Applying a Taylor series WITH RESPECT TO … and AROUND…Taylor / Maclaurin series expansion origin.Another question about $x_0$ in the Taylor seriesLimitations of fractional derivative approximation with Taylor seriesConvergence of a Taylor series expansion involving $tan^-1$Understanding why the Maclaurin series for $sin theta$ converges to $sin theta$ and not something elseIto's lemma and taylor seriesUse the Taylor series for sin(x) to find a Taylor series for $f(x) = p(sin x) $ around point x=0, where $p(x) = 4x^3 - 3x. $Limits with Taylor seriesCauchy product of two different Taylor series $e^x$ and $cos x$Taylor expansion with two variables
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Applying a Taylor series WITH RESPECT TO … and AROUND…
Taylor / Maclaurin series expansion origin.Another question about $x_0$ in the Taylor seriesLimitations of fractional derivative approximation with Taylor seriesConvergence of a Taylor series expansion involving $tan^-1$Understanding why the Maclaurin series for $sin theta$ converges to $sin theta$ and not something elseIto's lemma and taylor seriesUse the Taylor series for sin(x) to find a Taylor series for $f(x) = p(sin x) $ around point x=0, where $p(x) = 4x^3 - 3x. $Limits with Taylor seriesCauchy product of two different Taylor series $e^x$ and $cos x$Taylor expansion with two variables
.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty margin-bottom:0;
$begingroup$
My question is what does it mean
applying a Taylor series with respect to something and around a point.
What is the difference?
Please explain it with the following example:
Apply a Taylor series expansion to $r'$ with respect to $a/r$ around $0$.
Only until the squared terms (inclusive).
Where $$r'^2 = a^2 + r^2 +2arsin(theta)cos(alpha-phi)$$
taylor-expansion
edited 2 hours ago
nmasanta
3,0682 gold badges8 silver badges24 bronze badges
asked 8 hours ago
nachofestnachofest
205 bronze badges
$endgroup$
add a comment |
$begingroup$
My question is what does it mean
applying a Taylor series with respect to something and around a point.
What is the difference?
Please explain it with the following example:
Apply a Taylor series expansion to $r'$ with respect to $a/r$ around $0$.
Only until the squared terms (inclusive).
Where $$r'^2 = a^2 + r^2 +2arsin(theta)cos(alpha-phi)$$
taylor-expansion
edited 2 hours ago
nmasanta
3,0682 gold badges8 silver badges24 bronze badges
asked 8 hours ago
nachofestnachofest
205 bronze badges
$endgroup$
$begingroup$
This is a polynomial in $a$, so its Taylor series is fairly straightforward.
$endgroup$
– eyeballfrog
8 hours ago
$begingroup$
As far as I know, they mean the same thing.
$endgroup$
– The Count
4 hours ago
add a comment |
$begingroup$
My question is what does it mean
applying a Taylor series with respect to something and around a point.
What is the difference?
Please explain it with the following example:
Apply a Taylor series expansion to $r'$ with respect to $a/r$ around $0$.
Only until the squared terms (inclusive).
Where $$r'^2 = a^2 + r^2 +2arsin(theta)cos(alpha-phi)$$
taylor-expansion
edited 2 hours ago
nmasanta
3,0682 gold badges8 silver badges24 bronze badges
asked 8 hours ago
nachofestnachofest
205 bronze badges
$endgroup$
My question is what does it mean
applying a Taylor series with respect to something and around a point.
What is the difference?
Please explain it with the following example:
Apply a Taylor series expansion to $r'$ with respect to $a/r$ around $0$.
Only until the squared terms (inclusive).
Where $$r'^2 = a^2 + r^2 +2arsin(theta)cos(alpha-phi)$$
taylor-expansion
taylor-expansion
edited 2 hours ago
nmasanta
3,0682 gold badges8 silver badges24 bronze badges
asked 8 hours ago
nachofestnachofest
205 bronze badges
edited 2 hours ago
nmasanta
3,0682 gold badges8 silver badges24 bronze badges
asked 8 hours ago
nachofestnachofest
205 bronze badges
edited 2 hours ago
nmasanta
3,0682 gold badges8 silver badges24 bronze badges
edited 2 hours ago
nmasanta
3,0682 gold badges8 silver badges24 bronze badges
edited 2 hours ago
nmasanta
3,0682 gold badges8 silver badges24 bronze badges
3,0682 gold badges8 silver badges24 bronze badges
asked 8 hours ago
nachofestnachofest
205 bronze badges
asked 8 hours ago
nachofestnachofest
205 bronze badges
asked 8 hours ago
nachofestnachofest
205 bronze badges
205 bronze badges
$begingroup$
This is a polynomial in $a$, so its Taylor series is fairly straightforward.
$endgroup$
– eyeballfrog
8 hours ago
$begingroup$
As far as I know, they mean the same thing.
$endgroup$
– The Count
4 hours ago
add a comment |
$begingroup$
This is a polynomial in $a$, so its Taylor series is fairly straightforward.
$endgroup$
– eyeballfrog
8 hours ago
$begingroup$
As far as I know, they mean the same thing.
$endgroup$
– The Count
4 hours ago
$begingroup$
This is a polynomial in $a$, so its Taylor series is fairly straightforward.
$endgroup$
– eyeballfrog
8 hours ago
$begingroup$
This is a polynomial in $a$, so its Taylor series is fairly straightforward.
$endgroup$
– eyeballfrog
8 hours ago
$begingroup$
As far as I know, they mean the same thing.
$endgroup$
– The Count
4 hours ago
$begingroup$
As far as I know, they mean the same thing.
$endgroup$
– The Count
4 hours ago
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
For an easier example, let $f(x) = sin(x)$. The Taylor series expansion of (or, I guess, in your above language 'to') $f$ with respect to $x$ around $0$ is the familar
$$ f(x) = x - fracx^33! + fracx^55! - fracx^77! + dots$$
So you want to rewrite $r'$ in terms of $a/r$ and expand around 0.
answered 8 hours ago
bouncebackbounceback
1,0683 silver badges14 bronze badges
$endgroup$
$begingroup$
thank you @bounceback for your answer, specially for specifying the idea about rewriting it, that was they key point.
$endgroup$
– nachofest
8 hours ago
add a comment |
$begingroup$
The "with respect to" specifies the variable whose powers appear in the series. "around" is the point near which you want an approximation. So your answer will look like
$$
c_o + c_1(a/r - 0) + c_2(a/r - 0)^2 + text higher order terms.
$$
(Of course the zeroes go away. I just put them in to illustrate "around $0$".)
answered 8 hours ago
Ethan BolkerEthan Bolker
53.3k5 gold badges61 silver badges131 bronze badges
$endgroup$
$begingroup$
thank you @ethan-bolker for the example!
$endgroup$
– nachofest
8 hours ago
$begingroup$
by the way, in your answer is it r/a or you meant a/r ? thanks
$endgroup$
– nachofest
7 hours ago
$begingroup$
@nachofest Yes,. Fixed thanks.
$endgroup$
– Ethan Bolker
6 hours ago
add a comment |
Your Answer
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
For an easier example, let $f(x) = sin(x)$. The Taylor series expansion of (or, I guess, in your above language 'to') $f$ with respect to $x$ around $0$ is the familar
$$ f(x) = x - fracx^33! + fracx^55! - fracx^77! + dots$$
So you want to rewrite $r'$ in terms of $a/r$ and expand around 0.
answered 8 hours ago
bouncebackbounceback
1,0683 silver badges14 bronze badges
$endgroup$
$begingroup$
thank you @bounceback for your answer, specially for specifying the idea about rewriting it, that was they key point.
$endgroup$
– nachofest
8 hours ago
add a comment |
$begingroup$
For an easier example, let $f(x) = sin(x)$. The Taylor series expansion of (or, I guess, in your above language 'to') $f$ with respect to $x$ around $0$ is the familar
$$ f(x) = x - fracx^33! + fracx^55! - fracx^77! + dots$$
So you want to rewrite $r'$ in terms of $a/r$ and expand around 0.
answered 8 hours ago
bouncebackbounceback
1,0683 silver badges14 bronze badges
$endgroup$
$begingroup$
thank you @bounceback for your answer, specially for specifying the idea about rewriting it, that was they key point.
$endgroup$
– nachofest
8 hours ago
add a comment |
$begingroup$
For an easier example, let $f(x) = sin(x)$. The Taylor series expansion of (or, I guess, in your above language 'to') $f$ with respect to $x$ around $0$ is the familar
$$ f(x) = x - fracx^33! + fracx^55! - fracx^77! + dots$$
So you want to rewrite $r'$ in terms of $a/r$ and expand around 0.
answered 8 hours ago
bouncebackbounceback
1,0683 silver badges14 bronze badges
$endgroup$
For an easier example, let $f(x) = sin(x)$. The Taylor series expansion of (or, I guess, in your above language 'to') $f$ with respect to $x$ around $0$ is the familar
$$ f(x) = x - fracx^33! + fracx^55! - fracx^77! + dots$$
So you want to rewrite $r'$ in terms of $a/r$ and expand around 0.
answered 8 hours ago
bouncebackbounceback
1,0683 silver badges14 bronze badges
answered 8 hours ago
bouncebackbounceback
1,0683 silver badges14 bronze badges
answered 8 hours ago
bouncebackbounceback
1,0683 silver badges14 bronze badges
answered 8 hours ago
bouncebackbounceback
1,0683 silver badges14 bronze badges
1,0683 silver badges14 bronze badges
$begingroup$
thank you @bounceback for your answer, specially for specifying the idea about rewriting it, that was they key point.
$endgroup$
– nachofest
8 hours ago
add a comment |
$begingroup$
thank you @bounceback for your answer, specially for specifying the idea about rewriting it, that was they key point.
$endgroup$
– nachofest
8 hours ago
$begingroup$
thank you @bounceback for your answer, specially for specifying the idea about rewriting it, that was they key point.
$endgroup$
– nachofest
8 hours ago
$begingroup$
thank you @bounceback for your answer, specially for specifying the idea about rewriting it, that was they key point.
$endgroup$
– nachofest
8 hours ago
add a comment |
$begingroup$
The "with respect to" specifies the variable whose powers appear in the series. "around" is the point near which you want an approximation. So your answer will look like
$$
c_o + c_1(a/r - 0) + c_2(a/r - 0)^2 + text higher order terms.
$$
(Of course the zeroes go away. I just put them in to illustrate "around $0$".)
answered 8 hours ago
Ethan BolkerEthan Bolker
53.3k5 gold badges61 silver badges131 bronze badges
$endgroup$
$begingroup$
thank you @ethan-bolker for the example!
$endgroup$
– nachofest
8 hours ago
$begingroup$
by the way, in your answer is it r/a or you meant a/r ? thanks
$endgroup$
– nachofest
7 hours ago
$begingroup$
@nachofest Yes,. Fixed thanks.
$endgroup$
– Ethan Bolker
6 hours ago
add a comment |
$begingroup$
The "with respect to" specifies the variable whose powers appear in the series. "around" is the point near which you want an approximation. So your answer will look like
$$
c_o + c_1(a/r - 0) + c_2(a/r - 0)^2 + text higher order terms.
$$
(Of course the zeroes go away. I just put them in to illustrate "around $0$".)
answered 8 hours ago
Ethan BolkerEthan Bolker
53.3k5 gold badges61 silver badges131 bronze badges
$endgroup$
$begingroup$
thank you @ethan-bolker for the example!
$endgroup$
– nachofest
8 hours ago
$begingroup$
by the way, in your answer is it r/a or you meant a/r ? thanks
$endgroup$
– nachofest
7 hours ago
$begingroup$
@nachofest Yes,. Fixed thanks.
$endgroup$
– Ethan Bolker
6 hours ago
add a comment |
$begingroup$
The "with respect to" specifies the variable whose powers appear in the series. "around" is the point near which you want an approximation. So your answer will look like
$$
c_o + c_1(a/r - 0) + c_2(a/r - 0)^2 + text higher order terms.
$$
(Of course the zeroes go away. I just put them in to illustrate "around $0$".)
answered 8 hours ago
Ethan BolkerEthan Bolker
53.3k5 gold badges61 silver badges131 bronze badges
$endgroup$
The "with respect to" specifies the variable whose powers appear in the series. "around" is the point near which you want an approximation. So your answer will look like
$$
c_o + c_1(a/r - 0) + c_2(a/r - 0)^2 + text higher order terms.
$$
(Of course the zeroes go away. I just put them in to illustrate "around $0$".)
answered 8 hours ago
Ethan BolkerEthan Bolker
53.3k5 gold badges61 silver badges131 bronze badges
edited 6 hours ago
answered 8 hours ago
Ethan BolkerEthan Bolker
53.3k5 gold badges61 silver badges131 bronze badges
answered 8 hours ago
Ethan BolkerEthan Bolker
53.3k5 gold badges61 silver badges131 bronze badges
answered 8 hours ago
Ethan BolkerEthan Bolker
53.3k5 gold badges61 silver badges131 bronze badges
53.3k5 gold badges61 silver badges131 bronze badges
$begingroup$
thank you @ethan-bolker for the example!
$endgroup$
– nachofest
8 hours ago
$begingroup$
by the way, in your answer is it r/a or you meant a/r ? thanks
$endgroup$
– nachofest
7 hours ago
$begingroup$
@nachofest Yes,. Fixed thanks.
$endgroup$
– Ethan Bolker
6 hours ago
add a comment |
$begingroup$
thank you @ethan-bolker for the example!
$endgroup$
– nachofest
8 hours ago
$begingroup$
by the way, in your answer is it r/a or you meant a/r ? thanks
$endgroup$
– nachofest
7 hours ago
$begingroup$
@nachofest Yes,. Fixed thanks.
$endgroup$
– Ethan Bolker
6 hours ago
$begingroup$
thank you @ethan-bolker for the example!
$endgroup$
– nachofest
8 hours ago
$begingroup$
thank you @ethan-bolker for the example!
$endgroup$
– nachofest
8 hours ago
$begingroup$
by the way, in your answer is it r/a or you meant a/r ? thanks
$endgroup$
– nachofest
7 hours ago
$begingroup$
by the way, in your answer is it r/a or you meant a/r ? thanks
$endgroup$
– nachofest
7 hours ago
$begingroup$
@nachofest Yes,. Fixed thanks.
$endgroup$
– Ethan Bolker
6 hours ago
$begingroup$
@nachofest Yes,. Fixed thanks.
$endgroup$
– Ethan Bolker
6 hours ago
add a comment |
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$begingroup$
This is a polynomial in $a$, so its Taylor series is fairly straightforward.
$endgroup$
– eyeballfrog
8 hours ago
$begingroup$
As far as I know, they mean the same thing.
$endgroup$
– The Count
4 hours ago