Mixing
geometric interpretation of Taylor's theorem
$begingroup$
Deriving Taylor's theorem is not a problem. But I am curious if there is any nice
geometric interpretation of the theorem.
calculus real-analysis geometry
asked May 31 '14 at 6:54
abiabi
1917
$endgroup$
add a comment |
$begingroup$
Deriving Taylor's theorem is not a problem. But I am curious if there is any nice
geometric interpretation of the theorem.
calculus real-analysis geometry
asked May 31 '14 at 6:54
abiabi
1917
$endgroup$
$begingroup$
Projecting the series expansion on the terms of $x^j$ with $j equiv m pmod{n}$ extract rotational symmetries of order n from the function. For instance, the $0 pmod{2}$ terms extract even components and the $1 pmod{2}$ terms extract the odd symmetries.
$endgroup$
– ex0du5
May 31 '14 at 7:01
$begingroup$
The following link might be useful math.stackexchange.com/questions/9422/…
$endgroup$
– Vishesh
May 31 '14 at 7:07
add a comment |
$begingroup$
Deriving Taylor's theorem is not a problem. But I am curious if there is any nice
geometric interpretation of the theorem.
calculus real-analysis geometry
asked May 31 '14 at 6:54
abiabi
1917
$endgroup$
Deriving Taylor's theorem is not a problem. But I am curious if there is any nice
geometric interpretation of the theorem.
calculus real-analysis geometry
calculus real-analysis geometry
asked May 31 '14 at 6:54
abiabi
1917
asked May 31 '14 at 6:54
abiabi
1917
asked May 31 '14 at 6:54
abiabi
1917
asked May 31 '14 at 6:54
abiabi
1917
asked May 31 '14 at 6:54
abiabi
1917
1917
$begingroup$
Projecting the series expansion on the terms of $x^j$ with $j equiv m pmod{n}$ extract rotational symmetries of order n from the function. For instance, the $0 pmod{2}$ terms extract even components and the $1 pmod{2}$ terms extract the odd symmetries.
$endgroup$
– ex0du5
May 31 '14 at 7:01
$begingroup$
The following link might be useful math.stackexchange.com/questions/9422/…
$endgroup$
– Vishesh
May 31 '14 at 7:07
add a comment |
$begingroup$
Projecting the series expansion on the terms of $x^j$ with $j equiv m pmod{n}$ extract rotational symmetries of order n from the function. For instance, the $0 pmod{2}$ terms extract even components and the $1 pmod{2}$ terms extract the odd symmetries.
$endgroup$
– ex0du5
May 31 '14 at 7:01
$begingroup$
The following link might be useful math.stackexchange.com/questions/9422/…
$endgroup$
– Vishesh
May 31 '14 at 7:07
$begingroup$
Projecting the series expansion on the terms of $x^j$ with $j equiv m pmod{n}$ extract rotational symmetries of order n from the function. For instance, the $0 pmod{2}$ terms extract even components and the $1 pmod{2}$ terms extract the odd symmetries.
$endgroup$
– ex0du5
May 31 '14 at 7:01
$begingroup$
Projecting the series expansion on the terms of $x^j$ with $j equiv m pmod{n}$ extract rotational symmetries of order n from the function. For instance, the $0 pmod{2}$ terms extract even components and the $1 pmod{2}$ terms extract the odd symmetries.
$endgroup$
– ex0du5
May 31 '14 at 7:01
$begingroup$
The following link might be useful math.stackexchange.com/questions/9422/…
$endgroup$
– Vishesh
May 31 '14 at 7:07
$begingroup$
The following link might be useful math.stackexchange.com/questions/9422/…
$endgroup$
– Vishesh
May 31 '14 at 7:07
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
I think the answers in the link I added in the comment are really useful. But my own interpretation, possibly not so refined would be two fold.
1) If you consider a "reasonably good-mannered" geometric object, say a curve or a surface, then depending on where you truncate your Taylor series, you can compare it with other well behaved, reference objects locally at least. That is, say for a curve, you can see as to how far it is not a straight line or a parabola/circle and so on.
2) Another possible thread not much different from the first one would be using analyticity, in that the geometric counterparts of such functions(which have a valid Taylor series expansion) are "predictably smooth", in that the smoothness is as you would expect it.
I guess there are a bit too many loose/cheesy words in this answer, but I do hope it helps a little.
answered May 31 '14 at 7:20
VisheshVishesh
1,5681227
$endgroup$
add a comment |
$begingroup$
You're approximating the given function by a polynomial. The polynomial approximation is constructed so that its derivatives agree with the given function at one particular point.
Geometrically, a Taylor series with two terms is a straight-line approximation; the straight line is the tangent at the given point. One with three terms is a parabola approximation, whose tangent and curvature agree with the given function at the given point. And so on.
Given their myopic focus on a single point, the approximations produced this way are often less than ideal -- you often need an approximation that's good throughout some interval. But at least they're easy to construct.
answered May 31 '14 at 7:03
bubbabubba
30.4k33086
$endgroup$
$begingroup$
Dear bubba, I agree with your first part of the answer. But the 2nd part of the answer "The approximations produced this way are not very useful in practice,....." is not fully correct (sorry),. In physics, the truncated Taylor's series (mostly up to the 2nd order correction), under right circumstances, is routinely used to get approximate answers which are quite useful.
$endgroup$
– abi
Jun 2 '14 at 13:49
$begingroup$
@abi -- yes, maybe "not very useful" was too strong. I edited.
$endgroup$
– bubba
Jun 3 '14 at 0:25
add a comment |
$begingroup$
If a function is differentiable in a given domain, then, the functional value of the function at each point is the sum of the functional values of several polynomials at that point.
Visualise several graphs in the $R$ plane, then, their functional values all add up.
answered May 31 '14 at 6:59
MathManMathMan
3,65341970
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
I think the answers in the link I added in the comment are really useful. But my own interpretation, possibly not so refined would be two fold.
1) If you consider a "reasonably good-mannered" geometric object, say a curve or a surface, then depending on where you truncate your Taylor series, you can compare it with other well behaved, reference objects locally at least. That is, say for a curve, you can see as to how far it is not a straight line or a parabola/circle and so on.
2) Another possible thread not much different from the first one would be using analyticity, in that the geometric counterparts of such functions(which have a valid Taylor series expansion) are "predictably smooth", in that the smoothness is as you would expect it.
I guess there are a bit too many loose/cheesy words in this answer, but I do hope it helps a little.
answered May 31 '14 at 7:20
VisheshVishesh
1,5681227
$endgroup$
add a comment |
$begingroup$
I think the answers in the link I added in the comment are really useful. But my own interpretation, possibly not so refined would be two fold.
1) If you consider a "reasonably good-mannered" geometric object, say a curve or a surface, then depending on where you truncate your Taylor series, you can compare it with other well behaved, reference objects locally at least. That is, say for a curve, you can see as to how far it is not a straight line or a parabola/circle and so on.
2) Another possible thread not much different from the first one would be using analyticity, in that the geometric counterparts of such functions(which have a valid Taylor series expansion) are "predictably smooth", in that the smoothness is as you would expect it.
I guess there are a bit too many loose/cheesy words in this answer, but I do hope it helps a little.
answered May 31 '14 at 7:20
VisheshVishesh
1,5681227
$endgroup$
add a comment |
$begingroup$
I think the answers in the link I added in the comment are really useful. But my own interpretation, possibly not so refined would be two fold.
1) If you consider a "reasonably good-mannered" geometric object, say a curve or a surface, then depending on where you truncate your Taylor series, you can compare it with other well behaved, reference objects locally at least. That is, say for a curve, you can see as to how far it is not a straight line or a parabola/circle and so on.
2) Another possible thread not much different from the first one would be using analyticity, in that the geometric counterparts of such functions(which have a valid Taylor series expansion) are "predictably smooth", in that the smoothness is as you would expect it.
I guess there are a bit too many loose/cheesy words in this answer, but I do hope it helps a little.
answered May 31 '14 at 7:20
VisheshVishesh
1,5681227
$endgroup$
I think the answers in the link I added in the comment are really useful. But my own interpretation, possibly not so refined would be two fold.
1) If you consider a "reasonably good-mannered" geometric object, say a curve or a surface, then depending on where you truncate your Taylor series, you can compare it with other well behaved, reference objects locally at least. That is, say for a curve, you can see as to how far it is not a straight line or a parabola/circle and so on.
2) Another possible thread not much different from the first one would be using analyticity, in that the geometric counterparts of such functions(which have a valid Taylor series expansion) are "predictably smooth", in that the smoothness is as you would expect it.
I guess there are a bit too many loose/cheesy words in this answer, but I do hope it helps a little.
answered May 31 '14 at 7:20
VisheshVishesh
1,5681227
answered May 31 '14 at 7:20
VisheshVishesh
1,5681227
answered May 31 '14 at 7:20
VisheshVishesh
1,5681227
answered May 31 '14 at 7:20
VisheshVishesh
1,5681227
1,5681227
add a comment |
add a comment |
$begingroup$
You're approximating the given function by a polynomial. The polynomial approximation is constructed so that its derivatives agree with the given function at one particular point.
Geometrically, a Taylor series with two terms is a straight-line approximation; the straight line is the tangent at the given point. One with three terms is a parabola approximation, whose tangent and curvature agree with the given function at the given point. And so on.
Given their myopic focus on a single point, the approximations produced this way are often less than ideal -- you often need an approximation that's good throughout some interval. But at least they're easy to construct.
answered May 31 '14 at 7:03
bubbabubba
30.4k33086
$endgroup$
$begingroup$
Dear bubba, I agree with your first part of the answer. But the 2nd part of the answer "The approximations produced this way are not very useful in practice,....." is not fully correct (sorry),. In physics, the truncated Taylor's series (mostly up to the 2nd order correction), under right circumstances, is routinely used to get approximate answers which are quite useful.
$endgroup$
– abi
Jun 2 '14 at 13:49
$begingroup$
@abi -- yes, maybe "not very useful" was too strong. I edited.
$endgroup$
– bubba
Jun 3 '14 at 0:25
add a comment |
$begingroup$
You're approximating the given function by a polynomial. The polynomial approximation is constructed so that its derivatives agree with the given function at one particular point.
Geometrically, a Taylor series with two terms is a straight-line approximation; the straight line is the tangent at the given point. One with three terms is a parabola approximation, whose tangent and curvature agree with the given function at the given point. And so on.
Given their myopic focus on a single point, the approximations produced this way are often less than ideal -- you often need an approximation that's good throughout some interval. But at least they're easy to construct.
answered May 31 '14 at 7:03
bubbabubba
30.4k33086
$endgroup$
$begingroup$
Dear bubba, I agree with your first part of the answer. But the 2nd part of the answer "The approximations produced this way are not very useful in practice,....." is not fully correct (sorry),. In physics, the truncated Taylor's series (mostly up to the 2nd order correction), under right circumstances, is routinely used to get approximate answers which are quite useful.
$endgroup$
– abi
Jun 2 '14 at 13:49
$begingroup$
@abi -- yes, maybe "not very useful" was too strong. I edited.
$endgroup$
– bubba
Jun 3 '14 at 0:25
add a comment |
$begingroup$
You're approximating the given function by a polynomial. The polynomial approximation is constructed so that its derivatives agree with the given function at one particular point.
Geometrically, a Taylor series with two terms is a straight-line approximation; the straight line is the tangent at the given point. One with three terms is a parabola approximation, whose tangent and curvature agree with the given function at the given point. And so on.
Given their myopic focus on a single point, the approximations produced this way are often less than ideal -- you often need an approximation that's good throughout some interval. But at least they're easy to construct.
answered May 31 '14 at 7:03
bubbabubba
30.4k33086
$endgroup$
You're approximating the given function by a polynomial. The polynomial approximation is constructed so that its derivatives agree with the given function at one particular point.
Geometrically, a Taylor series with two terms is a straight-line approximation; the straight line is the tangent at the given point. One with three terms is a parabola approximation, whose tangent and curvature agree with the given function at the given point. And so on.
Given their myopic focus on a single point, the approximations produced this way are often less than ideal -- you often need an approximation that's good throughout some interval. But at least they're easy to construct.
answered May 31 '14 at 7:03
bubbabubba
30.4k33086
edited Jun 3 '14 at 0:24
answered May 31 '14 at 7:03
bubbabubba
30.4k33086
answered May 31 '14 at 7:03
bubbabubba
30.4k33086
answered May 31 '14 at 7:03
bubbabubba
30.4k33086
30.4k33086
$begingroup$
Dear bubba, I agree with your first part of the answer. But the 2nd part of the answer "The approximations produced this way are not very useful in practice,....." is not fully correct (sorry),. In physics, the truncated Taylor's series (mostly up to the 2nd order correction), under right circumstances, is routinely used to get approximate answers which are quite useful.
$endgroup$
– abi
Jun 2 '14 at 13:49
$begingroup$
@abi -- yes, maybe "not very useful" was too strong. I edited.
$endgroup$
– bubba
Jun 3 '14 at 0:25
add a comment |
$begingroup$
Dear bubba, I agree with your first part of the answer. But the 2nd part of the answer "The approximations produced this way are not very useful in practice,....." is not fully correct (sorry),. In physics, the truncated Taylor's series (mostly up to the 2nd order correction), under right circumstances, is routinely used to get approximate answers which are quite useful.
$endgroup$
– abi
Jun 2 '14 at 13:49
$begingroup$
@abi -- yes, maybe "not very useful" was too strong. I edited.
$endgroup$
– bubba
Jun 3 '14 at 0:25
$begingroup$
Dear bubba, I agree with your first part of the answer. But the 2nd part of the answer "The approximations produced this way are not very useful in practice,....." is not fully correct (sorry),. In physics, the truncated Taylor's series (mostly up to the 2nd order correction), under right circumstances, is routinely used to get approximate answers which are quite useful.
$endgroup$
– abi
Jun 2 '14 at 13:49
$begingroup$
Dear bubba, I agree with your first part of the answer. But the 2nd part of the answer "The approximations produced this way are not very useful in practice,....." is not fully correct (sorry),. In physics, the truncated Taylor's series (mostly up to the 2nd order correction), under right circumstances, is routinely used to get approximate answers which are quite useful.
$endgroup$
– abi
Jun 2 '14 at 13:49
$begingroup$
@abi -- yes, maybe "not very useful" was too strong. I edited.
$endgroup$
– bubba
Jun 3 '14 at 0:25
$begingroup$
@abi -- yes, maybe "not very useful" was too strong. I edited.
$endgroup$
– bubba
Jun 3 '14 at 0:25
add a comment |
$begingroup$
If a function is differentiable in a given domain, then, the functional value of the function at each point is the sum of the functional values of several polynomials at that point.
Visualise several graphs in the $R$ plane, then, their functional values all add up.
answered May 31 '14 at 6:59
MathManMathMan
3,65341970
$endgroup$
add a comment |
$begingroup$
If a function is differentiable in a given domain, then, the functional value of the function at each point is the sum of the functional values of several polynomials at that point.
Visualise several graphs in the $R$ plane, then, their functional values all add up.
answered May 31 '14 at 6:59
MathManMathMan
3,65341970
$endgroup$
add a comment |
$begingroup$
If a function is differentiable in a given domain, then, the functional value of the function at each point is the sum of the functional values of several polynomials at that point.
Visualise several graphs in the $R$ plane, then, their functional values all add up.
answered May 31 '14 at 6:59
MathManMathMan
3,65341970
$endgroup$
If a function is differentiable in a given domain, then, the functional value of the function at each point is the sum of the functional values of several polynomials at that point.
Visualise several graphs in the $R$ plane, then, their functional values all add up.
answered May 31 '14 at 6:59
MathManMathMan
3,65341970
answered May 31 '14 at 6:59
MathManMathMan
3,65341970
answered May 31 '14 at 6:59
MathManMathMan
3,65341970
answered May 31 '14 at 6:59
MathManMathMan
3,65341970
3,65341970
add a comment |
add a comment |
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$begingroup$
Projecting the series expansion on the terms of $x^j$ with $j equiv m pmod{n}$ extract rotational symmetries of order n from the function. For instance, the $0 pmod{2}$ terms extract even components and the $1 pmod{2}$ terms extract the odd symmetries.
$endgroup$
– ex0du5
May 31 '14 at 7:01
$begingroup$
The following link might be useful math.stackexchange.com/questions/9422/…
$endgroup$
– Vishesh
May 31 '14 at 7:07