Radius of convergence $sum_{n=0}^{infty} a_n z^n$, where $a_n = 2a_{n-1} + 1$
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I want to show that the radius of convergence of $$g(z) = sum_{n=0}^{infty} a_n z^n$$ is $1/2$ when $a_0 = 1$ and $a_n = 2a_{n-1} + 1$ for $n geq 1$. Since $$g(z) = 1 + sum_{n=1}^{infty} a_n z^n = 1 + sum_{n=1}^{infty} (2a_{n-1} + 1)z^n = 2zg(z) + frac{1}{1-z}$$ $$iff g(z) = frac{1}{(1-2z)(1-z)}, $$
I am wondering if I can somehow use this closed form formula for $g(z)$. Since $1-2z = 0$ for $z=1/2$, that makes me suspect that somehow that is related to the radius of convergence, but I haven't been able to figure out how... Can anyone help explain it to me?
sequences-and-series complex-analysis power-series
asked Dec 13 '18 at 21:28
j.eeej.eee
787
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add a comment |
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I want to show that the radius of convergence of $$g(z) = sum_{n=0}^{infty} a_n z^n$$ is $1/2$ when $a_0 = 1$ and $a_n = 2a_{n-1} + 1$ for $n geq 1$. Since $$g(z) = 1 + sum_{n=1}^{infty} a_n z^n = 1 + sum_{n=1}^{infty} (2a_{n-1} + 1)z^n = 2zg(z) + frac{1}{1-z}$$ $$iff g(z) = frac{1}{(1-2z)(1-z)}, $$
I am wondering if I can somehow use this closed form formula for $g(z)$. Since $1-2z = 0$ for $z=1/2$, that makes me suspect that somehow that is related to the radius of convergence, but I haven't been able to figure out how... Can anyone help explain it to me?
sequences-and-series complex-analysis power-series
asked Dec 13 '18 at 21:28
j.eeej.eee
787
$endgroup$
1
$begingroup$
Can't you just use the ratio test? en.wikipedia.org/wiki/Ratio_test
$endgroup$
– SmileyCraft
Dec 13 '18 at 21:33
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Perhaps, but in the solutions manual they seem to do it through this formula, although they aren't very explicit about how they do it. If it is possible then it seems like the easier way. :)
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– j.eee
Dec 13 '18 at 21:45
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If you could show $g(z)$ has a positive radius of convergence, then the sum would have to be equal to this formula. Then, since this function extends to a complex function analytic off ${ 1/2, 1 }$, that would imply that $g$ has a Taylor series expansion with radius of convergence $frac{1}{2}$ - which then must have coefficients $a_n$. However, I don't see any way to bootstrap that initial step without a method which just directly proves the radius of convergence anyway...
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– Daniel Schepler
Dec 13 '18 at 22:28
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That makes some sense. :) Thank you. @DanielSchepler
$endgroup$
– j.eee
Dec 13 '18 at 22:41
add a comment |
$begingroup$
I want to show that the radius of convergence of $$g(z) = sum_{n=0}^{infty} a_n z^n$$ is $1/2$ when $a_0 = 1$ and $a_n = 2a_{n-1} + 1$ for $n geq 1$. Since $$g(z) = 1 + sum_{n=1}^{infty} a_n z^n = 1 + sum_{n=1}^{infty} (2a_{n-1} + 1)z^n = 2zg(z) + frac{1}{1-z}$$ $$iff g(z) = frac{1}{(1-2z)(1-z)}, $$
I am wondering if I can somehow use this closed form formula for $g(z)$. Since $1-2z = 0$ for $z=1/2$, that makes me suspect that somehow that is related to the radius of convergence, but I haven't been able to figure out how... Can anyone help explain it to me?
sequences-and-series complex-analysis power-series
asked Dec 13 '18 at 21:28
j.eeej.eee
787
$endgroup$
I want to show that the radius of convergence of $$g(z) = sum_{n=0}^{infty} a_n z^n$$ is $1/2$ when $a_0 = 1$ and $a_n = 2a_{n-1} + 1$ for $n geq 1$. Since $$g(z) = 1 + sum_{n=1}^{infty} a_n z^n = 1 + sum_{n=1}^{infty} (2a_{n-1} + 1)z^n = 2zg(z) + frac{1}{1-z}$$ $$iff g(z) = frac{1}{(1-2z)(1-z)}, $$
I am wondering if I can somehow use this closed form formula for $g(z)$. Since $1-2z = 0$ for $z=1/2$, that makes me suspect that somehow that is related to the radius of convergence, but I haven't been able to figure out how... Can anyone help explain it to me?
sequences-and-series complex-analysis power-series
sequences-and-series complex-analysis power-series
asked Dec 13 '18 at 21:28
j.eeej.eee
787
asked Dec 13 '18 at 21:28
j.eeej.eee
787
asked Dec 13 '18 at 21:28
j.eeej.eee
787
asked Dec 13 '18 at 21:28
j.eeej.eee
787
asked Dec 13 '18 at 21:28
j.eeej.eee
787
787
1
$begingroup$
Can't you just use the ratio test? en.wikipedia.org/wiki/Ratio_test
$endgroup$
– SmileyCraft
Dec 13 '18 at 21:33
$begingroup$
Perhaps, but in the solutions manual they seem to do it through this formula, although they aren't very explicit about how they do it. If it is possible then it seems like the easier way. :)
$endgroup$
– j.eee
Dec 13 '18 at 21:45
$begingroup$
If you could show $g(z)$ has a positive radius of convergence, then the sum would have to be equal to this formula. Then, since this function extends to a complex function analytic off ${ 1/2, 1 }$, that would imply that $g$ has a Taylor series expansion with radius of convergence $frac{1}{2}$ - which then must have coefficients $a_n$. However, I don't see any way to bootstrap that initial step without a method which just directly proves the radius of convergence anyway...
$endgroup$
– Daniel Schepler
Dec 13 '18 at 22:28
$begingroup$
That makes some sense. :) Thank you. @DanielSchepler
$endgroup$
– j.eee
Dec 13 '18 at 22:41
add a comment |
1
$begingroup$
Can't you just use the ratio test? en.wikipedia.org/wiki/Ratio_test
$endgroup$
– SmileyCraft
Dec 13 '18 at 21:33
$begingroup$
Perhaps, but in the solutions manual they seem to do it through this formula, although they aren't very explicit about how they do it. If it is possible then it seems like the easier way. :)
$endgroup$
– j.eee
Dec 13 '18 at 21:45
$begingroup$
If you could show $g(z)$ has a positive radius of convergence, then the sum would have to be equal to this formula. Then, since this function extends to a complex function analytic off ${ 1/2, 1 }$, that would imply that $g$ has a Taylor series expansion with radius of convergence $frac{1}{2}$ - which then must have coefficients $a_n$. However, I don't see any way to bootstrap that initial step without a method which just directly proves the radius of convergence anyway...
$endgroup$
– Daniel Schepler
Dec 13 '18 at 22:28
$begingroup$
That makes some sense. :) Thank you. @DanielSchepler
$endgroup$
– j.eee
Dec 13 '18 at 22:41
1
1
$begingroup$
Can't you just use the ratio test? en.wikipedia.org/wiki/Ratio_test
$endgroup$
– SmileyCraft
Dec 13 '18 at 21:33
$begingroup$
Can't you just use the ratio test? en.wikipedia.org/wiki/Ratio_test
$endgroup$
– SmileyCraft
Dec 13 '18 at 21:33
$begingroup$
Perhaps, but in the solutions manual they seem to do it through this formula, although they aren't very explicit about how they do it. If it is possible then it seems like the easier way. :)
$endgroup$
– j.eee
Dec 13 '18 at 21:45
$begingroup$
Perhaps, but in the solutions manual they seem to do it through this formula, although they aren't very explicit about how they do it. If it is possible then it seems like the easier way. :)
$endgroup$
– j.eee
Dec 13 '18 at 21:45
$begingroup$
If you could show $g(z)$ has a positive radius of convergence, then the sum would have to be equal to this formula. Then, since this function extends to a complex function analytic off ${ 1/2, 1 }$, that would imply that $g$ has a Taylor series expansion with radius of convergence $frac{1}{2}$ - which then must have coefficients $a_n$. However, I don't see any way to bootstrap that initial step without a method which just directly proves the radius of convergence anyway...
$endgroup$
– Daniel Schepler
Dec 13 '18 at 22:28
$begingroup$
If you could show $g(z)$ has a positive radius of convergence, then the sum would have to be equal to this formula. Then, since this function extends to a complex function analytic off ${ 1/2, 1 }$, that would imply that $g$ has a Taylor series expansion with radius of convergence $frac{1}{2}$ - which then must have coefficients $a_n$. However, I don't see any way to bootstrap that initial step without a method which just directly proves the radius of convergence anyway...
$endgroup$
– Daniel Schepler
Dec 13 '18 at 22:28
$begingroup$
That makes some sense. :) Thank you. @DanielSchepler
$endgroup$
– j.eee
Dec 13 '18 at 22:41
$begingroup$
That makes some sense. :) Thank you. @DanielSchepler
$endgroup$
– j.eee
Dec 13 '18 at 22:41
add a comment |
1 Answer
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You are nearly done:
$$frac{1}{(1-2z)(1-z)}=frac{2}{1-2z}-frac{1}{1-z}
=sumlimits_{n=0}2^{n+1}z^n-sumlimits_{n=0}z^n=\
sumlimits_{n=0}(2^{n+1}-1)z^n$$
from infinite geometric progression, or
$$a_n=2^{n+1}-1$$
This method is called method of generating functions for solving recurrences, here is another good resource. As it was mentioned in the comments, ratio test shows $r=frac{1}{2}$.
answered Dec 13 '18 at 22:52
rtybasertybase
11k21533
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add a comment |
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$begingroup$
You are nearly done:
$$frac{1}{(1-2z)(1-z)}=frac{2}{1-2z}-frac{1}{1-z}
=sumlimits_{n=0}2^{n+1}z^n-sumlimits_{n=0}z^n=\
sumlimits_{n=0}(2^{n+1}-1)z^n$$
from infinite geometric progression, or
$$a_n=2^{n+1}-1$$
This method is called method of generating functions for solving recurrences, here is another good resource. As it was mentioned in the comments, ratio test shows $r=frac{1}{2}$.
answered Dec 13 '18 at 22:52
rtybasertybase
11k21533
$endgroup$
add a comment |
$begingroup$
You are nearly done:
$$frac{1}{(1-2z)(1-z)}=frac{2}{1-2z}-frac{1}{1-z}
=sumlimits_{n=0}2^{n+1}z^n-sumlimits_{n=0}z^n=\
sumlimits_{n=0}(2^{n+1}-1)z^n$$
from infinite geometric progression, or
$$a_n=2^{n+1}-1$$
This method is called method of generating functions for solving recurrences, here is another good resource. As it was mentioned in the comments, ratio test shows $r=frac{1}{2}$.
answered Dec 13 '18 at 22:52
rtybasertybase
11k21533
$endgroup$
add a comment |
$begingroup$
You are nearly done:
$$frac{1}{(1-2z)(1-z)}=frac{2}{1-2z}-frac{1}{1-z}
=sumlimits_{n=0}2^{n+1}z^n-sumlimits_{n=0}z^n=\
sumlimits_{n=0}(2^{n+1}-1)z^n$$
from infinite geometric progression, or
$$a_n=2^{n+1}-1$$
This method is called method of generating functions for solving recurrences, here is another good resource. As it was mentioned in the comments, ratio test shows $r=frac{1}{2}$.
answered Dec 13 '18 at 22:52
rtybasertybase
11k21533
$endgroup$
You are nearly done:
$$frac{1}{(1-2z)(1-z)}=frac{2}{1-2z}-frac{1}{1-z}
=sumlimits_{n=0}2^{n+1}z^n-sumlimits_{n=0}z^n=\
sumlimits_{n=0}(2^{n+1}-1)z^n$$
from infinite geometric progression, or
$$a_n=2^{n+1}-1$$
This method is called method of generating functions for solving recurrences, here is another good resource. As it was mentioned in the comments, ratio test shows $r=frac{1}{2}$.
answered Dec 13 '18 at 22:52
rtybasertybase
11k21533
edited Dec 13 '18 at 23:39
answered Dec 13 '18 at 22:52
rtybasertybase
11k21533
answered Dec 13 '18 at 22:52
rtybasertybase
11k21533
answered Dec 13 '18 at 22:52
rtybasertybase
11k21533
11k21533
add a comment |
add a comment |
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$begingroup$
Can't you just use the ratio test? en.wikipedia.org/wiki/Ratio_test
$endgroup$
– SmileyCraft
Dec 13 '18 at 21:33
$begingroup$
Perhaps, but in the solutions manual they seem to do it through this formula, although they aren't very explicit about how they do it. If it is possible then it seems like the easier way. :)
$endgroup$
– j.eee
Dec 13 '18 at 21:45
$begingroup$
If you could show $g(z)$ has a positive radius of convergence, then the sum would have to be equal to this formula. Then, since this function extends to a complex function analytic off ${ 1/2, 1 }$, that would imply that $g$ has a Taylor series expansion with radius of convergence $frac{1}{2}$ - which then must have coefficients $a_n$. However, I don't see any way to bootstrap that initial step without a method which just directly proves the radius of convergence anyway...
$endgroup$
– Daniel Schepler
Dec 13 '18 at 22:28
$begingroup$
That makes some sense. :) Thank you. @DanielSchepler
$endgroup$
– j.eee
Dec 13 '18 at 22:41