How can I show a sequence of polynomials doesn't converge uniformly to sin(1/x)?

Question

Specifically, how can I show a sequence of polynomials doesn't converge uniformly on (0,1) to the function f(x) = sin(1/x)

Answer 1

Let's recall the following theorem:

Let f_n be a sequence of functions defined on a subset A of R^n that converges uniformly to a function f with range in R^m. Let a be an accumulation point of A such that, for every n, L_n = lim (x --> a) f(x) exists in R^m. Then, f has a limit at a and

lim (x --> a) f(x) = lim L_n

Polynomials are continuous on R and have a limit at each real number, specifically at 0, such limit being the independent term. Since 0 is an ccumulation point of (0,1) it follows that if a sequence of polynomials converge uniformly on (0,1) to a function f, then f has a limit at x =0.

But the function f(x) = sin(1/x) doesn't have a limit at x =0, so it can't be the uniform limit of a sequence of polynomials.

EDIT:
Ksoileau is right. f(x) = sin(1/x) is continuous on (0,1), so integrable over [a, 1], for every a in (0, 1). In addition, it is bounded on (0, 1). So, its improper integral exists and equals lim (a --> 0+) Integral (from a to 1) sin(1/x) dx


EDIT2: In order for Ksoileau's proof to be complete, he had to show the sequence of the derivatives of the polynomials converges to f'. The simple fact that a sequence f_n of differentiable functions converges uniformly to a function f does not imply f'_n --> f'. This is certainly true if the sequence of the derivatices converge uniformly.

So far, I think my proof is the simplest.

EDIT3

If a sequence of polynomials with bounded degree converges, then the limit function is a polynomial and the sequence of derivatives surely converges to the derivative of the limit function. But if the degres form an unbounded sequence, then the limit function doesn't need to be a polynomial and the the limit of the sequence of drivatives, if it exists, doesn't need to be the derivative of the limit function of the primitives.

Answer 2

@jeredwm: Actually the integral exists and equals .5040670619...

Suppose for a contradiction that the sequence P_n(x) converges uniformly to sin(1/x) on (0,1). Then the sequence of derivatives (which are again polynomials) P_n'(x) must converge to -1/x^2*cos(1/x). But since -1/x^2*cos(1/x) is unbounded on (0,1), it cannot be the uniform limit of polynomials there.

@Steiner: "The simple fact that a sequence f_n of differentiable functions converges uniformly to a function f does not imply f'_n --> f'."
True in general. However, if the f_n are polynomials, their derivative polynomials converge uniformly to the derivative f'.

Answer 3

it fairly is effect of right here theorem enable f_n be a chain of applications defined on a subset A of R^n that converges uniformly to a function f with variety in R^m. enable a be a shrink area of A such that, for each n, L_n = lim (x ? a) f_n(x) exists in R^m. Then, f has a shrink at a and lim (x ? a) f(x) = lim L_n (truthfully, this theorem holds customarily metric areas). this is, below uniform convergence, we are able to interchange the order the obstacles are taken: lim (n ? ?) lim (x ? a) f_n(x) = lim (x ? a) lim (n ? ?) f_n(x) Polynomials are non-provide up on R and, consequently, have a shrink at each and every actual extensive variety (purely the cost of the polynomial at this extensive variety). specially at 0, this shrink is the autonomous time era. in view that 0 is a shrink area of (0,a million) it follows that if a chain of polynomials converge uniformly on (0,a million) to a function f, then f has a shrink at 0+-. yet f(x) = sin(a million/x) does not have a shrink at 0+, it retains oscillating from -a million to a million and does not attitude any actual extensive variety. So, it won't be able to be the uniform shrink of a chain of polynomials defined on (0, a million).

Answer 4

Any polynomial on (0,1) has limited derivative
abs(p') < P

Consequently such polynomial cannot change more than PΔx from x to x+Δx.

Lets choose Δx = 1/F and two adjacent exterma of sin(1/x), which are closer than Δx.

At least one of those extrema (which are -1 and +1) is at lest 1 away from the sin(1/x):
|f(x)-p(x)| >= 1 .

Therefore any polynomial in the sequence has error of approximation of at least 1 somewhere.