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Weird result in complex limit
Limit of integral gives incorrect outputIntegrate returns unexpected resultLimit problem calculating directional derivativeWhy won't Limit evaluate, and what can be done about itLimit of an inverse functionDoes Mathematica implement Risch algorithm? If it does, in which cases?Limit problem no longer works in Mathematica 11.1.0Evaluating integral seems incorrectReal integral giving complex resultHow to apply NIntegrate three times
$begingroup$
I am trying to evaluate a limit:
gamma[w_] = Sqrt[-(u*e)w^2 + I*(u*s)w];
Limit[Re[gamma[x]], x -> DirectedInfinity[1]]
I calculated the limit by hand, and the correct answer is (I also checked numerically for some examples of $u,e,s$ using the software):
$qquad frac s2 sqrtfrac ue$
But for some reason, when using Limit, I get
DirectedInfinity[(Sign[e]^2 Sign[u]^2)^(1/4)]
So my questions are:
What is going here?
What issues should I be aware of when using Limit?
calculus-and-analysis complex
edited 1 hour ago
m_goldberg
89.3k873200
asked 4 hours ago
VillaVilla
1083
New contributor
Villa is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
I am trying to evaluate a limit:
gamma[w_] = Sqrt[-(u*e)w^2 + I*(u*s)w];
Limit[Re[gamma[x]], x -> DirectedInfinity[1]]
I calculated the limit by hand, and the correct answer is (I also checked numerically for some examples of $u,e,s$ using the software):
$qquad frac s2 sqrtfrac ue$
But for some reason, when using Limit, I get
DirectedInfinity[(Sign[e]^2 Sign[u]^2)^(1/4)]
So my questions are:
What is going here?
What issues should I be aware of when using Limit?
calculus-and-analysis complex
edited 1 hour ago
m_goldberg
89.3k873200
asked 4 hours ago
VillaVilla
1083
New contributor
Villa is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
I am trying to evaluate a limit:
gamma[w_] = Sqrt[-(u*e)w^2 + I*(u*s)w];
Limit[Re[gamma[x]], x -> DirectedInfinity[1]]
I calculated the limit by hand, and the correct answer is (I also checked numerically for some examples of $u,e,s$ using the software):
$qquad frac s2 sqrtfrac ue$
But for some reason, when using Limit, I get
DirectedInfinity[(Sign[e]^2 Sign[u]^2)^(1/4)]
So my questions are:
What is going here?
What issues should I be aware of when using Limit?
calculus-and-analysis complex
edited 1 hour ago
m_goldberg
89.3k873200
asked 4 hours ago
VillaVilla
1083
New contributor
Villa is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
I am trying to evaluate a limit:
gamma[w_] = Sqrt[-(u*e)w^2 + I*(u*s)w];
Limit[Re[gamma[x]], x -> DirectedInfinity[1]]
I calculated the limit by hand, and the correct answer is (I also checked numerically for some examples of $u,e,s$ using the software):
$qquad frac s2 sqrtfrac ue$
But for some reason, when using Limit, I get
DirectedInfinity[(Sign[e]^2 Sign[u]^2)^(1/4)]
So my questions are:
What is going here?
What issues should I be aware of when using Limit?
calculus-and-analysis complex
calculus-and-analysis complex
edited 1 hour ago
m_goldberg
89.3k873200
asked 4 hours ago
VillaVilla
1083
New contributor
Villa is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 1 hour ago
m_goldberg
89.3k873200
asked 4 hours ago
VillaVilla
1083
New contributor
Villa is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 1 hour ago
m_goldberg
89.3k873200
edited 1 hour ago
m_goldberg
89.3k873200
edited 1 hour ago
m_goldberg
89.3k873200
89.3k873200
asked 4 hours ago
VillaVilla
1083
New contributor
Villa is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 4 hours ago
VillaVilla
1083
asked 4 hours ago
VillaVilla
1083
1083
New contributor
Villa is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Villa is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Villa is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
I think it's worth reporting the issue to support. If you give appropriate assumptions, then you get your expected result:
Limit[Re[gamma[x]], x -> Infinity, Assumptions -> u>0 && e>0]
(s u)/(2 Sqrt[e u])
edited 31 mins ago
m_goldberg
89.3k873200
answered 4 hours ago
Carl WollCarl Woll
76.3k3100200
$endgroup$
$begingroup$
Thank you, it worked.
$endgroup$
– Villa
4 hours ago
$begingroup$
UsingAssumptions -> u >= 0 && e > 0gives the same form as the OP's hand calculation.
$endgroup$
– Bob Hanlon
3 hours ago
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
I think it's worth reporting the issue to support. If you give appropriate assumptions, then you get your expected result:
Limit[Re[gamma[x]], x -> Infinity, Assumptions -> u>0 && e>0]
(s u)/(2 Sqrt[e u])
edited 31 mins ago
m_goldberg
89.3k873200
answered 4 hours ago
Carl WollCarl Woll
76.3k3100200
$endgroup$
$begingroup$
Thank you, it worked.
$endgroup$
– Villa
4 hours ago
$begingroup$
UsingAssumptions -> u >= 0 && e > 0gives the same form as the OP's hand calculation.
$endgroup$
– Bob Hanlon
3 hours ago
add a comment |
$begingroup$
I think it's worth reporting the issue to support. If you give appropriate assumptions, then you get your expected result:
Limit[Re[gamma[x]], x -> Infinity, Assumptions -> u>0 && e>0]
(s u)/(2 Sqrt[e u])
edited 31 mins ago
m_goldberg
89.3k873200
answered 4 hours ago
Carl WollCarl Woll
76.3k3100200
$endgroup$
$begingroup$
Thank you, it worked.
$endgroup$
– Villa
4 hours ago
$begingroup$
UsingAssumptions -> u >= 0 && e > 0gives the same form as the OP's hand calculation.
$endgroup$
– Bob Hanlon
3 hours ago
add a comment |
$begingroup$
I think it's worth reporting the issue to support. If you give appropriate assumptions, then you get your expected result:
Limit[Re[gamma[x]], x -> Infinity, Assumptions -> u>0 && e>0]
(s u)/(2 Sqrt[e u])
edited 31 mins ago
m_goldberg
89.3k873200
answered 4 hours ago
Carl WollCarl Woll
76.3k3100200
$endgroup$
I think it's worth reporting the issue to support. If you give appropriate assumptions, then you get your expected result:
Limit[Re[gamma[x]], x -> Infinity, Assumptions -> u>0 && e>0]
(s u)/(2 Sqrt[e u])
edited 31 mins ago
m_goldberg
89.3k873200
answered 4 hours ago
Carl WollCarl Woll
76.3k3100200
edited 31 mins ago
m_goldberg
89.3k873200
edited 31 mins ago
m_goldberg
89.3k873200
edited 31 mins ago
m_goldberg
89.3k873200
89.3k873200
answered 4 hours ago
Carl WollCarl Woll
76.3k3100200
answered 4 hours ago
Carl WollCarl Woll
76.3k3100200
answered 4 hours ago
Carl WollCarl Woll
76.3k3100200
76.3k3100200
$begingroup$
Thank you, it worked.
$endgroup$
– Villa
4 hours ago
$begingroup$
UsingAssumptions -> u >= 0 && e > 0gives the same form as the OP's hand calculation.
$endgroup$
– Bob Hanlon
3 hours ago
add a comment |
$begingroup$
Thank you, it worked.
$endgroup$
– Villa
4 hours ago
$begingroup$
UsingAssumptions -> u >= 0 && e > 0gives the same form as the OP's hand calculation.
$endgroup$
– Bob Hanlon
3 hours ago
$begingroup$
Thank you, it worked.
$endgroup$
– Villa
4 hours ago
$begingroup$
Thank you, it worked.
$endgroup$
– Villa
4 hours ago
$begingroup$
Using
Assumptions -> u >= 0 && e > 0 gives the same form as the OP's hand calculation.$endgroup$
– Bob Hanlon
3 hours ago
$begingroup$
Using
Assumptions -> u >= 0 && e > 0 gives the same form as the OP's hand calculation.$endgroup$
– Bob Hanlon
3 hours ago
add a comment |
Villa is a new contributor. Be nice, and check out our Code of Conduct.
Villa is a new contributor. Be nice, and check out our Code of Conduct.
Villa is a new contributor. Be nice, and check out our Code of Conduct.
Villa is a new contributor. Be nice, and check out our Code of Conduct.
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