What is the limit of $phi_{t_k}(x_k,y_k) = 1 - frac{y_k}{sqrt{x_k^2 + y_k^2 + 2t_k + t_k^2}}$ when $y_k...












0












$begingroup$


Let ${x_k in [-1,1]} downarrow 0$, ${y_k in [0,1]} downarrow 0$ and ${t_k gt 0 } downarrow 0$ be sequences which satisfy the equality $y_k sqrt{x_k^2 + t_k^2} = t_k$. Now let $$phi_t(x,y) = 1 - frac{y}{sqrt{x^2 + y^2 + 2t + t^2}}$$
I want to show that $lim_{k to infty} phi_{t_k}(x_k,y_k) neq 0$. I do not know this is true for sure. My attempt so far has been to show that $lim_{k to infty} frac{y_k^2}{{x_k^2 + y_k^2 + 2t_k + t_k^2}} neq 1$ using the relation that $y_k^2 =frac{t_k^2}{x_k^2 + t_k^2}$ but I am still unable to come up with a way to show that the limit is not 0. If someone has any ideas or proofs for this I would be very thankful. Ideally, if someone knows what the exact limit of $phi_{t_k}(x_k,y_k)$ is that would also be great.






sequences-and-series limits







share|cite|improve this question









$endgroup$












  • $begingroup$
    Counter example $$x_k={1over k^3}\y_k={1over k}\t_k={1over k^3sqrt{k^2-1}}$$ for $kne 1$ and $t_1=0.1$.
    $endgroup$
    – Mostafa Ayaz
    Jan 9 at 18:25
















0












$begingroup$


Let ${x_k in [-1,1]} downarrow 0$, ${y_k in [0,1]} downarrow 0$ and ${t_k gt 0 } downarrow 0$ be sequences which satisfy the equality $y_k sqrt{x_k^2 + t_k^2} = t_k$. Now let $$phi_t(x,y) = 1 - frac{y}{sqrt{x^2 + y^2 + 2t + t^2}}$$
I want to show that $lim_{k to infty} phi_{t_k}(x_k,y_k) neq 0$. I do not know this is true for sure. My attempt so far has been to show that $lim_{k to infty} frac{y_k^2}{{x_k^2 + y_k^2 + 2t_k + t_k^2}} neq 1$ using the relation that $y_k^2 =frac{t_k^2}{x_k^2 + t_k^2}$ but I am still unable to come up with a way to show that the limit is not 0. If someone has any ideas or proofs for this I would be very thankful. Ideally, if someone knows what the exact limit of $phi_{t_k}(x_k,y_k)$ is that would also be great.






sequences-and-series limits







share|cite|improve this question









$endgroup$












  • $begingroup$
    Counter example $$x_k={1over k^3}\y_k={1over k}\t_k={1over k^3sqrt{k^2-1}}$$ for $kne 1$ and $t_1=0.1$.
    $endgroup$
    – Mostafa Ayaz
    Jan 9 at 18:25














0












0








0





$begingroup$


Let ${x_k in [-1,1]} downarrow 0$, ${y_k in [0,1]} downarrow 0$ and ${t_k gt 0 } downarrow 0$ be sequences which satisfy the equality $y_k sqrt{x_k^2 + t_k^2} = t_k$. Now let $$phi_t(x,y) = 1 - frac{y}{sqrt{x^2 + y^2 + 2t + t^2}}$$
I want to show that $lim_{k to infty} phi_{t_k}(x_k,y_k) neq 0$. I do not know this is true for sure. My attempt so far has been to show that $lim_{k to infty} frac{y_k^2}{{x_k^2 + y_k^2 + 2t_k + t_k^2}} neq 1$ using the relation that $y_k^2 =frac{t_k^2}{x_k^2 + t_k^2}$ but I am still unable to come up with a way to show that the limit is not 0. If someone has any ideas or proofs for this I would be very thankful. Ideally, if someone knows what the exact limit of $phi_{t_k}(x_k,y_k)$ is that would also be great.






sequences-and-series limits







share|cite|improve this question









$endgroup$




Let ${x_k in [-1,1]} downarrow 0$, ${y_k in [0,1]} downarrow 0$ and ${t_k gt 0 } downarrow 0$ be sequences which satisfy the equality $y_k sqrt{x_k^2 + t_k^2} = t_k$. Now let $$phi_t(x,y) = 1 - frac{y}{sqrt{x^2 + y^2 + 2t + t^2}}$$
I want to show that $lim_{k to infty} phi_{t_k}(x_k,y_k) neq 0$. I do not know this is true for sure. My attempt so far has been to show that $lim_{k to infty} frac{y_k^2}{{x_k^2 + y_k^2 + 2t_k + t_k^2}} neq 1$ using the relation that $y_k^2 =frac{t_k^2}{x_k^2 + t_k^2}$ but I am still unable to come up with a way to show that the limit is not 0. If someone has any ideas or proofs for this I would be very thankful. Ideally, if someone knows what the exact limit of $phi_{t_k}(x_k,y_k)$ is that would also be great.






sequences-and-series limits




sequences-and-series limits






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Jan 9 at 16:27









geo17geo17

967




967












  • $begingroup$
    Counter example $$x_k={1over k^3}\y_k={1over k}\t_k={1over k^3sqrt{k^2-1}}$$ for $kne 1$ and $t_1=0.1$.
    $endgroup$
    – Mostafa Ayaz
    Jan 9 at 18:25


















  • $begingroup$
    Counter example $$x_k={1over k^3}\y_k={1over k}\t_k={1over k^3sqrt{k^2-1}}$$ for $kne 1$ and $t_1=0.1$.
    $endgroup$
    – Mostafa Ayaz
    Jan 9 at 18:25
















$begingroup$
Counter example $$x_k={1over k^3}\y_k={1over k}\t_k={1over k^3sqrt{k^2-1}}$$ for $kne 1$ and $t_1=0.1$.
$endgroup$
– Mostafa Ayaz
Jan 9 at 18:25




$begingroup$
Counter example $$x_k={1over k^3}\y_k={1over k}\t_k={1over k^3sqrt{k^2-1}}$$ for $kne 1$ and $t_1=0.1$.
$endgroup$
– Mostafa Ayaz
Jan 9 at 18:25










1 Answer
1






active

oldest

votes


















0












$begingroup$

First we have $$phi_{t_k}(x_k,y_k)=1 - frac{y_k}{sqrt{x_k^2 + y_k^2 + 2t_k + t_k^2}}$$by substitution from $y_ksqrt{x_k^2+t_k^2}=t_k$ assuming $y_kne 0$ (if $y_k=0$ then $phi_{t_k}(x_k,y_k)=1$) we obtain $$phi_{t_k}(x_k,y_k){=1 - frac{y_k}{sqrt{y_k^2 + 2t_k + {t_k^2over y_k^2}}}\=1-{1over sqrt{1+{2t_kover y_k^2}+{t_k^2over y_k^4}}}}$$by defining $a_k={t_kover y_k^2}ge0$ we finally conclude that $$phi_{t_k}(x_k,y_k){=1-{1over sqrt{1+2a_k+a_k^2}}\=1-{1over sqrt{(1+a_k)^2}}\=1-{1over 1+a_k}\={a_kover 1+a_k}}$$therefore $$lim_{kto infty}phi_{t_k}(x_k,y_k)=lim_{kto infty}{a_kover 1+a_k}$$To proceed further we need an extra assumption $$lim_{kto infty}{t_kover y_k^2}$$unless lots of counter examples will drive out. As an example$$x_k={1over k^3}\y_k={1over k}$$therefore $$t_k={1over k}sqrt{t_k^2+{1over k^6}}$$from which we obtain $$t_k={1over k^3sqrt{k^2-1}}$$and we have $$lim_{ktoinfty}phi_{t_k}(x_k,y_k){=lim_{ktoinfty}1-{1over ksqrt{{1over k^6}+{1over k^2}+{2over k^3sqrt{k^2-1}}+{1over k^6(k^2-1)}}}\=lim_{ktoinfty}1-{1over sqrt{{1over k^4}+{1}+{2over ksqrt{k^2-1}}+{1over k^4(k^2-1)}}}\=0}$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    wow, thanks so much for providing this counter-example!
    $endgroup$
    – geo17
    Jan 9 at 20:50










  • $begingroup$
    ur welcome. Wish you luck!
    $endgroup$
    – Mostafa Ayaz
    Jan 9 at 20:53











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%2f3067656%2fwhat-is-the-limit-of-phi-t-kx-k-y-k-1-fracy-k-sqrtx-k2-y-k2%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









0












$begingroup$

First we have $$phi_{t_k}(x_k,y_k)=1 - frac{y_k}{sqrt{x_k^2 + y_k^2 + 2t_k + t_k^2}}$$by substitution from $y_ksqrt{x_k^2+t_k^2}=t_k$ assuming $y_kne 0$ (if $y_k=0$ then $phi_{t_k}(x_k,y_k)=1$) we obtain $$phi_{t_k}(x_k,y_k){=1 - frac{y_k}{sqrt{y_k^2 + 2t_k + {t_k^2over y_k^2}}}\=1-{1over sqrt{1+{2t_kover y_k^2}+{t_k^2over y_k^4}}}}$$by defining $a_k={t_kover y_k^2}ge0$ we finally conclude that $$phi_{t_k}(x_k,y_k){=1-{1over sqrt{1+2a_k+a_k^2}}\=1-{1over sqrt{(1+a_k)^2}}\=1-{1over 1+a_k}\={a_kover 1+a_k}}$$therefore $$lim_{kto infty}phi_{t_k}(x_k,y_k)=lim_{kto infty}{a_kover 1+a_k}$$To proceed further we need an extra assumption $$lim_{kto infty}{t_kover y_k^2}$$unless lots of counter examples will drive out. As an example$$x_k={1over k^3}\y_k={1over k}$$therefore $$t_k={1over k}sqrt{t_k^2+{1over k^6}}$$from which we obtain $$t_k={1over k^3sqrt{k^2-1}}$$and we have $$lim_{ktoinfty}phi_{t_k}(x_k,y_k){=lim_{ktoinfty}1-{1over ksqrt{{1over k^6}+{1over k^2}+{2over k^3sqrt{k^2-1}}+{1over k^6(k^2-1)}}}\=lim_{ktoinfty}1-{1over sqrt{{1over k^4}+{1}+{2over ksqrt{k^2-1}}+{1over k^4(k^2-1)}}}\=0}$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    wow, thanks so much for providing this counter-example!
    $endgroup$
    – geo17
    Jan 9 at 20:50










  • $begingroup$
    ur welcome. Wish you luck!
    $endgroup$
    – Mostafa Ayaz
    Jan 9 at 20:53
















0












$begingroup$

First we have $$phi_{t_k}(x_k,y_k)=1 - frac{y_k}{sqrt{x_k^2 + y_k^2 + 2t_k + t_k^2}}$$by substitution from $y_ksqrt{x_k^2+t_k^2}=t_k$ assuming $y_kne 0$ (if $y_k=0$ then $phi_{t_k}(x_k,y_k)=1$) we obtain $$phi_{t_k}(x_k,y_k){=1 - frac{y_k}{sqrt{y_k^2 + 2t_k + {t_k^2over y_k^2}}}\=1-{1over sqrt{1+{2t_kover y_k^2}+{t_k^2over y_k^4}}}}$$by defining $a_k={t_kover y_k^2}ge0$ we finally conclude that $$phi_{t_k}(x_k,y_k){=1-{1over sqrt{1+2a_k+a_k^2}}\=1-{1over sqrt{(1+a_k)^2}}\=1-{1over 1+a_k}\={a_kover 1+a_k}}$$therefore $$lim_{kto infty}phi_{t_k}(x_k,y_k)=lim_{kto infty}{a_kover 1+a_k}$$To proceed further we need an extra assumption $$lim_{kto infty}{t_kover y_k^2}$$unless lots of counter examples will drive out. As an example$$x_k={1over k^3}\y_k={1over k}$$therefore $$t_k={1over k}sqrt{t_k^2+{1over k^6}}$$from which we obtain $$t_k={1over k^3sqrt{k^2-1}}$$and we have $$lim_{ktoinfty}phi_{t_k}(x_k,y_k){=lim_{ktoinfty}1-{1over ksqrt{{1over k^6}+{1over k^2}+{2over k^3sqrt{k^2-1}}+{1over k^6(k^2-1)}}}\=lim_{ktoinfty}1-{1over sqrt{{1over k^4}+{1}+{2over ksqrt{k^2-1}}+{1over k^4(k^2-1)}}}\=0}$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    wow, thanks so much for providing this counter-example!
    $endgroup$
    – geo17
    Jan 9 at 20:50










  • $begingroup$
    ur welcome. Wish you luck!
    $endgroup$
    – Mostafa Ayaz
    Jan 9 at 20:53














0












0








0





$begingroup$

First we have $$phi_{t_k}(x_k,y_k)=1 - frac{y_k}{sqrt{x_k^2 + y_k^2 + 2t_k + t_k^2}}$$by substitution from $y_ksqrt{x_k^2+t_k^2}=t_k$ assuming $y_kne 0$ (if $y_k=0$ then $phi_{t_k}(x_k,y_k)=1$) we obtain $$phi_{t_k}(x_k,y_k){=1 - frac{y_k}{sqrt{y_k^2 + 2t_k + {t_k^2over y_k^2}}}\=1-{1over sqrt{1+{2t_kover y_k^2}+{t_k^2over y_k^4}}}}$$by defining $a_k={t_kover y_k^2}ge0$ we finally conclude that $$phi_{t_k}(x_k,y_k){=1-{1over sqrt{1+2a_k+a_k^2}}\=1-{1over sqrt{(1+a_k)^2}}\=1-{1over 1+a_k}\={a_kover 1+a_k}}$$therefore $$lim_{kto infty}phi_{t_k}(x_k,y_k)=lim_{kto infty}{a_kover 1+a_k}$$To proceed further we need an extra assumption $$lim_{kto infty}{t_kover y_k^2}$$unless lots of counter examples will drive out. As an example$$x_k={1over k^3}\y_k={1over k}$$therefore $$t_k={1over k}sqrt{t_k^2+{1over k^6}}$$from which we obtain $$t_k={1over k^3sqrt{k^2-1}}$$and we have $$lim_{ktoinfty}phi_{t_k}(x_k,y_k){=lim_{ktoinfty}1-{1over ksqrt{{1over k^6}+{1over k^2}+{2over k^3sqrt{k^2-1}}+{1over k^6(k^2-1)}}}\=lim_{ktoinfty}1-{1over sqrt{{1over k^4}+{1}+{2over ksqrt{k^2-1}}+{1over k^4(k^2-1)}}}\=0}$$






share|cite|improve this answer









$endgroup$



First we have $$phi_{t_k}(x_k,y_k)=1 - frac{y_k}{sqrt{x_k^2 + y_k^2 + 2t_k + t_k^2}}$$by substitution from $y_ksqrt{x_k^2+t_k^2}=t_k$ assuming $y_kne 0$ (if $y_k=0$ then $phi_{t_k}(x_k,y_k)=1$) we obtain $$phi_{t_k}(x_k,y_k){=1 - frac{y_k}{sqrt{y_k^2 + 2t_k + {t_k^2over y_k^2}}}\=1-{1over sqrt{1+{2t_kover y_k^2}+{t_k^2over y_k^4}}}}$$by defining $a_k={t_kover y_k^2}ge0$ we finally conclude that $$phi_{t_k}(x_k,y_k){=1-{1over sqrt{1+2a_k+a_k^2}}\=1-{1over sqrt{(1+a_k)^2}}\=1-{1over 1+a_k}\={a_kover 1+a_k}}$$therefore $$lim_{kto infty}phi_{t_k}(x_k,y_k)=lim_{kto infty}{a_kover 1+a_k}$$To proceed further we need an extra assumption $$lim_{kto infty}{t_kover y_k^2}$$unless lots of counter examples will drive out. As an example$$x_k={1over k^3}\y_k={1over k}$$therefore $$t_k={1over k}sqrt{t_k^2+{1over k^6}}$$from which we obtain $$t_k={1over k^3sqrt{k^2-1}}$$and we have $$lim_{ktoinfty}phi_{t_k}(x_k,y_k){=lim_{ktoinfty}1-{1over ksqrt{{1over k^6}+{1over k^2}+{2over k^3sqrt{k^2-1}}+{1over k^6(k^2-1)}}}\=lim_{ktoinfty}1-{1over sqrt{{1over k^4}+{1}+{2over ksqrt{k^2-1}}+{1over k^4(k^2-1)}}}\=0}$$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jan 9 at 19:58









Mostafa AyazMostafa Ayaz

15.5k3939




15.5k3939












  • $begingroup$
    wow, thanks so much for providing this counter-example!
    $endgroup$
    – geo17
    Jan 9 at 20:50










  • $begingroup$
    ur welcome. Wish you luck!
    $endgroup$
    – Mostafa Ayaz
    Jan 9 at 20:53


















  • $begingroup$
    wow, thanks so much for providing this counter-example!
    $endgroup$
    – geo17
    Jan 9 at 20:50










  • $begingroup$
    ur welcome. Wish you luck!
    $endgroup$
    – Mostafa Ayaz
    Jan 9 at 20:53
















$begingroup$
wow, thanks so much for providing this counter-example!
$endgroup$
– geo17
Jan 9 at 20:50




$begingroup$
wow, thanks so much for providing this counter-example!
$endgroup$
– geo17
Jan 9 at 20:50












$begingroup$
ur welcome. Wish you luck!
$endgroup$
– Mostafa Ayaz
Jan 9 at 20:53




$begingroup$
ur welcome. Wish you luck!
$endgroup$
– Mostafa Ayaz
Jan 9 at 20:53


















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%2f3067656%2fwhat-is-the-limit-of-phi-t-kx-k-y-k-1-fracy-k-sqrtx-k2-y-k2%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

Can a sorcerer learn a 5th-level spell early by creating spell slots using the Font of Magic feature?

Image
28 6 $begingroup$ Per the Font of Magic feature, sorcerer can use Flexible Casting to create 5th-level spell slots at level 7, even though they are not typically available until level 9. You can transform unexpended sorcery points into one spell slot as a bonus action on your turn. The Creating Spell Slots table shows the cost of creating a spell slot of [5th level is 7] If such a sorcerer levels up while still having this 5th-level spell slot, can they choose a 5th-level spell as the spell they gain upon levelling up? The Spellcasting feature states: Additionally, when you gain a level in this class, you can choose one of the sorcerer spells you know and replace it with another spell from the sorcerer spell list, which also must be of a level for which you have spell slots.
Read more

Does disintegrating a polymorphed enemy still kill it after the 2018 errata?

Image
up vote 13 down vote favorite 1 After the 2018 errata, the disintegrate spell description now reads: A creature targeted by this spell must make a Dexterity saving throw. On a failed save, the target takes 10d6 + 40 force damage. The target is disintegrated if this damage leaves it with 0 hit points. If you were to polymorph an enemy into a rat and then disintegrate it, would the enemy be disintegrated or would it just return to its original form? I know that for Druids, it’s not an instant kill anymore, but is this the case for polymorph as well? dnd-5e spells polymorph errata share | improve this question edited 18 hours ago
Read more

A Topological Invariant for $pi_3(U(n))$

Image
3 3 $begingroup$ Recently, I saw a construction of topological invariant for $pi_3(U(n))$ with $ngeq 2$ : $$ N=frac{1}{24pi^2}int_{S^3} d^3x epsilon^{ijk} Tr[(U^{-1}partial_{x_i}U)(U^{-1}partial_{x_j}U)(U^{-1}partial_{x_k}U)] , $$ where $Uin U(n)$ depends on $boldsymbol{x}=(x_1,x_2,x_3)in S^3$ , $epsilon^{ijk}$ is the Levi-Civita symbol, $i,j,k=1,2,3$ , and the duplicated indexes are summed over. It is claimed that $N$ is an integer, but why? Update 02/02/2019 I think I got an argument for $n=2$ . In this case, $U=e^{i varphi} q$ with $qin SU(2)$ . Due to the trace and the Levi-Civita symbol in $N$ , $varphi$ does not contribute to $N$ . As $Tr[q^{dagger}partial_i q]=0$ and $(q^{dagger}partial_i q)^{dagger}=-q^{dagger}partial_i q$ , $q^{dagger}partial_i q$ in geneeral has the form $q^{dagger}partia
Read more