Mixing
Proving that $C(r) = c_1r+c_2r^2 + c_3r^3 + cdots$ can be made as small as desired?
$begingroup$
The $c_i$ are all positive and $C(r)$ converges on some interval around $0$.
This seems like a sort of obvious fact, but it took me several minutes to come up with the idea of doing
$$C(r) = sum_{j=1} ^ {infty} c_j r^j= sum_{j=1}^N c_j r^j + sum_{J=N+1}^{infty} c_j r^j$$
First we can make the sum on the right as small as desired by picking $N$ large enough, and then we can make the finite sum as small as desired because the coefficients are now bounded.
My question is, is there another way to solve this? Since this seems like an obvious fact, it seems like it should have a simpler proof.
real-analysis analysis
asked Jan 27 at 15:37
OviOvi
12.4k1040114
$endgroup$
add a comment |
$begingroup$
The $c_i$ are all positive and $C(r)$ converges on some interval around $0$.
This seems like a sort of obvious fact, but it took me several minutes to come up with the idea of doing
$$C(r) = sum_{j=1} ^ {infty} c_j r^j= sum_{j=1}^N c_j r^j + sum_{J=N+1}^{infty} c_j r^j$$
First we can make the sum on the right as small as desired by picking $N$ large enough, and then we can make the finite sum as small as desired because the coefficients are now bounded.
My question is, is there another way to solve this? Since this seems like an obvious fact, it seems like it should have a simpler proof.
real-analysis analysis
asked Jan 27 at 15:37
OviOvi
12.4k1040114
$endgroup$
$begingroup$
Using the fact that $limsup |c_n|^{1/n}$ is finite, you can bound $|C(r)|$ by a geometric series.
$endgroup$
– Mike Earnest
Jan 27 at 17:58
add a comment |
$begingroup$
The $c_i$ are all positive and $C(r)$ converges on some interval around $0$.
This seems like a sort of obvious fact, but it took me several minutes to come up with the idea of doing
$$C(r) = sum_{j=1} ^ {infty} c_j r^j= sum_{j=1}^N c_j r^j + sum_{J=N+1}^{infty} c_j r^j$$
First we can make the sum on the right as small as desired by picking $N$ large enough, and then we can make the finite sum as small as desired because the coefficients are now bounded.
My question is, is there another way to solve this? Since this seems like an obvious fact, it seems like it should have a simpler proof.
real-analysis analysis
asked Jan 27 at 15:37
OviOvi
12.4k1040114
$endgroup$
The $c_i$ are all positive and $C(r)$ converges on some interval around $0$.
This seems like a sort of obvious fact, but it took me several minutes to come up with the idea of doing
$$C(r) = sum_{j=1} ^ {infty} c_j r^j= sum_{j=1}^N c_j r^j + sum_{J=N+1}^{infty} c_j r^j$$
First we can make the sum on the right as small as desired by picking $N$ large enough, and then we can make the finite sum as small as desired because the coefficients are now bounded.
My question is, is there another way to solve this? Since this seems like an obvious fact, it seems like it should have a simpler proof.
real-analysis analysis
real-analysis analysis
asked Jan 27 at 15:37
OviOvi
12.4k1040114
asked Jan 27 at 15:37
OviOvi
12.4k1040114
asked Jan 27 at 15:37
OviOvi
12.4k1040114
asked Jan 27 at 15:37
OviOvi
12.4k1040114
asked Jan 27 at 15:37
OviOvi
12.4k1040114
12.4k1040114
$begingroup$
Using the fact that $limsup |c_n|^{1/n}$ is finite, you can bound $|C(r)|$ by a geometric series.
$endgroup$
– Mike Earnest
Jan 27 at 17:58
add a comment |
$begingroup$
Using the fact that $limsup |c_n|^{1/n}$ is finite, you can bound $|C(r)|$ by a geometric series.
$endgroup$
– Mike Earnest
Jan 27 at 17:58
$begingroup$
Using the fact that $limsup |c_n|^{1/n}$ is finite, you can bound $|C(r)|$ by a geometric series.
$endgroup$
– Mike Earnest
Jan 27 at 17:58
$begingroup$
Using the fact that $limsup |c_n|^{1/n}$ is finite, you can bound $|C(r)|$ by a geometric series.
$endgroup$
– Mike Earnest
Jan 27 at 17:58
add a comment |
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$begingroup$
Using the fact that $limsup |c_n|^{1/n}$ is finite, you can bound $|C(r)|$ by a geometric series.
$endgroup$
– Mike Earnest
Jan 27 at 17:58