For each n epsilon N, set fn(x) = x + x/n sin nx, x epsilon R. Show that the seq

Question

For each n epsilon N, set fn(x) = x + x/n sin nx, x epsilon R. Show that the sequence {fn} converges uniformly to f(x) = x for all x epsilon [-a, a], a > 0. Does {fn} converges uniformly to f on R ?

Explanation / Answer

as n-> infinity ((x/n)*sin(nx))->0 therefore fn(x) converges to x pointwise for uniform convergence M=|fn(x)-f(x)| M=|(x/n)*sin(nx)| max of M occurs for tan(nx)=-nx max of M is M=|(x/n)*(-nx/((1+(nx)^2)^0.5)| M=((1/n)^2*((nx)^2/(1+(nx)^2)^0.5)) where nx is obtained from solution of tan(nx)=-nx as n->infinity it can be seen that M->0 therefore f is uniformly convergent

Related Questions

Navigate

• Browse (All)
• Browse F
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.