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Integrating the errorfunction using the method of undetermined coefficients
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$DeclareMathOperatorerfErf$
I am trying to solve using the method of undetermined coefficients:
$$interf(x)dx$$
With the method of undetermined coefficients one would start by "simply" guessing a general solution to the integral.
In the case of $interf(x)dx$ my calculusbook suggests we try:
$$interf(x)=P(x)erf(x)+Q(x)e^-x^2+C$$
Here P and Q are polynomials to be determined.
Question: Where on earth does this guess come from? I don't intuit it at all. Any help would be appreciated.
integration error-function
asked Jul 19 at 19:05
GambitSquared
1,2801034
add a comment |Â
up vote
0
down vote
favorite
$DeclareMathOperatorerfErf$
I am trying to solve using the method of undetermined coefficients:
$$interf(x)dx$$
With the method of undetermined coefficients one would start by "simply" guessing a general solution to the integral.
In the case of $interf(x)dx$ my calculusbook suggests we try:
$$interf(x)=P(x)erf(x)+Q(x)e^-x^2+C$$
Here P and Q are polynomials to be determined.
Question: Where on earth does this guess come from? I don't intuit it at all. Any help would be appreciated.
integration error-function
asked Jul 19 at 19:05
GambitSquared
1,2801034
It's called guessing for a reason ... :)
– Hagen von Eitzen
Jul 19 at 19:08
@HagenvonEitzen I think it's supposed to be an educated guess... but for now the "educated" part is beyond me...
– GambitSquared
Jul 19 at 19:09
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
$DeclareMathOperatorerfErf$
I am trying to solve using the method of undetermined coefficients:
$$interf(x)dx$$
With the method of undetermined coefficients one would start by "simply" guessing a general solution to the integral.
In the case of $interf(x)dx$ my calculusbook suggests we try:
$$interf(x)=P(x)erf(x)+Q(x)e^-x^2+C$$
Here P and Q are polynomials to be determined.
Question: Where on earth does this guess come from? I don't intuit it at all. Any help would be appreciated.
integration error-function
asked Jul 19 at 19:05
GambitSquared
1,2801034
$DeclareMathOperatorerfErf$
I am trying to solve using the method of undetermined coefficients:
$$interf(x)dx$$
With the method of undetermined coefficients one would start by "simply" guessing a general solution to the integral.
In the case of $interf(x)dx$ my calculusbook suggests we try:
$$interf(x)=P(x)erf(x)+Q(x)e^-x^2+C$$
Here P and Q are polynomials to be determined.
Question: Where on earth does this guess come from? I don't intuit it at all. Any help would be appreciated.
integration error-function
asked Jul 19 at 19:05
GambitSquared
1,2801034
asked Jul 19 at 19:05
GambitSquared
1,2801034
asked Jul 19 at 19:05
GambitSquared
1,2801034
asked Jul 19 at 19:05
GambitSquared
1,2801034
1,2801034
It's called guessing for a reason ... :)
– Hagen von Eitzen
Jul 19 at 19:08
@HagenvonEitzen I think it's supposed to be an educated guess... but for now the "educated" part is beyond me...
– GambitSquared
Jul 19 at 19:09
add a comment |Â
It's called guessing for a reason ... :)
– Hagen von Eitzen
Jul 19 at 19:08
@HagenvonEitzen I think it's supposed to be an educated guess... but for now the "educated" part is beyond me...
– GambitSquared
Jul 19 at 19:09
It's called guessing for a reason ... :)
– Hagen von Eitzen
Jul 19 at 19:08
It's called guessing for a reason ... :)
– Hagen von Eitzen
Jul 19 at 19:08
@HagenvonEitzen I think it's supposed to be an educated guess... but for now the "educated" part is beyond me...
– GambitSquared
Jul 19 at 19:09
@HagenvonEitzen I think it's supposed to be an educated guess... but for now the "educated" part is beyond me...
– GambitSquared
Jul 19 at 19:09
add a comment |Â
1 Answer
1
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What could be on the right hand side of
$$int f(x),mathrm dx = ldots? $$
After differentiating, the left hand side becomes $f(x)$, of course. So we want something that produces - initially perhaps with some additional "waste" - $f(x)$. By the product rule, $xf(x)$ does produce $f(x)$ (and also $xf'(x)$, with which we have to deal as well), so allowing $xf(x)$ would be a good idea. A slight generalization is to allow not just $x$ times $f(x)$, but in fact $P(x)f(x)$ for arbitrary polynomial $P$, as well as $Q(x)f'(x)$ in order to somehow hopefully cancel the "waste". We may want to continue with polynomial multiples of $f''(x)$ etc., but fortunately in our specific example, $f''$ is already a polynomial multiple of $f'$ ...
answered Jul 19 at 19:16
Hagen von Eitzen
265k20258477
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
What could be on the right hand side of
$$int f(x),mathrm dx = ldots? $$
After differentiating, the left hand side becomes $f(x)$, of course. So we want something that produces - initially perhaps with some additional "waste" - $f(x)$. By the product rule, $xf(x)$ does produce $f(x)$ (and also $xf'(x)$, with which we have to deal as well), so allowing $xf(x)$ would be a good idea. A slight generalization is to allow not just $x$ times $f(x)$, but in fact $P(x)f(x)$ for arbitrary polynomial $P$, as well as $Q(x)f'(x)$ in order to somehow hopefully cancel the "waste". We may want to continue with polynomial multiples of $f''(x)$ etc., but fortunately in our specific example, $f''$ is already a polynomial multiple of $f'$ ...
answered Jul 19 at 19:16
Hagen von Eitzen
265k20258477
add a comment |Â
up vote
1
down vote
accepted
What could be on the right hand side of
$$int f(x),mathrm dx = ldots? $$
After differentiating, the left hand side becomes $f(x)$, of course. So we want something that produces - initially perhaps with some additional "waste" - $f(x)$. By the product rule, $xf(x)$ does produce $f(x)$ (and also $xf'(x)$, with which we have to deal as well), so allowing $xf(x)$ would be a good idea. A slight generalization is to allow not just $x$ times $f(x)$, but in fact $P(x)f(x)$ for arbitrary polynomial $P$, as well as $Q(x)f'(x)$ in order to somehow hopefully cancel the "waste". We may want to continue with polynomial multiples of $f''(x)$ etc., but fortunately in our specific example, $f''$ is already a polynomial multiple of $f'$ ...
answered Jul 19 at 19:16
Hagen von Eitzen
265k20258477
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
What could be on the right hand side of
$$int f(x),mathrm dx = ldots? $$
After differentiating, the left hand side becomes $f(x)$, of course. So we want something that produces - initially perhaps with some additional "waste" - $f(x)$. By the product rule, $xf(x)$ does produce $f(x)$ (and also $xf'(x)$, with which we have to deal as well), so allowing $xf(x)$ would be a good idea. A slight generalization is to allow not just $x$ times $f(x)$, but in fact $P(x)f(x)$ for arbitrary polynomial $P$, as well as $Q(x)f'(x)$ in order to somehow hopefully cancel the "waste". We may want to continue with polynomial multiples of $f''(x)$ etc., but fortunately in our specific example, $f''$ is already a polynomial multiple of $f'$ ...
answered Jul 19 at 19:16
Hagen von Eitzen
265k20258477
What could be on the right hand side of
$$int f(x),mathrm dx = ldots? $$
After differentiating, the left hand side becomes $f(x)$, of course. So we want something that produces - initially perhaps with some additional "waste" - $f(x)$. By the product rule, $xf(x)$ does produce $f(x)$ (and also $xf'(x)$, with which we have to deal as well), so allowing $xf(x)$ would be a good idea. A slight generalization is to allow not just $x$ times $f(x)$, but in fact $P(x)f(x)$ for arbitrary polynomial $P$, as well as $Q(x)f'(x)$ in order to somehow hopefully cancel the "waste". We may want to continue with polynomial multiples of $f''(x)$ etc., but fortunately in our specific example, $f''$ is already a polynomial multiple of $f'$ ...
answered Jul 19 at 19:16
Hagen von Eitzen
265k20258477
answered Jul 19 at 19:16
Hagen von Eitzen
265k20258477
answered Jul 19 at 19:16
Hagen von Eitzen
265k20258477
answered Jul 19 at 19:16
Hagen von Eitzen
265k20258477
265k20258477
add a comment |Â
add a comment |Â
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It's called guessing for a reason ... :)
– Hagen von Eitzen
Jul 19 at 19:08
@HagenvonEitzen I think it's supposed to be an educated guess... but for now the "educated" part is beyond me...
– GambitSquared
Jul 19 at 19:09