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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;








4












$begingroup$


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?










share|improve this question











$endgroup$













  • $begingroup$
    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

















4












$begingroup$


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?










share|improve this question











$endgroup$













  • $begingroup$
    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













4












4








4





$begingroup$


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?










share|improve this question











$endgroup$




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






share|improve this question















share|improve this question













share|improve this question




share|improve this question








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 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
















  • $begingroup$
    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















$begingroup$
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










1 Answer
1






active

oldest

votes


















10












$begingroup$

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.







share|improve this answer











$endgroup$










  • 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






  • 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 MapThreadcommand?
    $endgroup$
    – 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













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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









10












$begingroup$

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.







share|improve this answer











$endgroup$










  • 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






  • 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 MapThreadcommand?
    $endgroup$
    – 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















10












$begingroup$

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.







share|improve this answer











$endgroup$










  • 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






  • 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 MapThreadcommand?
    $endgroup$
    – 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













10












10








10





$begingroup$

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.







share|improve this answer











$endgroup$



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.








share|improve this answer














share|improve this answer



share|improve this answer








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 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




    $begingroup$
    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 MapThreadcommand?
    $endgroup$
    – 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












  • 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






  • 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 MapThreadcommand?
    $endgroup$
    – 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







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 MapThreadcommand?
$endgroup$
– 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 MapThreadcommand?
$endgroup$
– 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

















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