In perturbation theory, does it make sense to consider added error to “shift” the function and alter no....
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In perturbation theory, does it make sense to consider added error to "shift" the function and alter no. of roots?
That is consider e.g. $f(x)=x^2+epsilon$. For $epsilon=0$ there's a single root $x=0$.
If $epsilon > 0$, then there would be no roots.
If $epsilon < 0$, then there would be two roots.
What confuses me is that I've not seen this kind of reasoning in the perturbation theory I've read so far. Rather thay seem to treat $epsilon$ as something that you plug in after you've done asymptotic expansion. And it spits out you an approximate value of the function near that $x(epsilon)$.
But for example if one considers:
$$f(x(epsilon))=(x_0+x_1epsilon+x_2epsilon^2+...)^2+epsilon$$
Then I don't see how $epsilon > 0, epsilon < 0$ would make the solution through perturbation technique any different. Like as if perturbation does not consider the number of roots, but only approximate solution, for root.
perturbation-theory
asked Jan 13 at 18:11
mavaviljmavavilj
2,85711138
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add a comment |
$begingroup$
In perturbation theory, does it make sense to consider added error to "shift" the function and alter no. of roots?
That is consider e.g. $f(x)=x^2+epsilon$. For $epsilon=0$ there's a single root $x=0$.
If $epsilon > 0$, then there would be no roots.
If $epsilon < 0$, then there would be two roots.
What confuses me is that I've not seen this kind of reasoning in the perturbation theory I've read so far. Rather thay seem to treat $epsilon$ as something that you plug in after you've done asymptotic expansion. And it spits out you an approximate value of the function near that $x(epsilon)$.
But for example if one considers:
$$f(x(epsilon))=(x_0+x_1epsilon+x_2epsilon^2+...)^2+epsilon$$
Then I don't see how $epsilon > 0, epsilon < 0$ would make the solution through perturbation technique any different. Like as if perturbation does not consider the number of roots, but only approximate solution, for root.
perturbation-theory
asked Jan 13 at 18:11
mavaviljmavavilj
2,85711138
$endgroup$
$begingroup$
Perturbation theory virtually always is applied in the case $0<epsilonll1$. In that case for $epsilon=0$ there is one solution and for $epsilon>0$ there are two solutions. This situation is handled pretty simply by perturbation theory (e.g. method of dominant balance).
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– David
Jan 14 at 18:05
add a comment |
$begingroup$
In perturbation theory, does it make sense to consider added error to "shift" the function and alter no. of roots?
That is consider e.g. $f(x)=x^2+epsilon$. For $epsilon=0$ there's a single root $x=0$.
If $epsilon > 0$, then there would be no roots.
If $epsilon < 0$, then there would be two roots.
What confuses me is that I've not seen this kind of reasoning in the perturbation theory I've read so far. Rather thay seem to treat $epsilon$ as something that you plug in after you've done asymptotic expansion. And it spits out you an approximate value of the function near that $x(epsilon)$.
But for example if one considers:
$$f(x(epsilon))=(x_0+x_1epsilon+x_2epsilon^2+...)^2+epsilon$$
Then I don't see how $epsilon > 0, epsilon < 0$ would make the solution through perturbation technique any different. Like as if perturbation does not consider the number of roots, but only approximate solution, for root.
perturbation-theory
asked Jan 13 at 18:11
mavaviljmavavilj
2,85711138
$endgroup$
In perturbation theory, does it make sense to consider added error to "shift" the function and alter no. of roots?
That is consider e.g. $f(x)=x^2+epsilon$. For $epsilon=0$ there's a single root $x=0$.
If $epsilon > 0$, then there would be no roots.
If $epsilon < 0$, then there would be two roots.
What confuses me is that I've not seen this kind of reasoning in the perturbation theory I've read so far. Rather thay seem to treat $epsilon$ as something that you plug in after you've done asymptotic expansion. And it spits out you an approximate value of the function near that $x(epsilon)$.
But for example if one considers:
$$f(x(epsilon))=(x_0+x_1epsilon+x_2epsilon^2+...)^2+epsilon$$
Then I don't see how $epsilon > 0, epsilon < 0$ would make the solution through perturbation technique any different. Like as if perturbation does not consider the number of roots, but only approximate solution, for root.
perturbation-theory
perturbation-theory
asked Jan 13 at 18:11
mavaviljmavavilj
2,85711138
asked Jan 13 at 18:11
mavaviljmavavilj
2,85711138
asked Jan 13 at 18:11
mavaviljmavavilj
2,85711138
asked Jan 13 at 18:11
mavaviljmavavilj
2,85711138
asked Jan 13 at 18:11
mavaviljmavavilj
2,85711138
2,85711138
$begingroup$
Perturbation theory virtually always is applied in the case $0<epsilonll1$. In that case for $epsilon=0$ there is one solution and for $epsilon>0$ there are two solutions. This situation is handled pretty simply by perturbation theory (e.g. method of dominant balance).
$endgroup$
– David
Jan 14 at 18:05
add a comment |
$begingroup$
Perturbation theory virtually always is applied in the case $0<epsilonll1$. In that case for $epsilon=0$ there is one solution and for $epsilon>0$ there are two solutions. This situation is handled pretty simply by perturbation theory (e.g. method of dominant balance).
$endgroup$
– David
Jan 14 at 18:05
$begingroup$
Perturbation theory virtually always is applied in the case $0<epsilonll1$. In that case for $epsilon=0$ there is one solution and for $epsilon>0$ there are two solutions. This situation is handled pretty simply by perturbation theory (e.g. method of dominant balance).
$endgroup$
– David
Jan 14 at 18:05
$begingroup$
Perturbation theory virtually always is applied in the case $0<epsilonll1$. In that case for $epsilon=0$ there is one solution and for $epsilon>0$ there are two solutions. This situation is handled pretty simply by perturbation theory (e.g. method of dominant balance).
$endgroup$
– David
Jan 14 at 18:05
add a comment |
1 Answer
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The two simple roots of eg. $x^2 - 1 +epsilon$ vary analytically with $epsilon$ around $epsilon = 0$ so you could use perturbation theory to estimate them. The roots of $x^2+epsilon$ do not vary analytically around $epsilon = 0$.
answered Jan 13 at 19:35
Keith McClaryKeith McClary
8481412
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The two simple roots of eg. $x^2 - 1 +epsilon$ vary analytically with $epsilon$ around $epsilon = 0$ so you could use perturbation theory to estimate them. The roots of $x^2+epsilon$ do not vary analytically around $epsilon = 0$.
answered Jan 13 at 19:35
Keith McClaryKeith McClary
8481412
$endgroup$
add a comment |
$begingroup$
The two simple roots of eg. $x^2 - 1 +epsilon$ vary analytically with $epsilon$ around $epsilon = 0$ so you could use perturbation theory to estimate them. The roots of $x^2+epsilon$ do not vary analytically around $epsilon = 0$.
answered Jan 13 at 19:35
Keith McClaryKeith McClary
8481412
$endgroup$
add a comment |
$begingroup$
The two simple roots of eg. $x^2 - 1 +epsilon$ vary analytically with $epsilon$ around $epsilon = 0$ so you could use perturbation theory to estimate them. The roots of $x^2+epsilon$ do not vary analytically around $epsilon = 0$.
answered Jan 13 at 19:35
Keith McClaryKeith McClary
8481412
$endgroup$
The two simple roots of eg. $x^2 - 1 +epsilon$ vary analytically with $epsilon$ around $epsilon = 0$ so you could use perturbation theory to estimate them. The roots of $x^2+epsilon$ do not vary analytically around $epsilon = 0$.
answered Jan 13 at 19:35
Keith McClaryKeith McClary
8481412
answered Jan 13 at 19:35
Keith McClaryKeith McClary
8481412
answered Jan 13 at 19:35
Keith McClaryKeith McClary
8481412
answered Jan 13 at 19:35
Keith McClaryKeith McClary
8481412
8481412
add a comment |
add a comment |
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Perturbation theory virtually always is applied in the case $0<epsilonll1$. In that case for $epsilon=0$ there is one solution and for $epsilon>0$ there are two solutions. This situation is handled pretty simply by perturbation theory (e.g. method of dominant balance).
$endgroup$
– David
Jan 14 at 18:05