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Even and odd proof

  • Thread starter quasar987
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  • #1
quasar987
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I'm puzzled by this question: Show that for all function f:R-->R. there exists an even function p and an odd function i such that f(x) = p(x) + i(x) forall x in R.

I got nothing.
 

Answers and Replies

  • #2
Physics Monkey
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Try looking at f(-x) and relating it to p(x) and i(x). Do you notice anything?
 
  • #3
quasar987
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But there is nothing to look at. f(-x) = ..........................?

The only thing would be SUPPOSING the result of the thorem is true, then it would implies that there exist p and i such that f(x) = p+i and hence f(-x) = p(x)-i(x), but that's as far as that goes. :grumpy:
 
  • #4
Hurkyl
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f(x) = p+i and hence f(-x) = p(x)-i(x), but that's as far as that goes.
No it's not.
 
  • #5
quasar987
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Oh I see. That was very insightful Hurkyl. :tongue:
 

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