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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?

evaluation

share|improve this question

edited Aug 2 at 7:56

J. M. is away♦

100k1010 gold badges316316 silver badges473473 bronze badges

asked Aug 1 at 22:43

mf67mf67

19066 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  

  
  
  
  

add a comment | 

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?

evaluation

share|improve this question

edited Aug 2 at 7:56

J. M. is away♦

100k1010 gold badges316316 silver badges473473 bronze badges

asked Aug 1 at 22:43

mf67mf67

19066 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  

  
  
  
  

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

share|improve this question

edited Aug 2 at 7:56

J. M. is away♦

100k1010 gold badges316316 silver badges473473 bronze badges

asked Aug 1 at 22:43

mf67mf67

19066 bronze badges

$endgroup$

y[x_] := 1/Sqrt[2] (5 - x - Sqrt[8 - (x + 1)^2])
x0, x1 = -1 - Sqrt[8], -1 + Sqrt[8];

y'[x0], y'[1], y'[x1]

share|improve this question

edited Aug 2 at 7:56

J. M. is away♦

100k1010 gold badges316316 silver badges473473 bronze badges

asked Aug 1 at 22:43

mf67mf67

19066 bronze badges

share|improve this question

edited Aug 2 at 7:56

J. M. is away♦

100k1010 gold badges316316 silver badges473473 bronze badges

asked Aug 1 at 22:43

mf67mf67

19066 bronze badges

edited Aug 2 at 7:56

J. M. is away♦

100k1010 gold badges316316 silver badges473473 bronze badges

edited Aug 2 at 7:56

J. M. is away♦

100k1010 gold badges316316 silver badges473473 bronze badges

asked Aug 1 at 22:43

mf67mf67

19066 bronze badges

asked Aug 1 at 22:43

mf67mf67

19066 bronze badges

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

edited Aug 2 at 9:50

answered Aug 1 at 23:20

kglrkglr

211k1010 gold badges241241 silver badges483483 bronze badges

$endgroup$

• 
  
  
  

  
  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
  

  
  
  
  

add a comment | 

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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

edited Aug 2 at 9:50

answered Aug 1 at 23:20

kglrkglr

211k1010 gold badges241241 silver badges483483 bronze badges

$endgroup$

• 
  
  
  

  
  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
  

  
  
  
  

add a comment | 

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

edited Aug 2 at 9:50

answered Aug 1 at 23:20

kglrkglr

211k1010 gold badges241241 silver badges483483 bronze badges

$endgroup$

• 
  
  
  

  
  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
  

  
  
  
  

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.

share|improve this answer

edited Aug 2 at 9:50

answered Aug 1 at 23:20

kglrkglr

211k1010 gold badges241241 silver badges483483 bronze badges

$endgroup$

MapThread[ Limit[y'[z], z -> #, Direction -> #2] &, 
 x0, 1, x1, "FromAbove", "TwoSided", "FromBelow"]

Limit[y'[z], z -> x0, 1, x1, Direction -> "FromAbove", "TwoSided", "FromBelow"]

$VersionNumber

share|improve this answer

edited Aug 2 at 9:50

answered Aug 1 at 23:20

kglrkglr

211k1010 gold badges241241 silver badges483483 bronze badges

answered Aug 1 at 23:20

kglrkglr

211k1010 gold badges241241 silver badges483483 bronze badges

answered Aug 1 at 23:20

kglrkglr

211k1010 gold badges241241 silver badges483483 bronze badges