Logarithm of exponentialHow to take the derivative w.r.t. an arbitrary function?How can I separate two parts of power of Exponential functions?Matrix exponential MatrixExp[] vs Sum[MatrixPower[]] doesn't match?Computing approximations of exponentialExtracting the Argument of a Complex ExponentialError for some specific values in Matrix exponentialEvaluate complex exponential expressionRe[“Complex Exponential”] does not return Cosine
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Logarithm of exponential
How to take the derivative w.r.t. an arbitrary function?How can I separate two parts of power of Exponential functions?Matrix exponential MatrixExp[] vs Sum[MatrixPower[]] doesn't match?Computing approximations of exponentialExtracting the Argument of a Complex ExponentialError for some specific values in Matrix exponentialEvaluate complex exponential expressionRe[“Complex Exponential”] does not return Cosine
$begingroup$
Ok, this is awkward.
E^Log[a]
gives a, which is what I expected, BUT,
Log[E^a]
is not evaluated at all (I expected the result to be a—this is actually an MWE, what I was trying to do was more complex).
Since I still struggle to understand some of Mathematica's concepts (like when expressions are/n't evaluated), I thought (wrongly, it seems) that it was an Evaluate or Assumptions issue, but now I think it wasn't. I tried:
Assuming[a ∈ Reals, Log[E^a]] // Evaluate
Assuming[a > 0, Log[Exp[a]]] // Simplify
and other variants—which in turn shows that I'm giving a shot in the dark.
What is the explanation for this behavior? How do I get to a? Am I missing an assumption? I don't really care about any weird special cases, a is real, not complex, nor a function, nor anything else.
It obviously seems that I'm missing something (possibly very simple and obvious), but I can't figure out what.
functions version-11
edited Jun 4 at 18:49
Carl Woll
83.3k3105216
asked Jun 4 at 2:42
RafaelRafael
257311
$endgroup$
add a comment |
$begingroup$
Ok, this is awkward.
E^Log[a]
gives a, which is what I expected, BUT,
Log[E^a]
is not evaluated at all (I expected the result to be a—this is actually an MWE, what I was trying to do was more complex).
Since I still struggle to understand some of Mathematica's concepts (like when expressions are/n't evaluated), I thought (wrongly, it seems) that it was an Evaluate or Assumptions issue, but now I think it wasn't. I tried:
Assuming[a ∈ Reals, Log[E^a]] // Evaluate
Assuming[a > 0, Log[Exp[a]]] // Simplify
and other variants—which in turn shows that I'm giving a shot in the dark.
What is the explanation for this behavior? How do I get to a? Am I missing an assumption? I don't really care about any weird special cases, a is real, not complex, nor a function, nor anything else.
It obviously seems that I'm missing something (possibly very simple and obvious), but I can't figure out what.
functions version-11
edited Jun 4 at 18:49
Carl Woll
83.3k3105216
asked Jun 4 at 2:42
RafaelRafael
257311
$endgroup$
add a comment |
$begingroup$
Ok, this is awkward.
E^Log[a]
gives a, which is what I expected, BUT,
Log[E^a]
is not evaluated at all (I expected the result to be a—this is actually an MWE, what I was trying to do was more complex).
Since I still struggle to understand some of Mathematica's concepts (like when expressions are/n't evaluated), I thought (wrongly, it seems) that it was an Evaluate or Assumptions issue, but now I think it wasn't. I tried:
Assuming[a ∈ Reals, Log[E^a]] // Evaluate
Assuming[a > 0, Log[Exp[a]]] // Simplify
and other variants—which in turn shows that I'm giving a shot in the dark.
What is the explanation for this behavior? How do I get to a? Am I missing an assumption? I don't really care about any weird special cases, a is real, not complex, nor a function, nor anything else.
It obviously seems that I'm missing something (possibly very simple and obvious), but I can't figure out what.
functions version-11
edited Jun 4 at 18:49
Carl Woll
83.3k3105216
asked Jun 4 at 2:42
RafaelRafael
257311
$endgroup$
Ok, this is awkward.
E^Log[a]
gives a, which is what I expected, BUT,
Log[E^a]
is not evaluated at all (I expected the result to be a—this is actually an MWE, what I was trying to do was more complex).
Since I still struggle to understand some of Mathematica's concepts (like when expressions are/n't evaluated), I thought (wrongly, it seems) that it was an Evaluate or Assumptions issue, but now I think it wasn't. I tried:
Assuming[a ∈ Reals, Log[E^a]] // Evaluate
Assuming[a > 0, Log[Exp[a]]] // Simplify
and other variants—which in turn shows that I'm giving a shot in the dark.
What is the explanation for this behavior? How do I get to a? Am I missing an assumption? I don't really care about any weird special cases, a is real, not complex, nor a function, nor anything else.
It obviously seems that I'm missing something (possibly very simple and obvious), but I can't figure out what.
functions version-11
functions version-11
edited Jun 4 at 18:49
Carl Woll
83.3k3105216
asked Jun 4 at 2:42
RafaelRafael
257311
edited Jun 4 at 18:49
Carl Woll
83.3k3105216
asked Jun 4 at 2:42
RafaelRafael
257311
edited Jun 4 at 18:49
Carl Woll
83.3k3105216
edited Jun 4 at 18:49
Carl Woll
83.3k3105216
edited Jun 4 at 18:49
Carl Woll
83.3k3105216
83.3k3105216
asked Jun 4 at 2:42
RafaelRafael
257311
asked Jun 4 at 2:42
RafaelRafael
257311
asked Jun 4 at 2:42
RafaelRafael
257311
257311
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Since Log is a multi-valued inverse function of Exp, Mathematica doesn't evaluate Log[Exp[a]] (or equivalently, Log[E^a]). If you want to simplify these, you need to provide an assumption on the domain of a, e.g.,
Simplify[Log[E^a], a ∈ Reals]
a
or
Simplify[Log[E^a], a > 0]
a
as suggested in the other answer. Another method is to use PowerExpand:
PowerExpand[Log[E^a], Assumptions -> True]
a + 2 I π Floor[1/2 - Im[a]/(2 π)]
Addendum
As an aside, Log behave exactly like ArcSin here:
ArcSin[Sin[x]]
ArcSin[Sin[x]]
Including a domain restriction:
Simplify[ArcSin[Sin[x]], -Pi/2 < x < Pi/2]
x
For larger domains, Simplify doesn't work:
Simplify[ArcSin[Sin[x]], 0 < x < 2 Pi]
ArcSin[Sin[x]]
Again, using PowerExpand is useful:
p = PowerExpand[ArcSin[Sin[x]], Assumptions -> 0 < x < 2Pi];
p //TeXForm
$begincases
pi -x & fracpi 2<xleq frac3 pi 2 \
x & xleq fracpi 2 \
x-2 pi & textTrue
endcases$
Visualization:
GraphicsColumn[
Plot[ArcSin[Sin[x]], x, 0, 2 Pi],
Plot[p, x, 0, 2 Pi]
]
answered Jun 4 at 3:08
Carl WollCarl Woll
83.3k3105216
$endgroup$
$begingroup$
Nice explanation! Is there/Shouldn't there be a standard, single valued version ofLog, just as the classical, ubiquitousasin,acos,atan? (I know this question may be beyond Mathematica). Is there a mathematical reason for the different treatment? Is it just for historic/convenience/arbitrary reasons?
$endgroup$
– Rafael
Jun 4 at 16:10
4
$begingroup$
Rafael Log is a standard single valued function when given numerical input. When given symbolic input, Log is not able to figure out the right branch to use without assumptions. This is exactly how ArcSin etc, behave, e.g., ArcSin[Sin[x]] returns it's input.
$endgroup$
– Carl Woll
Jun 4 at 16:17
$begingroup$
I see, silly me. Thanks
$endgroup$
– Rafael
Jun 4 at 16:20
add a comment |
$begingroup$
The assumption a > 0 is needed when Simplify is called:
Assuming[a > 0, Log[Exp[a]] // Simplify]
a
answered Jun 4 at 2:54
CoolwaterCoolwater
15.9k32554
$endgroup$
8
$begingroup$
a [Element] Realsworks, too. The problem with complexais thatLog[Exp[a + 2 Pi I]]andLog[Exp[a]]are equal.
$endgroup$
– Michael E2
Jun 4 at 2:59
2
$begingroup$
Thanks. Is there a good reason whyAssuming[a > 0, Log[Exp[a]]]won't just evaluate toa? I don't get it. (cc: @MichaelE2 )
$endgroup$
– Rafael
Jun 4 at 3:01
8
$begingroup$
@Rafael - The assumptions specified inAssumingonly affect functions that take the optionAssumptions. NeitherLognorExptake options. You need to includeSimplifyorFullSimplifywithin theAssumingfor the assumptions to be used.
$endgroup$
– Bob Hanlon
Jun 4 at 4:06
$begingroup$
@BobHanlon, Ok, I see. Thank you for your answer.
$endgroup$
– Rafael
Jun 4 at 4:40
add a comment |
$begingroup$
Logarithm is a problematic function for MMA. I do not know the reason. One can see, for example Log[a] + Log[b] // Simplify returns Log[a] + Log[b] though LeafCount[Log[a] + Log[b]] yields 5, while LeafCount[Log[a*b]] gives 4.
I do not think that in the case Log[E^a] we should use such conditions as a>0 or whatever, since the relation in question is universal. On the other hand, in the case Log[E^a]===a the expression a is doubtless much simpler and, therefore, preferable. It is not that clear what one prefers to have in the case Log[a] + Log[b]
The way around can be using rules. For example, in this case one may use
rule = Log[x_^y_] :> y*Log[x];
Then its application gives
In[5]:= Log[E^a] /. rule
(* Out[5]= a *)
Have fun!
answered Jun 4 at 11:49
Alexei BoulbitchAlexei Boulbitch
22.6k2775
$endgroup$
$begingroup$
Would you say you disagree with the other answers, then? I'd love to read more debate on the issue, if it is somehow contentious
$endgroup$
– Rafael
Jun 4 at 12:28
2
$begingroup$
@Rafael No, I do not. However, they used some conditions imposed onato achieve the desired results. We know, however, that the relationLog[Exp[a]]===aholds for anya. So, using the relation, say,a>0is to extent misleading. Indeed, in this case, we know that it holds for anya. However, in a more cumbersome case this condition may lead you into an erroneous assumption that such an inequality is necessary. Anyway, one should be careful.
$endgroup$
– Alexei Boulbitch
Jun 4 at 12:52
$begingroup$
@AlexeiBoulbitch the expression isn't general for complex values ofa.a > 0forcesainto the reals. Also, considera = -1 + I,b=-1; the sum of the logs is not the log of the product.
$endgroup$
– Davidmh
Jun 4 at 14:00
1
$begingroup$
@AlexeiBoulbitch sorry, I was unclear. I should have split my comment:Log[Exp[a]]===aisn't general for complex values ofa(in particular, whenAbs[Im[a]] > 2 Pi), buta > 0forcesainto the reals. Furthermore,Log[a] + Log[b]is not, in general,Log[a*b]ifaorbare complex.
$endgroup$
– Davidmh
Jun 4 at 14:39
4
$begingroup$
In[33]:= Log[Exp[a]] /. a->-1-5*I Out[33]= (-1 - 5 I) + (2 I) PiSo this claimed identity is simply wrong. As has been noted in the comment by @Davidmh
$endgroup$
– Daniel Lichtblau
Jun 4 at 16:17
|
show 1 more comment
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Since Log is a multi-valued inverse function of Exp, Mathematica doesn't evaluate Log[Exp[a]] (or equivalently, Log[E^a]). If you want to simplify these, you need to provide an assumption on the domain of a, e.g.,
Simplify[Log[E^a], a ∈ Reals]
a
or
Simplify[Log[E^a], a > 0]
a
as suggested in the other answer. Another method is to use PowerExpand:
PowerExpand[Log[E^a], Assumptions -> True]
a + 2 I π Floor[1/2 - Im[a]/(2 π)]
Addendum
As an aside, Log behave exactly like ArcSin here:
ArcSin[Sin[x]]
ArcSin[Sin[x]]
Including a domain restriction:
Simplify[ArcSin[Sin[x]], -Pi/2 < x < Pi/2]
x
For larger domains, Simplify doesn't work:
Simplify[ArcSin[Sin[x]], 0 < x < 2 Pi]
ArcSin[Sin[x]]
Again, using PowerExpand is useful:
p = PowerExpand[ArcSin[Sin[x]], Assumptions -> 0 < x < 2Pi];
p //TeXForm
$begincases
pi -x & fracpi 2<xleq frac3 pi 2 \
x & xleq fracpi 2 \
x-2 pi & textTrue
endcases$
Visualization:
GraphicsColumn[
Plot[ArcSin[Sin[x]], x, 0, 2 Pi],
Plot[p, x, 0, 2 Pi]
]
answered Jun 4 at 3:08
Carl WollCarl Woll
83.3k3105216
$endgroup$
$begingroup$
Nice explanation! Is there/Shouldn't there be a standard, single valued version ofLog, just as the classical, ubiquitousasin,acos,atan? (I know this question may be beyond Mathematica). Is there a mathematical reason for the different treatment? Is it just for historic/convenience/arbitrary reasons?
$endgroup$
– Rafael
Jun 4 at 16:10
4
$begingroup$
Rafael Log is a standard single valued function when given numerical input. When given symbolic input, Log is not able to figure out the right branch to use without assumptions. This is exactly how ArcSin etc, behave, e.g., ArcSin[Sin[x]] returns it's input.
$endgroup$
– Carl Woll
Jun 4 at 16:17
$begingroup$
I see, silly me. Thanks
$endgroup$
– Rafael
Jun 4 at 16:20
add a comment |
$begingroup$
Since Log is a multi-valued inverse function of Exp, Mathematica doesn't evaluate Log[Exp[a]] (or equivalently, Log[E^a]). If you want to simplify these, you need to provide an assumption on the domain of a, e.g.,
Simplify[Log[E^a], a ∈ Reals]
a
or
Simplify[Log[E^a], a > 0]
a
as suggested in the other answer. Another method is to use PowerExpand:
PowerExpand[Log[E^a], Assumptions -> True]
a + 2 I π Floor[1/2 - Im[a]/(2 π)]
Addendum
As an aside, Log behave exactly like ArcSin here:
ArcSin[Sin[x]]
ArcSin[Sin[x]]
Including a domain restriction:
Simplify[ArcSin[Sin[x]], -Pi/2 < x < Pi/2]
x
For larger domains, Simplify doesn't work:
Simplify[ArcSin[Sin[x]], 0 < x < 2 Pi]
ArcSin[Sin[x]]
Again, using PowerExpand is useful:
p = PowerExpand[ArcSin[Sin[x]], Assumptions -> 0 < x < 2Pi];
p //TeXForm
$begincases
pi -x & fracpi 2<xleq frac3 pi 2 \
x & xleq fracpi 2 \
x-2 pi & textTrue
endcases$
Visualization:
GraphicsColumn[
Plot[ArcSin[Sin[x]], x, 0, 2 Pi],
Plot[p, x, 0, 2 Pi]
]
answered Jun 4 at 3:08
Carl WollCarl Woll
83.3k3105216
$endgroup$
$begingroup$
Nice explanation! Is there/Shouldn't there be a standard, single valued version ofLog, just as the classical, ubiquitousasin,acos,atan? (I know this question may be beyond Mathematica). Is there a mathematical reason for the different treatment? Is it just for historic/convenience/arbitrary reasons?
$endgroup$
– Rafael
Jun 4 at 16:10
4
$begingroup$
Rafael Log is a standard single valued function when given numerical input. When given symbolic input, Log is not able to figure out the right branch to use without assumptions. This is exactly how ArcSin etc, behave, e.g., ArcSin[Sin[x]] returns it's input.
$endgroup$
– Carl Woll
Jun 4 at 16:17
$begingroup$
I see, silly me. Thanks
$endgroup$
– Rafael
Jun 4 at 16:20
add a comment |
$begingroup$
Since Log is a multi-valued inverse function of Exp, Mathematica doesn't evaluate Log[Exp[a]] (or equivalently, Log[E^a]). If you want to simplify these, you need to provide an assumption on the domain of a, e.g.,
Simplify[Log[E^a], a ∈ Reals]
a
or
Simplify[Log[E^a], a > 0]
a
as suggested in the other answer. Another method is to use PowerExpand:
PowerExpand[Log[E^a], Assumptions -> True]
a + 2 I π Floor[1/2 - Im[a]/(2 π)]
Addendum
As an aside, Log behave exactly like ArcSin here:
ArcSin[Sin[x]]
ArcSin[Sin[x]]
Including a domain restriction:
Simplify[ArcSin[Sin[x]], -Pi/2 < x < Pi/2]
x
For larger domains, Simplify doesn't work:
Simplify[ArcSin[Sin[x]], 0 < x < 2 Pi]
ArcSin[Sin[x]]
Again, using PowerExpand is useful:
p = PowerExpand[ArcSin[Sin[x]], Assumptions -> 0 < x < 2Pi];
p //TeXForm
$begincases
pi -x & fracpi 2<xleq frac3 pi 2 \
x & xleq fracpi 2 \
x-2 pi & textTrue
endcases$
Visualization:
GraphicsColumn[
Plot[ArcSin[Sin[x]], x, 0, 2 Pi],
Plot[p, x, 0, 2 Pi]
]
answered Jun 4 at 3:08
Carl WollCarl Woll
83.3k3105216
$endgroup$
Since Log is a multi-valued inverse function of Exp, Mathematica doesn't evaluate Log[Exp[a]] (or equivalently, Log[E^a]). If you want to simplify these, you need to provide an assumption on the domain of a, e.g.,
Simplify[Log[E^a], a ∈ Reals]
a
or
Simplify[Log[E^a], a > 0]
a
as suggested in the other answer. Another method is to use PowerExpand:
PowerExpand[Log[E^a], Assumptions -> True]
a + 2 I π Floor[1/2 - Im[a]/(2 π)]
Addendum
As an aside, Log behave exactly like ArcSin here:
ArcSin[Sin[x]]
ArcSin[Sin[x]]
Including a domain restriction:
Simplify[ArcSin[Sin[x]], -Pi/2 < x < Pi/2]
x
For larger domains, Simplify doesn't work:
Simplify[ArcSin[Sin[x]], 0 < x < 2 Pi]
ArcSin[Sin[x]]
Again, using PowerExpand is useful:
p = PowerExpand[ArcSin[Sin[x]], Assumptions -> 0 < x < 2Pi];
p //TeXForm
$begincases
pi -x & fracpi 2<xleq frac3 pi 2 \
x & xleq fracpi 2 \
x-2 pi & textTrue
endcases$
Visualization:
GraphicsColumn[
Plot[ArcSin[Sin[x]], x, 0, 2 Pi],
Plot[p, x, 0, 2 Pi]
]
answered Jun 4 at 3:08
Carl WollCarl Woll
83.3k3105216
edited Jun 4 at 18:48
answered Jun 4 at 3:08
Carl WollCarl Woll
83.3k3105216
answered Jun 4 at 3:08
Carl WollCarl Woll
83.3k3105216
answered Jun 4 at 3:08
Carl WollCarl Woll
83.3k3105216
83.3k3105216
$begingroup$
Nice explanation! Is there/Shouldn't there be a standard, single valued version ofLog, just as the classical, ubiquitousasin,acos,atan? (I know this question may be beyond Mathematica). Is there a mathematical reason for the different treatment? Is it just for historic/convenience/arbitrary reasons?
$endgroup$
– Rafael
Jun 4 at 16:10
4
$begingroup$
Rafael Log is a standard single valued function when given numerical input. When given symbolic input, Log is not able to figure out the right branch to use without assumptions. This is exactly how ArcSin etc, behave, e.g., ArcSin[Sin[x]] returns it's input.
$endgroup$
– Carl Woll
Jun 4 at 16:17
$begingroup$
I see, silly me. Thanks
$endgroup$
– Rafael
Jun 4 at 16:20
add a comment |
$begingroup$
Nice explanation! Is there/Shouldn't there be a standard, single valued version ofLog, just as the classical, ubiquitousasin,acos,atan? (I know this question may be beyond Mathematica). Is there a mathematical reason for the different treatment? Is it just for historic/convenience/arbitrary reasons?
$endgroup$
– Rafael
Jun 4 at 16:10
4
$begingroup$
Rafael Log is a standard single valued function when given numerical input. When given symbolic input, Log is not able to figure out the right branch to use without assumptions. This is exactly how ArcSin etc, behave, e.g., ArcSin[Sin[x]] returns it's input.
$endgroup$
– Carl Woll
Jun 4 at 16:17
$begingroup$
I see, silly me. Thanks
$endgroup$
– Rafael
Jun 4 at 16:20
$begingroup$
Nice explanation! Is there/Shouldn't there be a standard, single valued version of
Log, just as the classical, ubiquitous asin, acos, atan? (I know this question may be beyond Mathematica). Is there a mathematical reason for the different treatment? Is it just for historic/convenience/arbitrary reasons?$endgroup$
– Rafael
Jun 4 at 16:10
$begingroup$
Nice explanation! Is there/Shouldn't there be a standard, single valued version of
Log, just as the classical, ubiquitous asin, acos, atan? (I know this question may be beyond Mathematica). Is there a mathematical reason for the different treatment? Is it just for historic/convenience/arbitrary reasons?$endgroup$
– Rafael
Jun 4 at 16:10
4
4
$begingroup$
Rafael Log is a standard single valued function when given numerical input. When given symbolic input, Log is not able to figure out the right branch to use without assumptions. This is exactly how ArcSin etc, behave, e.g., ArcSin[Sin[x]] returns it's input.
$endgroup$
– Carl Woll
Jun 4 at 16:17
$begingroup$
Rafael Log is a standard single valued function when given numerical input. When given symbolic input, Log is not able to figure out the right branch to use without assumptions. This is exactly how ArcSin etc, behave, e.g., ArcSin[Sin[x]] returns it's input.
$endgroup$
– Carl Woll
Jun 4 at 16:17
$begingroup$
I see, silly me. Thanks
$endgroup$
– Rafael
Jun 4 at 16:20
$begingroup$
I see, silly me. Thanks
$endgroup$
– Rafael
Jun 4 at 16:20
add a comment |
$begingroup$
The assumption a > 0 is needed when Simplify is called:
Assuming[a > 0, Log[Exp[a]] // Simplify]
a
answered Jun 4 at 2:54
CoolwaterCoolwater
15.9k32554
$endgroup$
8
$begingroup$
a [Element] Realsworks, too. The problem with complexais thatLog[Exp[a + 2 Pi I]]andLog[Exp[a]]are equal.
$endgroup$
– Michael E2
Jun 4 at 2:59
2
$begingroup$
Thanks. Is there a good reason whyAssuming[a > 0, Log[Exp[a]]]won't just evaluate toa? I don't get it. (cc: @MichaelE2 )
$endgroup$
– Rafael
Jun 4 at 3:01
8
$begingroup$
@Rafael - The assumptions specified inAssumingonly affect functions that take the optionAssumptions. NeitherLognorExptake options. You need to includeSimplifyorFullSimplifywithin theAssumingfor the assumptions to be used.
$endgroup$
– Bob Hanlon
Jun 4 at 4:06
$begingroup$
@BobHanlon, Ok, I see. Thank you for your answer.
$endgroup$
– Rafael
Jun 4 at 4:40
add a comment |
$begingroup$
The assumption a > 0 is needed when Simplify is called:
Assuming[a > 0, Log[Exp[a]] // Simplify]
a
answered Jun 4 at 2:54
CoolwaterCoolwater
15.9k32554
$endgroup$
8
$begingroup$
a [Element] Realsworks, too. The problem with complexais thatLog[Exp[a + 2 Pi I]]andLog[Exp[a]]are equal.
$endgroup$
– Michael E2
Jun 4 at 2:59
2
$begingroup$
Thanks. Is there a good reason whyAssuming[a > 0, Log[Exp[a]]]won't just evaluate toa? I don't get it. (cc: @MichaelE2 )
$endgroup$
– Rafael
Jun 4 at 3:01
8
$begingroup$
@Rafael - The assumptions specified inAssumingonly affect functions that take the optionAssumptions. NeitherLognorExptake options. You need to includeSimplifyorFullSimplifywithin theAssumingfor the assumptions to be used.
$endgroup$
– Bob Hanlon
Jun 4 at 4:06
$begingroup$
@BobHanlon, Ok, I see. Thank you for your answer.
$endgroup$
– Rafael
Jun 4 at 4:40
add a comment |
$begingroup$
The assumption a > 0 is needed when Simplify is called:
Assuming[a > 0, Log[Exp[a]] // Simplify]
a
answered Jun 4 at 2:54
CoolwaterCoolwater
15.9k32554
$endgroup$
The assumption a > 0 is needed when Simplify is called:
Assuming[a > 0, Log[Exp[a]] // Simplify]
a
answered Jun 4 at 2:54
CoolwaterCoolwater
15.9k32554
answered Jun 4 at 2:54
CoolwaterCoolwater
15.9k32554
answered Jun 4 at 2:54
CoolwaterCoolwater
15.9k32554
answered Jun 4 at 2:54
CoolwaterCoolwater
15.9k32554
15.9k32554
8
$begingroup$
a [Element] Realsworks, too. The problem with complexais thatLog[Exp[a + 2 Pi I]]andLog[Exp[a]]are equal.
$endgroup$
– Michael E2
Jun 4 at 2:59
2
$begingroup$
Thanks. Is there a good reason whyAssuming[a > 0, Log[Exp[a]]]won't just evaluate toa? I don't get it. (cc: @MichaelE2 )
$endgroup$
– Rafael
Jun 4 at 3:01
8
$begingroup$
@Rafael - The assumptions specified inAssumingonly affect functions that take the optionAssumptions. NeitherLognorExptake options. You need to includeSimplifyorFullSimplifywithin theAssumingfor the assumptions to be used.
$endgroup$
– Bob Hanlon
Jun 4 at 4:06
$begingroup$
@BobHanlon, Ok, I see. Thank you for your answer.
$endgroup$
– Rafael
Jun 4 at 4:40
add a comment |
8
$begingroup$
a [Element] Realsworks, too. The problem with complexais thatLog[Exp[a + 2 Pi I]]andLog[Exp[a]]are equal.
$endgroup$
– Michael E2
Jun 4 at 2:59
2
$begingroup$
Thanks. Is there a good reason whyAssuming[a > 0, Log[Exp[a]]]won't just evaluate toa? I don't get it. (cc: @MichaelE2 )
$endgroup$
– Rafael
Jun 4 at 3:01
8
$begingroup$
@Rafael - The assumptions specified inAssumingonly affect functions that take the optionAssumptions. NeitherLognorExptake options. You need to includeSimplifyorFullSimplifywithin theAssumingfor the assumptions to be used.
$endgroup$
– Bob Hanlon
Jun 4 at 4:06
$begingroup$
@BobHanlon, Ok, I see. Thank you for your answer.
$endgroup$
– Rafael
Jun 4 at 4:40
8
8
$begingroup$
a [Element] Reals works, too. The problem with complex a is that Log[Exp[a + 2 Pi I]] and Log[Exp[a]] are equal.$endgroup$
– Michael E2
Jun 4 at 2:59
$begingroup$
a [Element] Reals works, too. The problem with complex a is that Log[Exp[a + 2 Pi I]] and Log[Exp[a]] are equal.$endgroup$
– Michael E2
Jun 4 at 2:59
2
2
$begingroup$
Thanks. Is there a good reason why
Assuming[a > 0, Log[Exp[a]]] won't just evaluate to a? I don't get it. (cc: @MichaelE2 )$endgroup$
– Rafael
Jun 4 at 3:01
$begingroup$
Thanks. Is there a good reason why
Assuming[a > 0, Log[Exp[a]]] won't just evaluate to a? I don't get it. (cc: @MichaelE2 )$endgroup$
– Rafael
Jun 4 at 3:01
8
8
$begingroup$
@Rafael - The assumptions specified in
Assuming only affect functions that take the option Assumptions. Neither Log nor Exp take options. You need to include Simplify or FullSimplify within the Assuming for the assumptions to be used.$endgroup$
– Bob Hanlon
Jun 4 at 4:06
$begingroup$
@Rafael - The assumptions specified in
Assuming only affect functions that take the option Assumptions. Neither Log nor Exp take options. You need to include Simplify or FullSimplify within the Assuming for the assumptions to be used.$endgroup$
– Bob Hanlon
Jun 4 at 4:06
$begingroup$
@BobHanlon, Ok, I see. Thank you for your answer.
$endgroup$
– Rafael
Jun 4 at 4:40
$begingroup$
@BobHanlon, Ok, I see. Thank you for your answer.
$endgroup$
– Rafael
Jun 4 at 4:40
add a comment |
$begingroup$
Logarithm is a problematic function for MMA. I do not know the reason. One can see, for example Log[a] + Log[b] // Simplify returns Log[a] + Log[b] though LeafCount[Log[a] + Log[b]] yields 5, while LeafCount[Log[a*b]] gives 4.
I do not think that in the case Log[E^a] we should use such conditions as a>0 or whatever, since the relation in question is universal. On the other hand, in the case Log[E^a]===a the expression a is doubtless much simpler and, therefore, preferable. It is not that clear what one prefers to have in the case Log[a] + Log[b]
The way around can be using rules. For example, in this case one may use
rule = Log[x_^y_] :> y*Log[x];
Then its application gives
In[5]:= Log[E^a] /. rule
(* Out[5]= a *)
Have fun!
answered Jun 4 at 11:49
Alexei BoulbitchAlexei Boulbitch
22.6k2775
$endgroup$
$begingroup$
Would you say you disagree with the other answers, then? I'd love to read more debate on the issue, if it is somehow contentious
$endgroup$
– Rafael
Jun 4 at 12:28
2
$begingroup$
@Rafael No, I do not. However, they used some conditions imposed onato achieve the desired results. We know, however, that the relationLog[Exp[a]]===aholds for anya. So, using the relation, say,a>0is to extent misleading. Indeed, in this case, we know that it holds for anya. However, in a more cumbersome case this condition may lead you into an erroneous assumption that such an inequality is necessary. Anyway, one should be careful.
$endgroup$
– Alexei Boulbitch
Jun 4 at 12:52
$begingroup$
@AlexeiBoulbitch the expression isn't general for complex values ofa.a > 0forcesainto the reals. Also, considera = -1 + I,b=-1; the sum of the logs is not the log of the product.
$endgroup$
– Davidmh
Jun 4 at 14:00
1
$begingroup$
@AlexeiBoulbitch sorry, I was unclear. I should have split my comment:Log[Exp[a]]===aisn't general for complex values ofa(in particular, whenAbs[Im[a]] > 2 Pi), buta > 0forcesainto the reals. Furthermore,Log[a] + Log[b]is not, in general,Log[a*b]ifaorbare complex.
$endgroup$
– Davidmh
Jun 4 at 14:39
4
$begingroup$
In[33]:= Log[Exp[a]] /. a->-1-5*I Out[33]= (-1 - 5 I) + (2 I) PiSo this claimed identity is simply wrong. As has been noted in the comment by @Davidmh
$endgroup$
– Daniel Lichtblau
Jun 4 at 16:17
|
show 1 more comment
$begingroup$
Logarithm is a problematic function for MMA. I do not know the reason. One can see, for example Log[a] + Log[b] // Simplify returns Log[a] + Log[b] though LeafCount[Log[a] + Log[b]] yields 5, while LeafCount[Log[a*b]] gives 4.
I do not think that in the case Log[E^a] we should use such conditions as a>0 or whatever, since the relation in question is universal. On the other hand, in the case Log[E^a]===a the expression a is doubtless much simpler and, therefore, preferable. It is not that clear what one prefers to have in the case Log[a] + Log[b]
The way around can be using rules. For example, in this case one may use
rule = Log[x_^y_] :> y*Log[x];
Then its application gives
In[5]:= Log[E^a] /. rule
(* Out[5]= a *)
Have fun!
answered Jun 4 at 11:49
Alexei BoulbitchAlexei Boulbitch
22.6k2775
$endgroup$
$begingroup$
Would you say you disagree with the other answers, then? I'd love to read more debate on the issue, if it is somehow contentious
$endgroup$
– Rafael
Jun 4 at 12:28
2
$begingroup$
@Rafael No, I do not. However, they used some conditions imposed onato achieve the desired results. We know, however, that the relationLog[Exp[a]]===aholds for anya. So, using the relation, say,a>0is to extent misleading. Indeed, in this case, we know that it holds for anya. However, in a more cumbersome case this condition may lead you into an erroneous assumption that such an inequality is necessary. Anyway, one should be careful.
$endgroup$
– Alexei Boulbitch
Jun 4 at 12:52
$begingroup$
@AlexeiBoulbitch the expression isn't general for complex values ofa.a > 0forcesainto the reals. Also, considera = -1 + I,b=-1; the sum of the logs is not the log of the product.
$endgroup$
– Davidmh
Jun 4 at 14:00
1
$begingroup$
@AlexeiBoulbitch sorry, I was unclear. I should have split my comment:Log[Exp[a]]===aisn't general for complex values ofa(in particular, whenAbs[Im[a]] > 2 Pi), buta > 0forcesainto the reals. Furthermore,Log[a] + Log[b]is not, in general,Log[a*b]ifaorbare complex.
$endgroup$
– Davidmh
Jun 4 at 14:39
4
$begingroup$
In[33]:= Log[Exp[a]] /. a->-1-5*I Out[33]= (-1 - 5 I) + (2 I) PiSo this claimed identity is simply wrong. As has been noted in the comment by @Davidmh
$endgroup$
– Daniel Lichtblau
Jun 4 at 16:17
|
show 1 more comment
$begingroup$
Logarithm is a problematic function for MMA. I do not know the reason. One can see, for example Log[a] + Log[b] // Simplify returns Log[a] + Log[b] though LeafCount[Log[a] + Log[b]] yields 5, while LeafCount[Log[a*b]] gives 4.
I do not think that in the case Log[E^a] we should use such conditions as a>0 or whatever, since the relation in question is universal. On the other hand, in the case Log[E^a]===a the expression a is doubtless much simpler and, therefore, preferable. It is not that clear what one prefers to have in the case Log[a] + Log[b]
The way around can be using rules. For example, in this case one may use
rule = Log[x_^y_] :> y*Log[x];
Then its application gives
In[5]:= Log[E^a] /. rule
(* Out[5]= a *)
Have fun!
answered Jun 4 at 11:49
Alexei BoulbitchAlexei Boulbitch
22.6k2775
$endgroup$
Logarithm is a problematic function for MMA. I do not know the reason. One can see, for example Log[a] + Log[b] // Simplify returns Log[a] + Log[b] though LeafCount[Log[a] + Log[b]] yields 5, while LeafCount[Log[a*b]] gives 4.
I do not think that in the case Log[E^a] we should use such conditions as a>0 or whatever, since the relation in question is universal. On the other hand, in the case Log[E^a]===a the expression a is doubtless much simpler and, therefore, preferable. It is not that clear what one prefers to have in the case Log[a] + Log[b]
The way around can be using rules. For example, in this case one may use
rule = Log[x_^y_] :> y*Log[x];
Then its application gives
In[5]:= Log[E^a] /. rule
(* Out[5]= a *)
Have fun!
answered Jun 4 at 11:49
Alexei BoulbitchAlexei Boulbitch
22.6k2775
answered Jun 4 at 11:49
Alexei BoulbitchAlexei Boulbitch
22.6k2775
answered Jun 4 at 11:49
Alexei BoulbitchAlexei Boulbitch
22.6k2775
answered Jun 4 at 11:49
Alexei BoulbitchAlexei Boulbitch
22.6k2775
22.6k2775
$begingroup$
Would you say you disagree with the other answers, then? I'd love to read more debate on the issue, if it is somehow contentious
$endgroup$
– Rafael
Jun 4 at 12:28
2
$begingroup$
@Rafael No, I do not. However, they used some conditions imposed onato achieve the desired results. We know, however, that the relationLog[Exp[a]]===aholds for anya. So, using the relation, say,a>0is to extent misleading. Indeed, in this case, we know that it holds for anya. However, in a more cumbersome case this condition may lead you into an erroneous assumption that such an inequality is necessary. Anyway, one should be careful.
$endgroup$
– Alexei Boulbitch
Jun 4 at 12:52
$begingroup$
@AlexeiBoulbitch the expression isn't general for complex values ofa.a > 0forcesainto the reals. Also, considera = -1 + I,b=-1; the sum of the logs is not the log of the product.
$endgroup$
– Davidmh
Jun 4 at 14:00
1
$begingroup$
@AlexeiBoulbitch sorry, I was unclear. I should have split my comment:Log[Exp[a]]===aisn't general for complex values ofa(in particular, whenAbs[Im[a]] > 2 Pi), buta > 0forcesainto the reals. Furthermore,Log[a] + Log[b]is not, in general,Log[a*b]ifaorbare complex.
$endgroup$
– Davidmh
Jun 4 at 14:39
4
$begingroup$
In[33]:= Log[Exp[a]] /. a->-1-5*I Out[33]= (-1 - 5 I) + (2 I) PiSo this claimed identity is simply wrong. As has been noted in the comment by @Davidmh
$endgroup$
– Daniel Lichtblau
Jun 4 at 16:17
|
show 1 more comment
$begingroup$
Would you say you disagree with the other answers, then? I'd love to read more debate on the issue, if it is somehow contentious
$endgroup$
– Rafael
Jun 4 at 12:28
2
$begingroup$
@Rafael No, I do not. However, they used some conditions imposed onato achieve the desired results. We know, however, that the relationLog[Exp[a]]===aholds for anya. So, using the relation, say,a>0is to extent misleading. Indeed, in this case, we know that it holds for anya. However, in a more cumbersome case this condition may lead you into an erroneous assumption that such an inequality is necessary. Anyway, one should be careful.
$endgroup$
– Alexei Boulbitch
Jun 4 at 12:52
$begingroup$
@AlexeiBoulbitch the expression isn't general for complex values ofa.a > 0forcesainto the reals. Also, considera = -1 + I,b=-1; the sum of the logs is not the log of the product.
$endgroup$
– Davidmh
Jun 4 at 14:00
1
$begingroup$
@AlexeiBoulbitch sorry, I was unclear. I should have split my comment:Log[Exp[a]]===aisn't general for complex values ofa(in particular, whenAbs[Im[a]] > 2 Pi), buta > 0forcesainto the reals. Furthermore,Log[a] + Log[b]is not, in general,Log[a*b]ifaorbare complex.
$endgroup$
– Davidmh
Jun 4 at 14:39
4
$begingroup$
In[33]:= Log[Exp[a]] /. a->-1-5*I Out[33]= (-1 - 5 I) + (2 I) PiSo this claimed identity is simply wrong. As has been noted in the comment by @Davidmh
$endgroup$
– Daniel Lichtblau
Jun 4 at 16:17
$begingroup$
Would you say you disagree with the other answers, then? I'd love to read more debate on the issue, if it is somehow contentious
$endgroup$
– Rafael
Jun 4 at 12:28
$begingroup$
Would you say you disagree with the other answers, then? I'd love to read more debate on the issue, if it is somehow contentious
$endgroup$
– Rafael
Jun 4 at 12:28
2
2
$begingroup$
@Rafael No, I do not. However, they used some conditions imposed on
a to achieve the desired results. We know, however, that the relation Log[Exp[a]]===a holds for any a. So, using the relation, say, a>0 is to extent misleading. Indeed, in this case, we know that it holds for any a. However, in a more cumbersome case this condition may lead you into an erroneous assumption that such an inequality is necessary. Anyway, one should be careful.$endgroup$
– Alexei Boulbitch
Jun 4 at 12:52
$begingroup$
@Rafael No, I do not. However, they used some conditions imposed on
a to achieve the desired results. We know, however, that the relation Log[Exp[a]]===a holds for any a. So, using the relation, say, a>0 is to extent misleading. Indeed, in this case, we know that it holds for any a. However, in a more cumbersome case this condition may lead you into an erroneous assumption that such an inequality is necessary. Anyway, one should be careful.$endgroup$
– Alexei Boulbitch
Jun 4 at 12:52
$begingroup$
@AlexeiBoulbitch the expression isn't general for complex values of
a. a > 0 forces a into the reals. Also, consider a = -1 + I, b=-1; the sum of the logs is not the log of the product.$endgroup$
– Davidmh
Jun 4 at 14:00
$begingroup$
@AlexeiBoulbitch the expression isn't general for complex values of
a. a > 0 forces a into the reals. Also, consider a = -1 + I, b=-1; the sum of the logs is not the log of the product.$endgroup$
– Davidmh
Jun 4 at 14:00
1
1
$begingroup$
@AlexeiBoulbitch sorry, I was unclear. I should have split my comment:
Log[Exp[a]]===a isn't general for complex values of a (in particular, when Abs[Im[a]] > 2 Pi), but a > 0 forces a into the reals. Furthermore, Log[a] + Log[b] is not, in general, Log[a*b] if a or b are complex.$endgroup$
– Davidmh
Jun 4 at 14:39
$begingroup$
@AlexeiBoulbitch sorry, I was unclear. I should have split my comment:
Log[Exp[a]]===a isn't general for complex values of a (in particular, when Abs[Im[a]] > 2 Pi), but a > 0 forces a into the reals. Furthermore, Log[a] + Log[b] is not, in general, Log[a*b] if a or b are complex.$endgroup$
– Davidmh
Jun 4 at 14:39
4
4
$begingroup$
In[33]:= Log[Exp[a]] /. a->-1-5*I Out[33]= (-1 - 5 I) + (2 I) Pi So this claimed identity is simply wrong. As has been noted in the comment by @Davidmh$endgroup$
– Daniel Lichtblau
Jun 4 at 16:17
$begingroup$
In[33]:= Log[Exp[a]] /. a->-1-5*I Out[33]= (-1 - 5 I) + (2 I) Pi So this claimed identity is simply wrong. As has been noted in the comment by @Davidmh$endgroup$
– Daniel Lichtblau
Jun 4 at 16:17
|
show 1 more comment
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