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Evaluate an expression at singular points
How to evaluate the $0/0$ type limit in Mathematica?Evaluate an expression at a specific pointHow to evaluate only arguments, but leave topmost expression unevaluated?Failed to use N[%] for a infinite seriesEvaluating summations involving Fibonacci numbers in terms of Fibonacci numbersDid Fibonacci slow down?How to make Mathematica evaluate this integralExtracting order n coefficients from a non-closed form summationHow to evaluate limit of a user-defined functionFirst ReplaceAll, then evaluate an expression
.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty margin-bottom:0;
$begingroup$
I have
y[x_] := 1/Sqrt[2] (5 - x - Sqrt[8 - (x + 1)^2])
x0, x1 = -1 - Sqrt[8], -1 + Sqrt[8];
and I want to calculate
y'[x0], y'[1], y'[x1]
but Mathematica cannot evaluate the left and right values, nor can Limit
.
What is the proper syntax/method to get the answer -Infty, 0, Infty
?
evaluation
$endgroup$
add a comment |
$begingroup$
I have
y[x_] := 1/Sqrt[2] (5 - x - Sqrt[8 - (x + 1)^2])
x0, x1 = -1 - Sqrt[8], -1 + Sqrt[8];
and I want to calculate
y'[x0], y'[1], y'[x1]
but Mathematica cannot evaluate the left and right values, nor can Limit
.
What is the proper syntax/method to get the answer -Infty, 0, Infty
?
evaluation
$endgroup$
$begingroup$
Limit[y'[z], z -> #] & /@ x0, 1, x1
gives-[Infinity],0,(-I) [Infinity]
$endgroup$
– kglr
Aug 1 at 22:52
$begingroup$
I getIndeterminate, 0, Indeterminate
for the above command. I use version 11.2.0.0.
$endgroup$
– mf67
Aug 1 at 23:06
$begingroup$
mf67, it works in v9. In v12 I also getIndeterminate, 0, Indeterminate
$endgroup$
– kglr
Aug 1 at 23:09
$begingroup$
So I might have to wait for next major update to get a v9-result? Usingy'[x0 + 10.^-5], y'[1], y'[x0 - 10.^-5]
to try to "go around" the problem I get-266.621, 0, -0.707107 + 265.915 I
. Is there some way to have only real values returned?
$endgroup$
– mf67
Aug 1 at 23:16
$begingroup$
mf67, i posted an answer that works in v12.
$endgroup$
– kglr
Aug 1 at 23:21
add a comment |
$begingroup$
I have
y[x_] := 1/Sqrt[2] (5 - x - Sqrt[8 - (x + 1)^2])
x0, x1 = -1 - Sqrt[8], -1 + Sqrt[8];
and I want to calculate
y'[x0], y'[1], y'[x1]
but Mathematica cannot evaluate the left and right values, nor can Limit
.
What is the proper syntax/method to get the answer -Infty, 0, Infty
?
evaluation
$endgroup$
I have
y[x_] := 1/Sqrt[2] (5 - x - Sqrt[8 - (x + 1)^2])
x0, x1 = -1 - Sqrt[8], -1 + Sqrt[8];
and I want to calculate
y'[x0], y'[1], y'[x1]
but Mathematica cannot evaluate the left and right values, nor can Limit
.
What is the proper syntax/method to get the answer -Infty, 0, Infty
?
evaluation
evaluation
edited Aug 2 at 7:56
J. M. is away♦
100k10 gold badges316 silver badges473 bronze badges
100k10 gold badges316 silver badges473 bronze badges
asked Aug 1 at 22:43
mf67mf67
1906 bronze badges
1906 bronze badges
$begingroup$
Limit[y'[z], z -> #] & /@ x0, 1, x1
gives-[Infinity],0,(-I) [Infinity]
$endgroup$
– kglr
Aug 1 at 22:52
$begingroup$
I getIndeterminate, 0, Indeterminate
for the above command. I use version 11.2.0.0.
$endgroup$
– mf67
Aug 1 at 23:06
$begingroup$
mf67, it works in v9. In v12 I also getIndeterminate, 0, Indeterminate
$endgroup$
– kglr
Aug 1 at 23:09
$begingroup$
So I might have to wait for next major update to get a v9-result? Usingy'[x0 + 10.^-5], y'[1], y'[x0 - 10.^-5]
to try to "go around" the problem I get-266.621, 0, -0.707107 + 265.915 I
. Is there some way to have only real values returned?
$endgroup$
– mf67
Aug 1 at 23:16
$begingroup$
mf67, i posted an answer that works in v12.
$endgroup$
– kglr
Aug 1 at 23:21
add a comment |
$begingroup$
Limit[y'[z], z -> #] & /@ x0, 1, x1
gives-[Infinity],0,(-I) [Infinity]
$endgroup$
– kglr
Aug 1 at 22:52
$begingroup$
I getIndeterminate, 0, Indeterminate
for the above command. I use version 11.2.0.0.
$endgroup$
– mf67
Aug 1 at 23:06
$begingroup$
mf67, it works in v9. In v12 I also getIndeterminate, 0, Indeterminate
$endgroup$
– kglr
Aug 1 at 23:09
$begingroup$
So I might have to wait for next major update to get a v9-result? Usingy'[x0 + 10.^-5], y'[1], y'[x0 - 10.^-5]
to try to "go around" the problem I get-266.621, 0, -0.707107 + 265.915 I
. Is there some way to have only real values returned?
$endgroup$
– mf67
Aug 1 at 23:16
$begingroup$
mf67, i posted an answer that works in v12.
$endgroup$
– kglr
Aug 1 at 23:21
$begingroup$
Limit[y'[z], z -> #] & /@ x0, 1, x1
gives -[Infinity],0,(-I) [Infinity]
$endgroup$
– kglr
Aug 1 at 22:52
$begingroup$
Limit[y'[z], z -> #] & /@ x0, 1, x1
gives -[Infinity],0,(-I) [Infinity]
$endgroup$
– kglr
Aug 1 at 22:52
$begingroup$
I get
Indeterminate, 0, Indeterminate
for the above command. I use version 11.2.0.0.$endgroup$
– mf67
Aug 1 at 23:06
$begingroup$
I get
Indeterminate, 0, Indeterminate
for the above command. I use version 11.2.0.0.$endgroup$
– mf67
Aug 1 at 23:06
$begingroup$
mf67, it works in v9. In v12 I also get
Indeterminate, 0, Indeterminate
$endgroup$
– kglr
Aug 1 at 23:09
$begingroup$
mf67, it works in v9. In v12 I also get
Indeterminate, 0, Indeterminate
$endgroup$
– kglr
Aug 1 at 23:09
$begingroup$
So I might have to wait for next major update to get a v9-result? Using
y'[x0 + 10.^-5], y'[1], y'[x0 - 10.^-5]
to try to "go around" the problem I get -266.621, 0, -0.707107 + 265.915 I
. Is there some way to have only real values returned?$endgroup$
– mf67
Aug 1 at 23:16
$begingroup$
So I might have to wait for next major update to get a v9-result? Using
y'[x0 + 10.^-5], y'[1], y'[x0 - 10.^-5]
to try to "go around" the problem I get -266.621, 0, -0.707107 + 265.915 I
. Is there some way to have only real values returned?$endgroup$
– mf67
Aug 1 at 23:16
$begingroup$
mf67, i posted an answer that works in v12.
$endgroup$
– kglr
Aug 1 at 23:21
$begingroup$
mf67, i posted an answer that works in v12.
$endgroup$
– kglr
Aug 1 at 23:21
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
You can use Limit
with the option Direction
:
MapThread[ Limit[y'[z], z -> #, Direction -> #2] &,
x0, 1, x1, "FromAbove", "TwoSided", "FromBelow"]
-∞, 0, ∞
Much simpler form (from mf67's comment below):
Limit[y'[z], z -> x0, 1, x1, Direction -> "FromAbove", "TwoSided", "FromBelow"]
same result
$VersionNumber
12.
$endgroup$
1
$begingroup$
The older syntax (for people using older versions) goes likeMapThread[Limit[y'[z], z -> #, Direction -> #2] &, x0, 1, x1, -1, 0, 1]
$endgroup$
– J. M. is away♦
Aug 2 at 7:58
1
$begingroup$
I'm trying to understand the command. I got the same result withLimit[y'[z], z -> x0, 1, x1, Direction -> "FromAbove", "TwoSided", "FromBelow"]
. What are the differences compared to the originalMapThread
command?
$endgroup$
– mf67
Aug 2 at 9:46
$begingroup$
@mf67, the difference is my ignorance of the fact thatLimit
isListable
:)
$endgroup$
– kglr
Aug 2 at 9:49
add a comment |
Your Answer
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
You can use Limit
with the option Direction
:
MapThread[ Limit[y'[z], z -> #, Direction -> #2] &,
x0, 1, x1, "FromAbove", "TwoSided", "FromBelow"]
-∞, 0, ∞
Much simpler form (from mf67's comment below):
Limit[y'[z], z -> x0, 1, x1, Direction -> "FromAbove", "TwoSided", "FromBelow"]
same result
$VersionNumber
12.
$endgroup$
1
$begingroup$
The older syntax (for people using older versions) goes likeMapThread[Limit[y'[z], z -> #, Direction -> #2] &, x0, 1, x1, -1, 0, 1]
$endgroup$
– J. M. is away♦
Aug 2 at 7:58
1
$begingroup$
I'm trying to understand the command. I got the same result withLimit[y'[z], z -> x0, 1, x1, Direction -> "FromAbove", "TwoSided", "FromBelow"]
. What are the differences compared to the originalMapThread
command?
$endgroup$
– mf67
Aug 2 at 9:46
$begingroup$
@mf67, the difference is my ignorance of the fact thatLimit
isListable
:)
$endgroup$
– kglr
Aug 2 at 9:49
add a comment |
$begingroup$
You can use Limit
with the option Direction
:
MapThread[ Limit[y'[z], z -> #, Direction -> #2] &,
x0, 1, x1, "FromAbove", "TwoSided", "FromBelow"]
-∞, 0, ∞
Much simpler form (from mf67's comment below):
Limit[y'[z], z -> x0, 1, x1, Direction -> "FromAbove", "TwoSided", "FromBelow"]
same result
$VersionNumber
12.
$endgroup$
1
$begingroup$
The older syntax (for people using older versions) goes likeMapThread[Limit[y'[z], z -> #, Direction -> #2] &, x0, 1, x1, -1, 0, 1]
$endgroup$
– J. M. is away♦
Aug 2 at 7:58
1
$begingroup$
I'm trying to understand the command. I got the same result withLimit[y'[z], z -> x0, 1, x1, Direction -> "FromAbove", "TwoSided", "FromBelow"]
. What are the differences compared to the originalMapThread
command?
$endgroup$
– mf67
Aug 2 at 9:46
$begingroup$
@mf67, the difference is my ignorance of the fact thatLimit
isListable
:)
$endgroup$
– kglr
Aug 2 at 9:49
add a comment |
$begingroup$
You can use Limit
with the option Direction
:
MapThread[ Limit[y'[z], z -> #, Direction -> #2] &,
x0, 1, x1, "FromAbove", "TwoSided", "FromBelow"]
-∞, 0, ∞
Much simpler form (from mf67's comment below):
Limit[y'[z], z -> x0, 1, x1, Direction -> "FromAbove", "TwoSided", "FromBelow"]
same result
$VersionNumber
12.
$endgroup$
You can use Limit
with the option Direction
:
MapThread[ Limit[y'[z], z -> #, Direction -> #2] &,
x0, 1, x1, "FromAbove", "TwoSided", "FromBelow"]
-∞, 0, ∞
Much simpler form (from mf67's comment below):
Limit[y'[z], z -> x0, 1, x1, Direction -> "FromAbove", "TwoSided", "FromBelow"]
same result
$VersionNumber
12.
edited Aug 2 at 9:50
answered Aug 1 at 23:20
kglrkglr
211k10 gold badges241 silver badges483 bronze badges
211k10 gold badges241 silver badges483 bronze badges
1
$begingroup$
The older syntax (for people using older versions) goes likeMapThread[Limit[y'[z], z -> #, Direction -> #2] &, x0, 1, x1, -1, 0, 1]
$endgroup$
– J. M. is away♦
Aug 2 at 7:58
1
$begingroup$
I'm trying to understand the command. I got the same result withLimit[y'[z], z -> x0, 1, x1, Direction -> "FromAbove", "TwoSided", "FromBelow"]
. What are the differences compared to the originalMapThread
command?
$endgroup$
– mf67
Aug 2 at 9:46
$begingroup$
@mf67, the difference is my ignorance of the fact thatLimit
isListable
:)
$endgroup$
– kglr
Aug 2 at 9:49
add a comment |
1
$begingroup$
The older syntax (for people using older versions) goes likeMapThread[Limit[y'[z], z -> #, Direction -> #2] &, x0, 1, x1, -1, 0, 1]
$endgroup$
– J. M. is away♦
Aug 2 at 7:58
1
$begingroup$
I'm trying to understand the command. I got the same result withLimit[y'[z], z -> x0, 1, x1, Direction -> "FromAbove", "TwoSided", "FromBelow"]
. What are the differences compared to the originalMapThread
command?
$endgroup$
– mf67
Aug 2 at 9:46
$begingroup$
@mf67, the difference is my ignorance of the fact thatLimit
isListable
:)
$endgroup$
– kglr
Aug 2 at 9:49
1
1
$begingroup$
The older syntax (for people using older versions) goes like
MapThread[Limit[y'[z], z -> #, Direction -> #2] &, x0, 1, x1, -1, 0, 1]
$endgroup$
– J. M. is away♦
Aug 2 at 7:58
$begingroup$
The older syntax (for people using older versions) goes like
MapThread[Limit[y'[z], z -> #, Direction -> #2] &, x0, 1, x1, -1, 0, 1]
$endgroup$
– J. M. is away♦
Aug 2 at 7:58
1
1
$begingroup$
I'm trying to understand the command. I got the same result with
Limit[y'[z], z -> x0, 1, x1, Direction -> "FromAbove", "TwoSided", "FromBelow"]
. What are the differences compared to the original MapThread
command?$endgroup$
– mf67
Aug 2 at 9:46
$begingroup$
I'm trying to understand the command. I got the same result with
Limit[y'[z], z -> x0, 1, x1, Direction -> "FromAbove", "TwoSided", "FromBelow"]
. What are the differences compared to the original MapThread
command?$endgroup$
– mf67
Aug 2 at 9:46
$begingroup$
@mf67, the difference is my ignorance of the fact that
Limit
is Listable
:)$endgroup$
– kglr
Aug 2 at 9:49
$begingroup$
@mf67, the difference is my ignorance of the fact that
Limit
is Listable
:)$endgroup$
– kglr
Aug 2 at 9:49
add a comment |
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$begingroup$
Limit[y'[z], z -> #] & /@ x0, 1, x1
gives-[Infinity],0,(-I) [Infinity]
$endgroup$
– kglr
Aug 1 at 22:52
$begingroup$
I get
Indeterminate, 0, Indeterminate
for the above command. I use version 11.2.0.0.$endgroup$
– mf67
Aug 1 at 23:06
$begingroup$
mf67, it works in v9. In v12 I also get
Indeterminate, 0, Indeterminate
$endgroup$
– kglr
Aug 1 at 23:09
$begingroup$
So I might have to wait for next major update to get a v9-result? Using
y'[x0 + 10.^-5], y'[1], y'[x0 - 10.^-5]
to try to "go around" the problem I get-266.621, 0, -0.707107 + 265.915 I
. Is there some way to have only real values returned?$endgroup$
– mf67
Aug 1 at 23:16
$begingroup$
mf67, i posted an answer that works in v12.
$endgroup$
– kglr
Aug 1 at 23:21