Mixing
Descartes' rule of signs for infinite series
Clash Royale CLAN TAG#URR8PPP
up vote
7
down vote
favorite
Consider the function given by
$$f(x)=1-a_1x-a_2x^2-a_3x^3-cdots$$
where each $a_kgeq0$ and some $a_j>0$. If $f(x)$ is a polynomial then Descartes' Rule of signs tells us there is exactly one positive zero, i.e. root of $f(x)=0$.
Assume $f(x)$ is a (real) power series with radius of convergence $0<R<infty$.
Question. For which class or classes of such $f$ can we ensure that there is only one positive real root? This is asking for imposing condition(s).
real-analysis sequences-and-series power-series
asked Sep 2 at 4:16
T. Amdeberhan
15.7k225119
add a comment |Â
up vote
7
down vote
favorite
Consider the function given by
$$f(x)=1-a_1x-a_2x^2-a_3x^3-cdots$$
where each $a_kgeq0$ and some $a_j>0$. If $f(x)$ is a polynomial then Descartes' Rule of signs tells us there is exactly one positive zero, i.e. root of $f(x)=0$.
Assume $f(x)$ is a (real) power series with radius of convergence $0<R<infty$.
Question. For which class or classes of such $f$ can we ensure that there is only one positive real root? This is asking for imposing condition(s).
real-analysis sequences-and-series power-series
asked Sep 2 at 4:16
T. Amdeberhan
15.7k225119
add a comment |Â
up vote
7
down vote
favorite
up vote
7
down vote
favorite
Consider the function given by
$$f(x)=1-a_1x-a_2x^2-a_3x^3-cdots$$
where each $a_kgeq0$ and some $a_j>0$. If $f(x)$ is a polynomial then Descartes' Rule of signs tells us there is exactly one positive zero, i.e. root of $f(x)=0$.
Assume $f(x)$ is a (real) power series with radius of convergence $0<R<infty$.
Question. For which class or classes of such $f$ can we ensure that there is only one positive real root? This is asking for imposing condition(s).
real-analysis sequences-and-series power-series
asked Sep 2 at 4:16
T. Amdeberhan
15.7k225119
Consider the function given by
$$f(x)=1-a_1x-a_2x^2-a_3x^3-cdots$$
where each $a_kgeq0$ and some $a_j>0$. If $f(x)$ is a polynomial then Descartes' Rule of signs tells us there is exactly one positive zero, i.e. root of $f(x)=0$.
Assume $f(x)$ is a (real) power series with radius of convergence $0<R<infty$.
Question. For which class or classes of such $f$ can we ensure that there is only one positive real root? This is asking for imposing condition(s).
real-analysis sequences-and-series power-series
asked Sep 2 at 4:16
T. Amdeberhan
15.7k225119
asked Sep 2 at 4:16
T. Amdeberhan
15.7k225119
asked Sep 2 at 4:16
T. Amdeberhan
15.7k225119
asked Sep 2 at 4:16
T. Amdeberhan
15.7k225119
15.7k225119
add a comment |Â
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
11
down vote
accepted
$f$ is strictly decreasing on $[0,R)$, so if there is any positive zero there is only one. There is a positive zero in $[0,R)$ iff $lim_x to R- f(x) < 0$, which may or may not be true. For an example where it is not, consider
$$ 1 - sum_n=2^infty fracx^nn^2$$
answered Sep 2 at 4:44
Robert Israel
40.3k46113
4
So the condition may be restated as $sum_k=1^+infty a_k R^k>1$ (which is automatically true if $R=+infty$)
– Pietro Majer
Sep 2 at 7:21
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
11
down vote
accepted
$f$ is strictly decreasing on $[0,R)$, so if there is any positive zero there is only one. There is a positive zero in $[0,R)$ iff $lim_x to R- f(x) < 0$, which may or may not be true. For an example where it is not, consider
$$ 1 - sum_n=2^infty fracx^nn^2$$
answered Sep 2 at 4:44
Robert Israel
40.3k46113
4
So the condition may be restated as $sum_k=1^+infty a_k R^k>1$ (which is automatically true if $R=+infty$)
– Pietro Majer
Sep 2 at 7:21
add a comment |Â
up vote
11
down vote
accepted
$f$ is strictly decreasing on $[0,R)$, so if there is any positive zero there is only one. There is a positive zero in $[0,R)$ iff $lim_x to R- f(x) < 0$, which may or may not be true. For an example where it is not, consider
$$ 1 - sum_n=2^infty fracx^nn^2$$
answered Sep 2 at 4:44
Robert Israel
40.3k46113
4
So the condition may be restated as $sum_k=1^+infty a_k R^k>1$ (which is automatically true if $R=+infty$)
– Pietro Majer
Sep 2 at 7:21
add a comment |Â
up vote
11
down vote
accepted
up vote
11
down vote
accepted
$f$ is strictly decreasing on $[0,R)$, so if there is any positive zero there is only one. There is a positive zero in $[0,R)$ iff $lim_x to R- f(x) < 0$, which may or may not be true. For an example where it is not, consider
$$ 1 - sum_n=2^infty fracx^nn^2$$
answered Sep 2 at 4:44
Robert Israel
40.3k46113
$f$ is strictly decreasing on $[0,R)$, so if there is any positive zero there is only one. There is a positive zero in $[0,R)$ iff $lim_x to R- f(x) < 0$, which may or may not be true. For an example where it is not, consider
$$ 1 - sum_n=2^infty fracx^nn^2$$
answered Sep 2 at 4:44
Robert Israel
40.3k46113
answered Sep 2 at 4:44
Robert Israel
40.3k46113
answered Sep 2 at 4:44
Robert Israel
40.3k46113
answered Sep 2 at 4:44
Robert Israel
40.3k46113
40.3k46113
4
So the condition may be restated as $sum_k=1^+infty a_k R^k>1$ (which is automatically true if $R=+infty$)
– Pietro Majer
Sep 2 at 7:21
add a comment |Â
4
So the condition may be restated as $sum_k=1^+infty a_k R^k>1$ (which is automatically true if $R=+infty$)
– Pietro Majer
Sep 2 at 7:21
4
4
So the condition may be restated as $sum_k=1^+infty a_k R^k>1$ (which is automatically true if $R=+infty$)
– Pietro Majer
Sep 2 at 7:21
So the condition may be restated as $sum_k=1^+infty a_k R^k>1$ (which is automatically true if $R=+infty$)
– Pietro Majer
Sep 2 at 7:21
add a comment |Â
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
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmathoverflow.net%2fquestions%2f309645%2fdescartes-rule-of-signs-for-infinite-series%23new-answer', 'question_page');
);
Post as a guest
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
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
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
