Mixing
Using Rouché's theorem to bound zeroes of partial sums of $e^z$
$begingroup$
I'm having trouble with a question from an exam I'm studying for:
"
Prove that all roots of the polynomial $sumlimits_{k=0}^n frac{z^k}{k!}$ (for $ngeq 1$) are in the annulus ${z: frac{n}{e}<vert zvert<2n }$.
Hint: use the inequality $n^ne^{-n}<n!$"
These kind of questions are usually solved by using Rouché's theorem, so I assume that's the case here as well, but I'm having trouble choosing the right functions to use and bounding their norms correctly. I feel like I should be comparing $e^z$ and $left(e^z - sumlimits_{k=0}^n frac{z^k}{k!}right)$ on the circle with radius $frac{n}{e}$ since e^z has no roots there, so maybe that will imply the polynomial also has no roots there. And then maybe using some other argument to show that all the polynomial's roots are inside the larger disk of radius $2n$.
complex-analysis rouches-theorem
edited Jan 23 at 22:53
Bernard
123k741116
asked Jan 23 at 22:42
Oded CarmonOded Carmon
235
$endgroup$
add a comment |
$begingroup$
I'm having trouble with a question from an exam I'm studying for:
"
Prove that all roots of the polynomial $sumlimits_{k=0}^n frac{z^k}{k!}$ (for $ngeq 1$) are in the annulus ${z: frac{n}{e}<vert zvert<2n }$.
Hint: use the inequality $n^ne^{-n}<n!$"
These kind of questions are usually solved by using Rouché's theorem, so I assume that's the case here as well, but I'm having trouble choosing the right functions to use and bounding their norms correctly. I feel like I should be comparing $e^z$ and $left(e^z - sumlimits_{k=0}^n frac{z^k}{k!}right)$ on the circle with radius $frac{n}{e}$ since e^z has no roots there, so maybe that will imply the polynomial also has no roots there. And then maybe using some other argument to show that all the polynomial's roots are inside the larger disk of radius $2n$.
complex-analysis rouches-theorem
edited Jan 23 at 22:53
Bernard
123k741116
asked Jan 23 at 22:42
Oded CarmonOded Carmon
235
$endgroup$
2
$begingroup$
Possible duplicate of Complex zeros of the polynomials $sum_{k=0}^{n} z^k/k!$, inside balls, see also Roots of the incomplete gamma function
$endgroup$
– LutzL
Jan 24 at 0:14
add a comment |
$begingroup$
I'm having trouble with a question from an exam I'm studying for:
"
Prove that all roots of the polynomial $sumlimits_{k=0}^n frac{z^k}{k!}$ (for $ngeq 1$) are in the annulus ${z: frac{n}{e}<vert zvert<2n }$.
Hint: use the inequality $n^ne^{-n}<n!$"
These kind of questions are usually solved by using Rouché's theorem, so I assume that's the case here as well, but I'm having trouble choosing the right functions to use and bounding their norms correctly. I feel like I should be comparing $e^z$ and $left(e^z - sumlimits_{k=0}^n frac{z^k}{k!}right)$ on the circle with radius $frac{n}{e}$ since e^z has no roots there, so maybe that will imply the polynomial also has no roots there. And then maybe using some other argument to show that all the polynomial's roots are inside the larger disk of radius $2n$.
complex-analysis rouches-theorem
edited Jan 23 at 22:53
Bernard
123k741116
asked Jan 23 at 22:42
Oded CarmonOded Carmon
235
$endgroup$
I'm having trouble with a question from an exam I'm studying for:
"
Prove that all roots of the polynomial $sumlimits_{k=0}^n frac{z^k}{k!}$ (for $ngeq 1$) are in the annulus ${z: frac{n}{e}<vert zvert<2n }$.
Hint: use the inequality $n^ne^{-n}<n!$"
These kind of questions are usually solved by using Rouché's theorem, so I assume that's the case here as well, but I'm having trouble choosing the right functions to use and bounding their norms correctly. I feel like I should be comparing $e^z$ and $left(e^z - sumlimits_{k=0}^n frac{z^k}{k!}right)$ on the circle with radius $frac{n}{e}$ since e^z has no roots there, so maybe that will imply the polynomial also has no roots there. And then maybe using some other argument to show that all the polynomial's roots are inside the larger disk of radius $2n$.
complex-analysis rouches-theorem
complex-analysis rouches-theorem
edited Jan 23 at 22:53
Bernard
123k741116
asked Jan 23 at 22:42
Oded CarmonOded Carmon
235
edited Jan 23 at 22:53
Bernard
123k741116
asked Jan 23 at 22:42
Oded CarmonOded Carmon
235
edited Jan 23 at 22:53
Bernard
123k741116
edited Jan 23 at 22:53
Bernard
123k741116
edited Jan 23 at 22:53
Bernard
123k741116
123k741116
asked Jan 23 at 22:42
Oded CarmonOded Carmon
235
asked Jan 23 at 22:42
Oded CarmonOded Carmon
235
asked Jan 23 at 22:42
Oded CarmonOded Carmon
235
235
2
$begingroup$
Possible duplicate of Complex zeros of the polynomials $sum_{k=0}^{n} z^k/k!$, inside balls, see also Roots of the incomplete gamma function
$endgroup$
– LutzL
Jan 24 at 0:14
add a comment |
2
$begingroup$
Possible duplicate of Complex zeros of the polynomials $sum_{k=0}^{n} z^k/k!$, inside balls, see also Roots of the incomplete gamma function
$endgroup$
– LutzL
Jan 24 at 0:14
2
2
$begingroup$
Possible duplicate of Complex zeros of the polynomials $sum_{k=0}^{n} z^k/k!$, inside balls, see also Roots of the incomplete gamma function
$endgroup$
– LutzL
Jan 24 at 0:14
$begingroup$
Possible duplicate of Complex zeros of the polynomials $sum_{k=0}^{n} z^k/k!$, inside balls, see also Roots of the incomplete gamma function
$endgroup$
– LutzL
Jan 24 at 0:14
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Showing that all roots are inside the disk of radius $2n$ is easy - that's a simple application of Rouche's theorem, with the $frac{z^n}{n!}$ term dominating the sum on the circle.
Showing that no roots are inside the small circle - what you have looks like a good start. How about the Taylor's theorem error estimate to get a size for the difference? That leads to comparing $e^{-|z|}$ and $frac{e^{|z|}|z|^{n+1}}{(n+1)!}$ (Lagrange form), so we want $|z|^{n+1}e^{2|z|} < (n+1)!$. That's not true at $|z|=frac{n}{e}$, with the exponentials working against us - we get about $left(frac{n}{e}right)^{n+1}$ from the power of $z$ and $e$ times that from the factorial, so multiplying by the growing exponential breaks it. It's plausible that the result might be true, but this estimate isn't strong enough to prove it.
The thing about this one? The lower bound on the size of the roots is pretty tight. Running it numerically for $n=9$, there's a root at about $-3.33$, and the bound $frac{9}{e}$ says there are no roots with absolute value less than about $3.31$. We'll need a very strong estimate of some kind to make that inner circle work. I'm not sure what that will be right now - I'll have to think on it some more.
Update: following the links in the comment above leads to the Szegő curve $|zexp(1-z)|=1$. For $frac ne$ to be a lower bound for the size of the roots, the circle $|z|=frac1e$ would have to lie entirely inside the curve, where $|zexp(1-z)| < 1$. At $-e^{-1}$, we get $|-e^{-1}exp(1+e^{-1})|=exp(e^{-1})>1$. That point is outside the curve. The proposition you were asked to prove is false.
Indeed, for $n=11$, the negative real root is at approximately $-3.90545$, while $frac{11}{e}approx 4.04667$.
answered Jan 23 at 23:40
jmerryjmerry
14.1k1629
$endgroup$
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%2f3085182%2fusing-rouch%25c3%25a9s-theorem-to-bound-zeroes-of-partial-sums-of-ez%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
$begingroup$
Showing that all roots are inside the disk of radius $2n$ is easy - that's a simple application of Rouche's theorem, with the $frac{z^n}{n!}$ term dominating the sum on the circle.
Showing that no roots are inside the small circle - what you have looks like a good start. How about the Taylor's theorem error estimate to get a size for the difference? That leads to comparing $e^{-|z|}$ and $frac{e^{|z|}|z|^{n+1}}{(n+1)!}$ (Lagrange form), so we want $|z|^{n+1}e^{2|z|} < (n+1)!$. That's not true at $|z|=frac{n}{e}$, with the exponentials working against us - we get about $left(frac{n}{e}right)^{n+1}$ from the power of $z$ and $e$ times that from the factorial, so multiplying by the growing exponential breaks it. It's plausible that the result might be true, but this estimate isn't strong enough to prove it.
The thing about this one? The lower bound on the size of the roots is pretty tight. Running it numerically for $n=9$, there's a root at about $-3.33$, and the bound $frac{9}{e}$ says there are no roots with absolute value less than about $3.31$. We'll need a very strong estimate of some kind to make that inner circle work. I'm not sure what that will be right now - I'll have to think on it some more.
Update: following the links in the comment above leads to the Szegő curve $|zexp(1-z)|=1$. For $frac ne$ to be a lower bound for the size of the roots, the circle $|z|=frac1e$ would have to lie entirely inside the curve, where $|zexp(1-z)| < 1$. At $-e^{-1}$, we get $|-e^{-1}exp(1+e^{-1})|=exp(e^{-1})>1$. That point is outside the curve. The proposition you were asked to prove is false.
Indeed, for $n=11$, the negative real root is at approximately $-3.90545$, while $frac{11}{e}approx 4.04667$.
answered Jan 23 at 23:40
jmerryjmerry
14.1k1629
$endgroup$
add a comment |
$begingroup$
Showing that all roots are inside the disk of radius $2n$ is easy - that's a simple application of Rouche's theorem, with the $frac{z^n}{n!}$ term dominating the sum on the circle.
Showing that no roots are inside the small circle - what you have looks like a good start. How about the Taylor's theorem error estimate to get a size for the difference? That leads to comparing $e^{-|z|}$ and $frac{e^{|z|}|z|^{n+1}}{(n+1)!}$ (Lagrange form), so we want $|z|^{n+1}e^{2|z|} < (n+1)!$. That's not true at $|z|=frac{n}{e}$, with the exponentials working against us - we get about $left(frac{n}{e}right)^{n+1}$ from the power of $z$ and $e$ times that from the factorial, so multiplying by the growing exponential breaks it. It's plausible that the result might be true, but this estimate isn't strong enough to prove it.
The thing about this one? The lower bound on the size of the roots is pretty tight. Running it numerically for $n=9$, there's a root at about $-3.33$, and the bound $frac{9}{e}$ says there are no roots with absolute value less than about $3.31$. We'll need a very strong estimate of some kind to make that inner circle work. I'm not sure what that will be right now - I'll have to think on it some more.
Update: following the links in the comment above leads to the Szegő curve $|zexp(1-z)|=1$. For $frac ne$ to be a lower bound for the size of the roots, the circle $|z|=frac1e$ would have to lie entirely inside the curve, where $|zexp(1-z)| < 1$. At $-e^{-1}$, we get $|-e^{-1}exp(1+e^{-1})|=exp(e^{-1})>1$. That point is outside the curve. The proposition you were asked to prove is false.
Indeed, for $n=11$, the negative real root is at approximately $-3.90545$, while $frac{11}{e}approx 4.04667$.
answered Jan 23 at 23:40
jmerryjmerry
14.1k1629
$endgroup$
add a comment |
$begingroup$
Showing that all roots are inside the disk of radius $2n$ is easy - that's a simple application of Rouche's theorem, with the $frac{z^n}{n!}$ term dominating the sum on the circle.
Showing that no roots are inside the small circle - what you have looks like a good start. How about the Taylor's theorem error estimate to get a size for the difference? That leads to comparing $e^{-|z|}$ and $frac{e^{|z|}|z|^{n+1}}{(n+1)!}$ (Lagrange form), so we want $|z|^{n+1}e^{2|z|} < (n+1)!$. That's not true at $|z|=frac{n}{e}$, with the exponentials working against us - we get about $left(frac{n}{e}right)^{n+1}$ from the power of $z$ and $e$ times that from the factorial, so multiplying by the growing exponential breaks it. It's plausible that the result might be true, but this estimate isn't strong enough to prove it.
The thing about this one? The lower bound on the size of the roots is pretty tight. Running it numerically for $n=9$, there's a root at about $-3.33$, and the bound $frac{9}{e}$ says there are no roots with absolute value less than about $3.31$. We'll need a very strong estimate of some kind to make that inner circle work. I'm not sure what that will be right now - I'll have to think on it some more.
Update: following the links in the comment above leads to the Szegő curve $|zexp(1-z)|=1$. For $frac ne$ to be a lower bound for the size of the roots, the circle $|z|=frac1e$ would have to lie entirely inside the curve, where $|zexp(1-z)| < 1$. At $-e^{-1}$, we get $|-e^{-1}exp(1+e^{-1})|=exp(e^{-1})>1$. That point is outside the curve. The proposition you were asked to prove is false.
Indeed, for $n=11$, the negative real root is at approximately $-3.90545$, while $frac{11}{e}approx 4.04667$.
answered Jan 23 at 23:40
jmerryjmerry
14.1k1629
$endgroup$
Showing that all roots are inside the disk of radius $2n$ is easy - that's a simple application of Rouche's theorem, with the $frac{z^n}{n!}$ term dominating the sum on the circle.
Showing that no roots are inside the small circle - what you have looks like a good start. How about the Taylor's theorem error estimate to get a size for the difference? That leads to comparing $e^{-|z|}$ and $frac{e^{|z|}|z|^{n+1}}{(n+1)!}$ (Lagrange form), so we want $|z|^{n+1}e^{2|z|} < (n+1)!$. That's not true at $|z|=frac{n}{e}$, with the exponentials working against us - we get about $left(frac{n}{e}right)^{n+1}$ from the power of $z$ and $e$ times that from the factorial, so multiplying by the growing exponential breaks it. It's plausible that the result might be true, but this estimate isn't strong enough to prove it.
The thing about this one? The lower bound on the size of the roots is pretty tight. Running it numerically for $n=9$, there's a root at about $-3.33$, and the bound $frac{9}{e}$ says there are no roots with absolute value less than about $3.31$. We'll need a very strong estimate of some kind to make that inner circle work. I'm not sure what that will be right now - I'll have to think on it some more.
Update: following the links in the comment above leads to the Szegő curve $|zexp(1-z)|=1$. For $frac ne$ to be a lower bound for the size of the roots, the circle $|z|=frac1e$ would have to lie entirely inside the curve, where $|zexp(1-z)| < 1$. At $-e^{-1}$, we get $|-e^{-1}exp(1+e^{-1})|=exp(e^{-1})>1$. That point is outside the curve. The proposition you were asked to prove is false.
Indeed, for $n=11$, the negative real root is at approximately $-3.90545$, while $frac{11}{e}approx 4.04667$.
answered Jan 23 at 23:40
jmerryjmerry
14.1k1629
edited Jan 24 at 7:35
answered Jan 23 at 23:40
jmerryjmerry
14.1k1629
answered Jan 23 at 23:40
jmerryjmerry
14.1k1629
answered Jan 23 at 23:40
jmerryjmerry
14.1k1629
14.1k1629
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%2f3085182%2fusing-rouch%25c3%25a9s-theorem-to-bound-zeroes-of-partial-sums-of-ez%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
2
$begingroup$
Possible duplicate of Complex zeros of the polynomials $sum_{k=0}^{n} z^k/k!$, inside balls, see also Roots of the incomplete gamma function
$endgroup$
– LutzL
Jan 24 at 0:14