Hölder's Theorem & Manipulating 1/p + 1/q = 1

  • Thread starter Thread starter zxh
  • Start date Start date
Click For Summary

Homework Help Overview

The discussion revolves around the relationship between the equations 1/p + 1/q = 1 and (p-1)(q-1) = 1, within the context of Hölder's Theorem. Participants explore the manipulation of these equations and their connections to logarithmic properties.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss methods to manipulate the initial equation to derive the second equation, with some suggesting multiplication and factoring as potential steps. There is also a mention of the relevance of logarithmic properties, although some participants question this connection.

Discussion Status

The discussion is active, with various approaches being explored. Some participants have provided guidance on how to manipulate the equations, while others are questioning the relevance of logarithms to the problem at hand. There is no explicit consensus on the connections being made.

Contextual Notes

Participants are navigating the relationship between the two equations while considering the implications of Hölder's inequality. There is a suggestion that the original poster may be drawing parallels that are not directly related to the equations in question.

zxh
Messages
8
Reaction score
0

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" )

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!
 
Last edited by a moderator:
Physics news on Phys.org
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.
 
zxh said:
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" )

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
 
Last edited by a moderator:
Alright, makes me look tired, but i thought the similarity with the logarithm was too salient not to connect the two.
 

Similar threads

  • · Replies 7 ·
Replies
7
Views
1K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 12 ·
Replies
12
Views
3K
Replies
6
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 9 ·
Replies
9
Views
3K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K