P+q = pq Hölder

  • Thread starter zxh
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  • #1
zxh
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Homework Statement



How do you get from 1/p +1/q = 1 to (p-1)(q-1) = 1? (http://de.wikipedia.org/wiki/Konjugation_%28Reelle_Zahlen%29" [Broken])

Homework Equations



Is log(xy) = log(x) + log(y) in any way related to this, i.e the rewritten subject title?

The Attempt at a Solution



Manipulate[
ContourPlot[x + y == x*y, {x, -m, m}, {y, -m, m},
PlotRange -> Automatic], {m, -100, 100}]

TIA!
 
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Answers and Replies

  • #2
disregardthat
Science Advisor
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Multiply out and compare. Logarithms have nothing to do with the equation you mentioned, but you might be using the logarithm to prove some other aspect of Hölders inequality.
 
  • #3
ehild
Homework Helper
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How do you get from 1/p +1/q = 1 to (p-1)(q-1) = 1? (http://de.wikipedia.org/wiki/Konjugation_%28Reelle_Zahlen%29" [Broken])

You can get the second equation from the first one by multiplying both sides of the first equation by pq, then subtracting 1, and factoring. Try, it is fun. :smile:

ehild
 
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  • #4
zxh
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Alright, makes me look tired, but i thought the similarity with the logarithm was too salient not to connect the two.
 

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