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Around usage results

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Around usage results

9

1

$begingroup$

Why is

Around[10, 1]^2

100. +- 20

and

Around[10, 1] Around[10, 1]

giving 100 +- 14 ?

Just curious I would suspect them to be the same but I'm not an expert on this for sure.

core-language

share|improve this question

asked 8 hours ago

LouLou

1,6581323

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  If the intent for Around in Mathematica is for propagating errors, then this is almost certainly a bug. Interestingly enough, I think the second one is correct, whereas the first is not.
  $endgroup$
  – march
  8 hours ago
  
  
  

  
  
  
  

add a comment | 

9

1

$begingroup$

Why is

Around[10, 1]^2

100. +- 20

and

Around[10, 1] Around[10, 1]

giving 100 +- 14 ?

Just curious I would suspect them to be the same but I'm not an expert on this for sure.

core-language

share|improve this question

asked 8 hours ago

LouLou

1,6581323

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  If the intent for Around in Mathematica is for propagating errors, then this is almost certainly a bug. Interestingly enough, I think the second one is correct, whereas the first is not.
  $endgroup$
  – march
  8 hours ago
  
  
  

  
  
  
  

add a comment | 

$begingroup$

Why is

Around[10, 1]^2

100. +- 20

and

Around[10, 1] Around[10, 1]

giving 100 +- 14 ?

Just curious I would suspect them to be the same but I'm not an expert on this for sure.

core-language

share|improve this question

asked 8 hours ago

LouLou

1,6581323

$endgroup$

Around[10, 1]^2

Around[10, 1] Around[10, 1]

share|improve this question

asked 8 hours ago

LouLou

1,6581323

share|improve this question

asked 8 hours ago

LouLou

1,6581323

asked 8 hours ago

LouLou

1,6581323

asked 8 hours ago

LouLou

1,6581323

1 Answer
1

active

oldest

votes

10

$begingroup$

There's a subtlety here.

When operating together multiple Arounds, the two uncertainties will be considered separate and independent (in the statistical sense). Thus the result will be different from squaring a single one.

Consider

N@StandardDeviation@

  TransformedDistribution[

    x y, 

   {x [Distributed] NormalDistribution[10, 1], 

    y [Distributed] NormalDistribution[10, 1]}]



(* 14.1774 *)

versus

N@StandardDeviation@TransformedDistribution[x^2, {x [Distributed] NormalDistribution[10, 1]}]



(* 20.0499 *)

The same can also be observed with addition:

Around[10, 1] + Around[10, 1]

(* Around[20., 1.4142135623730951`] *)



2 Around[10, 1]

(* Around[20., 2.] *)

---

To specify that all occurrences of an Around expression are the same, use AroundReplace.

For example, compare

x^2 + x /. x -> Around[1, .1]

(* Around[2., 0.223606797749979] *)

with

AroundReplace[x^2 + x, x -> Around[1, .1]]

(* Around[2., 0.30000000000000004`] *)

share|improve this answer

edited 7 hours ago

Carl Lange

5,36411242

answered 8 hours ago

SzabolcsSzabolcs

164k14449953

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Also, AroundReplace[ x1 x2, {{x1, x2} -> VectorAround[{10., 10.}, {{1., 1.}, 1.}]}] vs AroundReplace[ x1 x2, {{x1, x2} -> VectorAround[{10., 10.}, {{1., 1.}, 0.}]}].
  $endgroup$
  – b.gatessucks
  8 hours ago
  

  
  
  
  

add a comment | 

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10

$begingroup$

There's a subtlety here.

When operating together multiple Arounds, the two uncertainties will be considered separate and independent (in the statistical sense). Thus the result will be different from squaring a single one.

Consider

N@StandardDeviation@

  TransformedDistribution[

    x y, 

   {x [Distributed] NormalDistribution[10, 1], 

    y [Distributed] NormalDistribution[10, 1]}]



(* 14.1774 *)

versus

N@StandardDeviation@TransformedDistribution[x^2, {x [Distributed] NormalDistribution[10, 1]}]



(* 20.0499 *)

The same can also be observed with addition:

Around[10, 1] + Around[10, 1]

(* Around[20., 1.4142135623730951`] *)



2 Around[10, 1]

(* Around[20., 2.] *)

---

To specify that all occurrences of an Around expression are the same, use AroundReplace.

For example, compare

x^2 + x /. x -> Around[1, .1]

(* Around[2., 0.223606797749979] *)

with

AroundReplace[x^2 + x, x -> Around[1, .1]]

(* Around[2., 0.30000000000000004`] *)

share|improve this answer

edited 7 hours ago

Carl Lange

5,36411242

answered 8 hours ago

SzabolcsSzabolcs

164k14449953

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Also, AroundReplace[ x1 x2, {{x1, x2} -> VectorAround[{10., 10.}, {{1., 1.}, 1.}]}] vs AroundReplace[ x1 x2, {{x1, x2} -> VectorAround[{10., 10.}, {{1., 1.}, 0.}]}].
  $endgroup$
  – b.gatessucks
  8 hours ago
  

  
  
  
  

add a comment | 

10

$begingroup$

There's a subtlety here.

When operating together multiple Arounds, the two uncertainties will be considered separate and independent (in the statistical sense). Thus the result will be different from squaring a single one.

Consider

N@StandardDeviation@

  TransformedDistribution[

    x y, 

   {x [Distributed] NormalDistribution[10, 1], 

    y [Distributed] NormalDistribution[10, 1]}]



(* 14.1774 *)

versus

N@StandardDeviation@TransformedDistribution[x^2, {x [Distributed] NormalDistribution[10, 1]}]



(* 20.0499 *)

The same can also be observed with addition:

Around[10, 1] + Around[10, 1]

(* Around[20., 1.4142135623730951`] *)



2 Around[10, 1]

(* Around[20., 2.] *)

---

To specify that all occurrences of an Around expression are the same, use AroundReplace.

For example, compare

x^2 + x /. x -> Around[1, .1]

(* Around[2., 0.223606797749979] *)

with

AroundReplace[x^2 + x, x -> Around[1, .1]]

(* Around[2., 0.30000000000000004`] *)

share|improve this answer

edited 7 hours ago

Carl Lange

5,36411242

answered 8 hours ago

SzabolcsSzabolcs

164k14449953

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Also, AroundReplace[ x1 x2, {{x1, x2} -> VectorAround[{10., 10.}, {{1., 1.}, 1.}]}] vs AroundReplace[ x1 x2, {{x1, x2} -> VectorAround[{10., 10.}, {{1., 1.}, 0.}]}].
  $endgroup$
  – b.gatessucks
  8 hours ago
  

  
  
  
  

add a comment | 

$begingroup$

There's a subtlety here.

When operating together multiple Arounds, the two uncertainties will be considered separate and independent (in the statistical sense). Thus the result will be different from squaring a single one.

Consider

N@StandardDeviation@

  TransformedDistribution[

    x y, 

   {x [Distributed] NormalDistribution[10, 1], 

    y [Distributed] NormalDistribution[10, 1]}]



(* 14.1774 *)

versus

N@StandardDeviation@TransformedDistribution[x^2, {x [Distributed] NormalDistribution[10, 1]}]



(* 20.0499 *)

The same can also be observed with addition:

Around[10, 1] + Around[10, 1]

(* Around[20., 1.4142135623730951`] *)



2 Around[10, 1]

(* Around[20., 2.] *)

---

To specify that all occurrences of an Around expression are the same, use AroundReplace.

For example, compare

x^2 + x /. x -> Around[1, .1]

(* Around[2., 0.223606797749979] *)

with

AroundReplace[x^2 + x, x -> Around[1, .1]]

(* Around[2., 0.30000000000000004`] *)

share|improve this answer

edited 7 hours ago

Carl Lange

5,36411242

answered 8 hours ago

SzabolcsSzabolcs

164k14449953

$endgroup$

N@StandardDeviation@

  TransformedDistribution[

    x y, 

   {x [Distributed] NormalDistribution[10, 1], 

    y [Distributed] NormalDistribution[10, 1]}]



(* 14.1774 *)

N@StandardDeviation@TransformedDistribution[x^2, {x [Distributed] NormalDistribution[10, 1]}]



(* 20.0499 *)

Around[10, 1] + Around[10, 1]

(* Around[20., 1.4142135623730951`] *)



2 Around[10, 1]

(* Around[20., 2.] *)

x^2 + x /. x -> Around[1, .1]

(* Around[2., 0.223606797749979] *)

AroundReplace[x^2 + x, x -> Around[1, .1]]

(* Around[2., 0.30000000000000004`] *)

share|improve this answer

edited 7 hours ago

Carl Lange

5,36411242

answered 8 hours ago

SzabolcsSzabolcs

164k14449953

edited 7 hours ago

Carl Lange

5,36411242

edited 7 hours ago

Carl Lange

5,36411242

answered 8 hours ago

SzabolcsSzabolcs

164k14449953

answered 8 hours ago

SzabolcsSzabolcs

164k14449953