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flightwingsd2 2022-06-24 Answered
Solve the following system:
{ a x 0 2 = exp ( x 0 2 4 σ 2 ) + a r 2 exp ( x 0 2 4 σ 2 ) + 4 a σ 2 = 0
where the varaibles are x 0 , a. Take x 0 > 0
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Answers (1)

nuvolor8
Answered 2022-06-25 Author has 32 answers
The solution for | r | 2 σ is
{ x 0 = r 2 4 σ 2 > 0 a = 1 4 σ 2 exp ( r 2 4 σ 2 4 σ 2 ) = 1 4 σ 2 exp ( x 0 2 4 σ 2 ) .
Since exp ( x 0 2 4 σ 2 ) > 0, the second equation implies that a < 0. For convenience write
E = exp ( x 0 2 4 σ 2 ) .
The system is easily solvable, e.g. as follows:
{ a x 0 2 = E + a r 2 E + 4 a σ 2 = 0 { a x 0 2 = 4 a σ 2 + a r 2 E = 4 a σ 2 { x 0 2 = 4 σ 2 + r 2 a = 1 4 σ 2 E .
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