MHB Prove/Disprove: Uniform Continuity of sin(sin(x))

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SUMMARY

The discussion centers on proving or disproving the uniform continuity of the function sin(sin(x)) using the ε, δ definition. The ε, δ definition states that for every ε > 0, there exists a δ > 0 such that for all x, y in the domain of f(x), if |x - y| < δ, then |f(x) - f(y)| < ε. A hint provided by Dan suggests using the formula sin(x) - sin(y) = 2cos((x+y)/2)sin((x-y)/2) to analyze the continuity of the function.

PREREQUISITES
  • Understanding of the ε, δ definition of continuity
  • Familiarity with trigonometric identities, specifically sin(x) and cos(x)
  • Basic knowledge of limits and continuity in real analysis
  • Ability to manipulate inequalities involving real functions
NEXT STEPS
  • Study the properties of uniform continuity in real analysis
  • Learn how to apply the ε, δ definition to various functions
  • Explore trigonometric identities and their applications in proofs
  • Investigate examples of uniformly continuous functions for comparison
USEFUL FOR

Students of real analysis, mathematicians, and anyone interested in understanding the concepts of continuity and trigonometric functions.

solakis1
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prove or disprove if the following function is uniformly continuous:

$$sin(sin(x)) $$ using the ε,δ definition
 
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Do you know what that means? What IS the "$\epsilon, \delta$ definition"?
 
Country Boy said:
Do you know what that means? What IS the "$\epsilon, \delta$ definition"?
This is a challenge problem. solaklis wants you to find the answer...

-Dan
 
hint
[sp]Use the formula :
sinx-siny =$2cos\frac{x+y}{2}sin\frac{x-y}{2}$[/sp]
 
Country Boy said:
Do you know what that means? What IS the "$\epsilon, \delta$ definition"?
The $\epsilon$ $\delta$ definition for real function f(x) is :

Given $\epsilon>0$ there exists a $\delta>0$ such that :
For all x,y belonging the to domain of f(x) and $|x-y|<\delta$ then $|f(x)-f(y)|<\epsilon$
 

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