Prove Injectivity of x^x Function?

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SUMMARY

The function f(x) = x^x is proven to be non-injective on the interval of positive real numbers. The analysis shows that while f(x) is monotonic on the intervals (0, 1/e) and (1/e, 1), it fails to be injective due to the existence of points where f(a) = f(b) for a ≠ b, specifically f(1/4) = f(1/2). The function is continuous and differentiable for x > 0, with a unique minimum at x = 1/e. The discussion concludes that f(x) is injective on negative integers but not on positive reals.

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  • Understanding of injective functions and their properties
  • Knowledge of calculus, specifically derivatives and monotonicity
  • Familiarity with the Mean Value Theorem (MVT) and Rolle's Theorem
  • Basic concepts of transcendental functions and their classifications
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  • Study the properties of monotonic functions and their implications for injectivity
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  • #31
mathwonk,this is strange to me ,but if that's really definition you're right (thank you!).
composition of standard elementary functions x,e^x,and ln(x) creates xx.Indeed,I thought elementary functions were those obtainable from standard elementary functions only by means of finite number of arithmetical operations (+,-,*,:) among them.
 
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  • #32
tehno said:
mathwonk,this is strange to me ,but if that's really definition you're right (thank you!).
composition of standard elementary functions x,e^x,and ln(x) creates xx.Indeed,I thought elementary functions were those obtainable from standard elementary functions only by means of finite number of arithmetical operations (+,-,*,:) among them.


throw in composition and you are there.

i.e. ln(ln(x)) is also elementary.

I guess the definition was made this way to try to encompass functions we usually try to antidifferentiate, like x^2 e^(x^3).


by the way in omitting trig functions, i was tacitly assuming the functions are complex valued, so that trig functions and their inverses are a special case of exponentials and logs.

e.g. arctan is the same as log except it twirls around i and -i instead of 0 and infinity. so if you compose with an automorphism like (1-i)/(1+i) they become almost the same.


to see this, just think of log as path integral of 1/z along paths that avoid 0 and infinity, and arctan as path integral of 1/(1+z^2) along paths that avoid i and -i.
 
Last edited:
  • #33
You pro-mathematicians certainly know *all rules * of your game.And We computer scientists must obey it without objections :smile:.
Maybe,I'll be hang for telling you this but most of us consider only
polynomials over Q-field truly elementary functions :wink:
 

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