
    $endgroup$
    – William Sun
    Dec 6 '18 at 22:55










  • $begingroup$
    one last question: the given sequence isn't convergent since it is not bounded right?
    $endgroup$
    – Zest
    Dec 6 '18 at 23:06






  • 1




    $begingroup$
    @Zest The answer is Yes assuming you are working in $mathbb{R}$, in which case all convergent sequences are bounded.
    $endgroup$
    – William Sun
    Dec 6 '18 at 23:08






  • 1




    $begingroup$
    @Zest $L=-1$ is not a limit for the recursively defined sequence in my example. The reasoning is “If the limit of the sequence exists, then it must be $-1$.” But this does not provide with any clue whether the limit exists or not.
    $endgroup$
    – William Sun
    Dec 6 '18 at 23:19
















4












$begingroup$

Take a sequence $(a_n)$ defined by $a_0=0$ and $a_n=2a_{n-1}+1$. If you assume that the limit $L=lim_{ntoinfty}(a_n)$ exists then by taking the limit of both sides you get $L=-1$. Does this answer your question?






share|cite|improve this answer











$endgroup$













  • $begingroup$
    this is exactly what i was looking for. i really appreciate every comment and your answer. thank you very much.
    $endgroup$
    – Zest
    Dec 6 '18 at 22:54






  • 1




    $begingroup$
    @Zest You are welcome! I’m glad to help.
    $endgroup$
    – William Sun
    Dec 6 '18 at 22:55










  • $begingroup$
    one last question: the given sequence isn't convergent since it is not bounded right?
    $endgroup$
    – Zest
    Dec 6 '18 at 23:06






  • 1




    $begingroup$
    @Zest The answer is Yes assuming you are working in $mathbb{R}$, in which case all convergent sequences are bounded.
    $endgroup$
    – William Sun
    Dec 6 '18 at 23:08






  • 1




    $begingroup$
    @Zest $L=-1$ is not a limit for the recursively defined sequence in my example. The reasoning is “If the limit of the sequence exists, then it must be $-1$.” But this does not provide with any clue whether the limit exists or not.
    $endgroup$
    – William Sun
    Dec 6 '18 at 23:19














4












4








4





$begingroup$

Take a sequence $(a_n)$ defined by $a_0=0$ and $a_n=2a_{n-1}+1$. If you assume that the limit $L=lim_{ntoinfty}(a_n)$ exists then by taking the limit of both sides you get $L=-1$. Does this answer your question?






share|cite|improve this answer











$endgroup$



Take a sequence $(a_n)$ defined by $a_0=0$ and $a_n=2a_{n-1}+1$. If you assume that the limit $L=lim_{ntoinfty}(a_n)$ exists then by taking the limit of both sides you get $L=-1$. Does this answer your question?







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 6 '18 at 22:58

























answered Dec 6 '18 at 22:51









William SunWilliam Sun

471111




471111












  • $begingroup$
    this is exactly what i was looking for. i really appreciate every comment and your answer. thank you very much.
    $endgroup$
    – Zest
    Dec 6 '18 at 22:54






  • 1




    $begingroup$
    @Zest You are welcome! I’m glad to help.
    $endgroup$
    – William Sun
    Dec 6 '18 at 22:55










  • $begingroup$
    one last question: the given sequence isn't convergent since it is not bounded right?
    $endgroup$
    – Zest
    Dec 6 '18 at 23:06






  • 1




    $begingroup$
    @Zest The answer is Yes assuming you are working in $mathbb{R}$, in which case all convergent sequences are bounded.
    $endgroup$
    – William Sun
    Dec 6 '18 at 23:08






  • 1




    $begingroup$
    @Zest $L=-1$ is not a limit for the recursively defined sequence in my example. The reasoning is “If the limit of the sequence exists, then it must be $-1$.” But this does not provide with any clue whether the limit exists or not.
    $endgroup$
    – William Sun
    Dec 6 '18 at 23:19


















  • $begingroup$
    this is exactly what i was looking for. i really appreciate every comment and your answer. thank you very much.
    $endgroup$
    – Zest
    Dec 6 '18 at 22:54






  • 1




    $begingroup$
    @Zest You are welcome! I’m glad to help.
    $endgroup$
    – William Sun
    Dec 6 '18 at 22:55










  • $begingroup$
    one last question: the given sequence isn't convergent since it is not bounded right?
    $endgroup$
    – Zest
    Dec 6 '18 at 23:06






  • 1




    $begingroup$
    @Zest The answer is Yes assuming you are working in $mathbb{R}$, in which case all convergent sequences are bounded.
    $endgroup$
    – William Sun
    Dec 6 '18 at 23:08






  • 1




    $begingroup$
    @Zest $L=-1$ is not a limit for the recursively defined sequence in my example. The reasoning is “If the limit of the sequence exists, then it must be $-1$.” But this does not provide with any clue whether the limit exists or not.
    $endgroup$
    – William Sun
    Dec 6 '18 at 23:19
















$begingroup$
this is exactly what i was looking for. i really appreciate every comment and your answer. thank you very much.
$endgroup$
– Zest
Dec 6 '18 at 22:54




$begingroup$
this is exactly what i was looking for. i really appreciate every comment and your answer. thank you very much.
$endgroup$
– Zest
Dec 6 '18 at 22:54




1




1




$begingroup$
@Zest You are welcome! I’m glad to help.
$endgroup$
– William Sun
Dec 6 '18 at 22:55




$begingroup$
@Zest You are welcome! I’m glad to help.
$endgroup$
– William Sun
Dec 6 '18 at 22:55












$begingroup$
one last question: the given sequence isn't convergent since it is not bounded right?
$endgroup$
– Zest
Dec 6 '18 at 23:06




$begingroup$
one last question: the given sequence isn't convergent since it is not bounded right?
$endgroup$
– Zest
Dec 6 '18 at 23:06




1




1




$begingroup$
@Zest The answer is Yes assuming you are working in $mathbb{R}$, in which case all convergent sequences are bounded.
$endgroup$
– William Sun
Dec 6 '18 at 23:08




$begingroup$
@Zest The answer is Yes assuming you are working in $mathbb{R}$, in which case all convergent sequences are bounded.
$endgroup$
– William Sun
Dec 6 '18 at 23:08




1




1




$begingroup$
@Zest $L=-1$ is not a limit for the recursively defined sequence in my example. The reasoning is “If the limit of the sequence exists, then it must be $-1$.” But this does not provide with any clue whether the limit exists or not.
$endgroup$
– William Sun
Dec 6 '18 at 23:19




$begingroup$
@Zest $L=-1$ is not a limit for the recursively defined sequence in my example. The reasoning is “If the limit of the sequence exists, then it must be $-1$.” But this does not provide with any clue whether the limit exists or not.
$endgroup$
– William Sun
Dec 6 '18 at 23:19


















draft saved

draft discarded




















































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.




draft saved


draft discarded














StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3029129%2fdoes-calculating-the-limit-of-a-given-real-valued-recursive-sequence-already-i%23new-answer', 'question_page');
}
);

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







-

Popular posts from this blog

Berounka

Image
Berounka.mw-parser-output .entete.map{background-image:url("//upload.wikimedia.org/wikipedia/commons/7/7a/Picto_infobox_map.png")} Caractéristiques Longueur 139  km Bassin 8 861  km 2 Bassin collecteur Elbe Débit moyen ? Cours Source Confluence de la Mže et de la Radbuza · Localisation Pilsen · Altitude 305  m · Coordonnées 49° 45′ 13″ N, 13° 23′ 24″ E Confluence Vltava · Localisation Prague · Altitude 187  m · Coordonnées 49° 59′ 43″ N, 14° 24′ 04″ E Géographie Pays traversés   République tchèque modifier   La Berounka est une rivière de la République tchèque. Elle porte le nom de Mže depuis sa source en Allemagne près de la frontière jusqu'à sa confluence avec la Radbuza à Plzeň. Elle continue ensuite sous le nom de Berounka jusqu'à sa confluence avec la Vltava, près de Prague. La rivière est une destination prisée pour les amateurs de canoë, qui appréc...
Read more

Sphinx de Gizeh

Image
Pour les articles homonymes, voir Sphinx. Sphinx de Gizeh.mw-parser-output .entete.egypte-antique{background-image:url("//upload.wikimedia.org/wikipedia/commons/c/cc/Picto_infobox_ancient_Egypt.png")} Site d'Égypte antique Le visage du sphinx de Gizeh. Noms en arabe أبو الهول , Abou al-Hôl Localisation Région Basse-Égypte Coordonnées 29° 58′ 31″ nord, 31° 08′ 16″ est Géolocalisation sur la carte : Égypte Sphinx de Gizeh modifier   Le sphinx de Gizeh est la statue thérianthrope qui se dresse devant les grandes pyramides du plateau de Gizeh, en Basse-Égypte. Sculpture monumentale monolithique la plus grande du monde [ 1 ] avec 73,5 mètres de longueur, 14 mètres de largeur et 20,22 mètres de hauteur, elle représente un sphinx couchant. Réalisée vers -2500, elle serait attribuée à Djédefrê, l'un des pharaons de la IV e  dynastie, représentant son père le pharaon Khéops. Sommaire 1 Étymologie 2 Desc...
Read more

Different font size/position of beamer's navigation symbols template's content depending on regular/plain...

Image
3 Let's suppose I'd like to divert the navigation symbols template (e.g. in order replace them by the frame numbers, see this bug report ;) I would be facing troubles regarding the font size/position, depending on regular/plain frames, as shown by the following MCE. How could I get rid of these troubles? documentclass{beamer} usetheme{PaloAlto} setbeamertemplate{navigation symbols}{insertframenumber/inserttotalframenumber} begin{document} begin{frame} frametitle{A regular frame} end{frame} begin{frame}[plain] frametitle{A plain frame} end{frame} end{document} beamer header-footer share | improve this question asked Dec 10 '18 a...
Read more