Mixing
Showing given function is real analytic on R
I was reading about real analytic functions and I understood that to show a function is real analytic on $mathbb{R}$, we need to show that there is a power series centered at $x_0$ which converges to an open interval around $x_0$ $forall x_0 in mathbb{R}$
The example I am working on is the function $$f(x) = frac{1}{1+x^2}$$
I can show that it converges for expansion around 0 by expanding it as a geometric series and this converges in (-1,1) when centered at x=0. How do I show it for other x $in mathbb{R}$ ?
I think I remember a fact that if power series centered around x=0 converges in (-1,1) then power series centered at any point in this interval would also converge(can someone give verify if this is indeed true?). But that still leaves us with the rest of the number line.
real-analysis analysis convergence
asked Nov 21 '18 at 5:28
Red Floyd
273314
add a comment |
I was reading about real analytic functions and I understood that to show a function is real analytic on $mathbb{R}$, we need to show that there is a power series centered at $x_0$ which converges to an open interval around $x_0$ $forall x_0 in mathbb{R}$
The example I am working on is the function $$f(x) = frac{1}{1+x^2}$$
I can show that it converges for expansion around 0 by expanding it as a geometric series and this converges in (-1,1) when centered at x=0. How do I show it for other x $in mathbb{R}$ ?
I think I remember a fact that if power series centered around x=0 converges in (-1,1) then power series centered at any point in this interval would also converge(can someone give verify if this is indeed true?). But that still leaves us with the rest of the number line.
real-analysis analysis convergence
asked Nov 21 '18 at 5:28
Red Floyd
273314
Have you learned about Taylor Series?
– JavaMan
Nov 21 '18 at 5:38
Yes, I have. Yea, maybe we can use taylor series to find the expansion. How do we show it converges? Also, is my statement correct about convergence about a different point in the interval?
– Red Floyd
Nov 21 '18 at 6:22
1
To show a Taylor Series converges, you have to show that $R_n(x) = sum_{k=n+1}^{infty} frac{f^{(k)}(a) (x-a)^k}{k!} to 0$. By Lagrange's Thm $R_n(x) = frac{f^{(n+1)}(c) (x-a)^{n+1}}{(n+1)!}$ for some $c$ between $a$ and $x$. Now, the $n$th derivative is something like $f(x) leq C x^{-n-2}$. Hence, $f^{(n)}(x) to 0$ as $n to infty$. In particular $f^{(n+1)}(c)$ is bounded. Finally, $(x-a)^{n+1}$ is polynomial, and factorials go to zero faster than any polynomial (this follows since $x^n/n! to 0$ as $n to infty$), which is demonstrated by convergence of Taylor series for $e^x$.
– JavaMan
Nov 22 '18 at 6:21
add a comment |
I was reading about real analytic functions and I understood that to show a function is real analytic on $mathbb{R}$, we need to show that there is a power series centered at $x_0$ which converges to an open interval around $x_0$ $forall x_0 in mathbb{R}$
The example I am working on is the function $$f(x) = frac{1}{1+x^2}$$
I can show that it converges for expansion around 0 by expanding it as a geometric series and this converges in (-1,1) when centered at x=0. How do I show it for other x $in mathbb{R}$ ?
I think I remember a fact that if power series centered around x=0 converges in (-1,1) then power series centered at any point in this interval would also converge(can someone give verify if this is indeed true?). But that still leaves us with the rest of the number line.
real-analysis analysis convergence
asked Nov 21 '18 at 5:28
Red Floyd
273314
I was reading about real analytic functions and I understood that to show a function is real analytic on $mathbb{R}$, we need to show that there is a power series centered at $x_0$ which converges to an open interval around $x_0$ $forall x_0 in mathbb{R}$
The example I am working on is the function $$f(x) = frac{1}{1+x^2}$$
I can show that it converges for expansion around 0 by expanding it as a geometric series and this converges in (-1,1) when centered at x=0. How do I show it for other x $in mathbb{R}$ ?
I think I remember a fact that if power series centered around x=0 converges in (-1,1) then power series centered at any point in this interval would also converge(can someone give verify if this is indeed true?). But that still leaves us with the rest of the number line.
real-analysis analysis convergence
real-analysis analysis convergence
asked Nov 21 '18 at 5:28
Red Floyd
273314
asked Nov 21 '18 at 5:28
Red Floyd
273314
asked Nov 21 '18 at 5:28
Red Floyd
273314
asked Nov 21 '18 at 5:28
Red Floyd
273314
asked Nov 21 '18 at 5:28
Red Floyd
273314
273314
Have you learned about Taylor Series?
– JavaMan
Nov 21 '18 at 5:38
Yes, I have. Yea, maybe we can use taylor series to find the expansion. How do we show it converges? Also, is my statement correct about convergence about a different point in the interval?
– Red Floyd
Nov 21 '18 at 6:22
1
To show a Taylor Series converges, you have to show that $R_n(x) = sum_{k=n+1}^{infty} frac{f^{(k)}(a) (x-a)^k}{k!} to 0$. By Lagrange's Thm $R_n(x) = frac{f^{(n+1)}(c) (x-a)^{n+1}}{(n+1)!}$ for some $c$ between $a$ and $x$. Now, the $n$th derivative is something like $f(x) leq C x^{-n-2}$. Hence, $f^{(n)}(x) to 0$ as $n to infty$. In particular $f^{(n+1)}(c)$ is bounded. Finally, $(x-a)^{n+1}$ is polynomial, and factorials go to zero faster than any polynomial (this follows since $x^n/n! to 0$ as $n to infty$), which is demonstrated by convergence of Taylor series for $e^x$.
– JavaMan
Nov 22 '18 at 6:21
add a comment |
Have you learned about Taylor Series?
– JavaMan
Nov 21 '18 at 5:38
Yes, I have. Yea, maybe we can use taylor series to find the expansion. How do we show it converges? Also, is my statement correct about convergence about a different point in the interval?
– Red Floyd
Nov 21 '18 at 6:22
1
To show a Taylor Series converges, you have to show that $R_n(x) = sum_{k=n+1}^{infty} frac{f^{(k)}(a) (x-a)^k}{k!} to 0$. By Lagrange's Thm $R_n(x) = frac{f^{(n+1)}(c) (x-a)^{n+1}}{(n+1)!}$ for some $c$ between $a$ and $x$. Now, the $n$th derivative is something like $f(x) leq C x^{-n-2}$. Hence, $f^{(n)}(x) to 0$ as $n to infty$. In particular $f^{(n+1)}(c)$ is bounded. Finally, $(x-a)^{n+1}$ is polynomial, and factorials go to zero faster than any polynomial (this follows since $x^n/n! to 0$ as $n to infty$), which is demonstrated by convergence of Taylor series for $e^x$.
– JavaMan
Nov 22 '18 at 6:21
Have you learned about Taylor Series?
– JavaMan
Nov 21 '18 at 5:38
Have you learned about Taylor Series?
– JavaMan
Nov 21 '18 at 5:38
Yes, I have. Yea, maybe we can use taylor series to find the expansion. How do we show it converges? Also, is my statement correct about convergence about a different point in the interval?
– Red Floyd
Nov 21 '18 at 6:22
Yes, I have. Yea, maybe we can use taylor series to find the expansion. How do we show it converges? Also, is my statement correct about convergence about a different point in the interval?
– Red Floyd
Nov 21 '18 at 6:22
1
1
To show a Taylor Series converges, you have to show that $R_n(x) = sum_{k=n+1}^{infty} frac{f^{(k)}(a) (x-a)^k}{k!} to 0$. By Lagrange's Thm $R_n(x) = frac{f^{(n+1)}(c) (x-a)^{n+1}}{(n+1)!}$ for some $c$ between $a$ and $x$. Now, the $n$th derivative is something like $f(x) leq C x^{-n-2}$. Hence, $f^{(n)}(x) to 0$ as $n to infty$. In particular $f^{(n+1)}(c)$ is bounded. Finally, $(x-a)^{n+1}$ is polynomial, and factorials go to zero faster than any polynomial (this follows since $x^n/n! to 0$ as $n to infty$), which is demonstrated by convergence of Taylor series for $e^x$.
– JavaMan
Nov 22 '18 at 6:21
To show a Taylor Series converges, you have to show that $R_n(x) = sum_{k=n+1}^{infty} frac{f^{(k)}(a) (x-a)^k}{k!} to 0$. By Lagrange's Thm $R_n(x) = frac{f^{(n+1)}(c) (x-a)^{n+1}}{(n+1)!}$ for some $c$ between $a$ and $x$. Now, the $n$th derivative is something like $f(x) leq C x^{-n-2}$. Hence, $f^{(n)}(x) to 0$ as $n to infty$. In particular $f^{(n+1)}(c)$ is bounded. Finally, $(x-a)^{n+1}$ is polynomial, and factorials go to zero faster than any polynomial (this follows since $x^n/n! to 0$ as $n to infty$), which is demonstrated by convergence of Taylor series for $e^x$.
– JavaMan
Nov 22 '18 at 6:21
add a comment |
1 Answer
1
active
oldest
votes
Try just centering a geometric series around a different point. For example, take $$frac{1}{1-(-(x-x_0)^2))}$$
If this does have power series that converges, what would it look like? This would mean that you can get a radius 1 of converge around any point that you desire.
answered Nov 21 '18 at 6:34
Alex
1323
1
Please check your solution again.
– Jacky Chong
Nov 21 '18 at 6:39
unless I'm mistaken, we should get the series $sum_{i=1}^{infty} (-1)^i (x-x_0)^{2i}$ which converges whenever $|x-x_0|<1$?
– Alex
Nov 21 '18 at 6:43
2
I think the OP wants to consider $(1+x^2)^{-1}$.
– Jacky Chong
Nov 21 '18 at 6:46
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%2f3007301%2fshowing-given-function-is-real-analytic-on-r%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
Try just centering a geometric series around a different point. For example, take $$frac{1}{1-(-(x-x_0)^2))}$$
If this does have power series that converges, what would it look like? This would mean that you can get a radius 1 of converge around any point that you desire.
answered Nov 21 '18 at 6:34
Alex
1323
1
Please check your solution again.
– Jacky Chong
Nov 21 '18 at 6:39
unless I'm mistaken, we should get the series $sum_{i=1}^{infty} (-1)^i (x-x_0)^{2i}$ which converges whenever $|x-x_0|<1$?
– Alex
Nov 21 '18 at 6:43
2
I think the OP wants to consider $(1+x^2)^{-1}$.
– Jacky Chong
Nov 21 '18 at 6:46
add a comment |
Try just centering a geometric series around a different point. For example, take $$frac{1}{1-(-(x-x_0)^2))}$$
If this does have power series that converges, what would it look like? This would mean that you can get a radius 1 of converge around any point that you desire.
answered Nov 21 '18 at 6:34
Alex
1323
1
Please check your solution again.
– Jacky Chong
Nov 21 '18 at 6:39
unless I'm mistaken, we should get the series $sum_{i=1}^{infty} (-1)^i (x-x_0)^{2i}$ which converges whenever $|x-x_0|<1$?
– Alex
Nov 21 '18 at 6:43
2
I think the OP wants to consider $(1+x^2)^{-1}$.
– Jacky Chong
Nov 21 '18 at 6:46
add a comment |
Try just centering a geometric series around a different point. For example, take $$frac{1}{1-(-(x-x_0)^2))}$$
If this does have power series that converges, what would it look like? This would mean that you can get a radius 1 of converge around any point that you desire.
answered Nov 21 '18 at 6:34
Alex
1323
Try just centering a geometric series around a different point. For example, take $$frac{1}{1-(-(x-x_0)^2))}$$
If this does have power series that converges, what would it look like? This would mean that you can get a radius 1 of converge around any point that you desire.
answered Nov 21 '18 at 6:34
Alex
1323
answered Nov 21 '18 at 6:34
Alex
1323
answered Nov 21 '18 at 6:34
Alex
1323
answered Nov 21 '18 at 6:34
Alex
1323
1323
1
Please check your solution again.
– Jacky Chong
Nov 21 '18 at 6:39
unless I'm mistaken, we should get the series $sum_{i=1}^{infty} (-1)^i (x-x_0)^{2i}$ which converges whenever $|x-x_0|<1$?
– Alex
Nov 21 '18 at 6:43
2
I think the OP wants to consider $(1+x^2)^{-1}$.
– Jacky Chong
Nov 21 '18 at 6:46
add a comment |
1
Please check your solution again.
– Jacky Chong
Nov 21 '18 at 6:39
unless I'm mistaken, we should get the series $sum_{i=1}^{infty} (-1)^i (x-x_0)^{2i}$ which converges whenever $|x-x_0|<1$?
– Alex
Nov 21 '18 at 6:43
2
I think the OP wants to consider $(1+x^2)^{-1}$.
– Jacky Chong
Nov 21 '18 at 6:46
1
1
Please check your solution again.
– Jacky Chong
Nov 21 '18 at 6:39
Please check your solution again.
– Jacky Chong
Nov 21 '18 at 6:39
unless I'm mistaken, we should get the series $sum_{i=1}^{infty} (-1)^i (x-x_0)^{2i}$ which converges whenever $|x-x_0|<1$?
– Alex
Nov 21 '18 at 6:43
unless I'm mistaken, we should get the series $sum_{i=1}^{infty} (-1)^i (x-x_0)^{2i}$ which converges whenever $|x-x_0|<1$?
– Alex
Nov 21 '18 at 6:43
2
2
I think the OP wants to consider $(1+x^2)^{-1}$.
– Jacky Chong
Nov 21 '18 at 6:46
I think the OP wants to consider $(1+x^2)^{-1}$.
– Jacky Chong
Nov 21 '18 at 6:46
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.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- 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.
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%2f3007301%2fshowing-given-function-is-real-analytic-on-r%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
Have you learned about Taylor Series?
– JavaMan
Nov 21 '18 at 5:38
Yes, I have. Yea, maybe we can use taylor series to find the expansion. How do we show it converges? Also, is my statement correct about convergence about a different point in the interval?
– Red Floyd
Nov 21 '18 at 6:22
1
To show a Taylor Series converges, you have to show that $R_n(x) = sum_{k=n+1}^{infty} frac{f^{(k)}(a) (x-a)^k}{k!} to 0$. By Lagrange's Thm $R_n(x) = frac{f^{(n+1)}(c) (x-a)^{n+1}}{(n+1)!}$ for some $c$ between $a$ and $x$. Now, the $n$th derivative is something like $f(x) leq C x^{-n-2}$. Hence, $f^{(n)}(x) to 0$ as $n to infty$. In particular $f^{(n+1)}(c)$ is bounded. Finally, $(x-a)^{n+1}$ is polynomial, and factorials go to zero faster than any polynomial (this follows since $x^n/n! to 0$ as $n to infty$), which is demonstrated by convergence of Taylor series for $e^x$.
– JavaMan
Nov 22 '18 at 6:21