Where Can I Find a Proof for the Wronskian Formula?

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    Formula Wronskian
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Discussion Overview

The discussion revolves around seeking a proof for the Wronskian formula, specifically in the context of complex functions and their derivatives. Participants explore mathematical expressions and relationships related to the Wronskian.

Discussion Character

  • Mathematical reasoning

Main Points Raised

  • One participant requests a proof for the Wronskian formula, presenting the formula as W[v,v*] = v'v* - vv*' = 2iIM(v', v*).
  • Another participant seeks clarification on the term 'IM(v', v*)', indicating a need for understanding the notation used.
  • A third participant provides a detailed derivation of the Wronskian, using specific variables and showing step-by-step calculations that lead to an expression involving the imaginary part of a complex product.
  • The same participant concludes their derivation by stating that the proof is already seen, implying confidence in their calculations.

Areas of Agreement / Disagreement

The discussion does not reach a consensus, as there is a request for clarification and varying levels of confidence in the proof presented. Some participants engage with the mathematical expressions while others seek further understanding.

Contextual Notes

The discussion includes assumptions about the definitions of terms and notation, which may not be universally understood. The steps in the mathematical proof are not fully resolved, leaving some aspects open to interpretation.

zaybu
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Can anyone point to a proof for the Wronskian formula:

W[v,v*] = v'v* - vv*' = 2iIM(v', v*)

Thanks
 
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What do you mean by 'IM(v', v*)'?
 
W= s z^{*}- s^{*}z
s=a+ib
z=c+id

W= (a+ib) (c-id) - (a-ib) (c+id)
W= ac -i ad +i bc -bd - ac -i ad +ibc+bd
W= -i ad +i bc -i ad +ibc
W= 2i bc - 2i ad = 2i (bc-ad)

sz^{*} = ac + bd + i (bc-ad)
so
Im(sz^{*})= bc-ad

The proof is already seen
 
Thanks
 

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