3
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Prove whether $f_n(x) = \frac{nx^4+1}{nx+1}$ is uniformly convergent on $[1,c]$ and $[1,\infty )$
I don't understand your notation, but your 'proof' is essentially correct.
Let me rewrite it more properly:
The sequence $f_n$ converges uniformly to $f$ on $[1,\infty)$ if
$$
\sup_{x \in [1,\infty)} |...
- 21.8k
3
votes
Accepted
Why do powers of $\operatorname{sinc}$ converge to a Gaussian function?
Your issue is explainable in a clean way with Fourier transform and spline functions.
Let us denote by $\chi$ the characteristic function $\chi_{[-\tfrac12,\tfrac12]}$ of interval $[-\tfrac12,\tfrac12]...
- 85.7k
3
votes
Accepted
a bounded measurable function is a uniform limit of increasing bounded simple functions
I think you've misunderstood. The statement "f is the uniform limit of bounded simple functions" means that there exists a sequence $f_n$ of bounded simple functions that converges ...
- 100k
2
votes
Accepted
Convergence of a sequence of function $t^n\sin(1-t)$
Hint: $0 \le t^{n}\sin (1-t)\le t^{n}(1-t)$. If $t >1-\epsilon$ then $t^{n}(1-t) \le 1-t <\epsilon$. For $t \le 1-\epsilon$ we have $t^{n}(1-t) \le t^{n}$ Can you finish?
- 41.5k
2
votes
Accepted
If $\lim_{x\to\infty}f(x)=1$, then $f_n(x)=f(nx)$ is not uniformly convergent
If $f$ is non-constant, then $f(\xi)\neq1$ for some $\xi\in(0,\infty)$. As $f_n(x)\to1$ as $n\to\infty$ for all $x\in(0,\infty)$, if it converged uniformly then the limit function would be the ...
- 11.1k
1
vote
Convergence of a sequence of function $t^n\sin(1-t)$
The sequence $x_n(t)$ is decreasing and convergent pointwise to $x(t)=t^2$ (a continuous function). By Dini's theorem the convergence is uniform.
Concerning an explicit estimate, with no use of ...
- 39.5k
1
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Determining if the function converges uniformly
How is $\bigg|\frac{nx}{\sqrt{1+n^2x^2}}-1\bigg| = \frac{nx-\sqrt{1+n^2x^2}}{\sqrt{1+n^2x^2}}$? What if $x$ is negative?
So, it is quite difficult to accept what you did. Firstly, notice that if
$$f(...
- 2,891
1
vote
Determining if the function converges uniformly
When we find the piecewise limit we get for $x\neq 0$
$$\lim_{n \to \infty} f_n(x) = \lim_{n\to \infty} \frac{nx} {\sqrt{1+n^2x^2}} = \frac{x}{|x|} = \text{sgn} (x) $$
while for $x=0$
$$\lim_{n \to \...
- 2,066
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