How Can I Prove Uniform Convergence of This Function as ρ→0?

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SUMMARY

The discussion focuses on proving the uniform convergence of the function \((\rho^4\cos^2{\theta} + \sin^3{\theta})^{\frac{1}{3}} - \sin{\theta}\) as \(\rho \to 0\). The key argument presented is that since \(|\sin \theta| \leq 1\) and the function is continuous, it suffices to show that the inner expression converges to \(\sin^3{\theta}\). Consequently, taking the cube root leads to the conclusion that the function approaches \(\sin{\theta}\), resulting in a limit of 0. The continuity of the cube root function is also a critical assumption in this proof.

PREREQUISITES
  • Understanding of polar coordinates in mathematical analysis
  • Knowledge of uniform convergence in the context of real analysis
  • Familiarity with continuity and limits of functions
  • Basic properties of trigonometric functions, specifically sine
NEXT STEPS
  • Study the concept of uniform convergence in real analysis
  • Explore the properties of continuous functions and their limits
  • Investigate the behavior of cube root functions near zero
  • Review examples of convergence proofs involving trigonometric functions
USEFUL FOR

Mathematics students, particularly those studying real analysis, and educators looking to deepen their understanding of uniform convergence and continuity in functions.

malachia
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Probably is a silly question, but how could I prove that the function (expressed in polar coordinates)

<br /> \left(\rho^4\cos^2{\theta} + \sin^3{\theta}\right)^{\frac{1}{3}} - \sin{\theta}<br />

converges to 0 as rho->0 uniformely in theta (if it is true, of course)?
 
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It should be straightforward enough, since |sin|≤ 1 and the angle domain is a finite interval.
 
Ok well your functions are continuous. So show that inside goes to sin^3(theta), then the cubed root is going to equal sin(theta), then subtract to get 0.

But this is assuming you have defined or can assume x^(1/3) is defined and is continuous.
 

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