MHB Does This Sequence Converge Uniformly?

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The sequence of functions \( f_n = \sin(x) - \frac{nx}{1+n^2} \) converges pointwise to \( f^{\star}(x) = \sin(x) \) as \( n \) approaches infinity. However, the discussion reveals that \( f_n \) does not converge uniformly, as the supremum for \( |f_n(x) - f^{\star}(x)| \) approaches infinity for \( x \in \mathbb{R} \). A clarification on grammar notes that "uniformly" should modify verbs or adjectives, not nouns. The conversation also touches on the importance of precise language in mathematical discussions. Overall, the sequence demonstrates pointwise convergence but fails to meet the criteria for uniform convergence.
mathmari
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Hey! :giggle:

We have the sequence of functions $$f_n=\sin (x)-\frac{nx}{1+n^2}$$ I want to check the pointwise andthe uniform convergence.

We have that $$f^{\star}(x)=\lim_{n\rightarrow \infty}f_n(x)=\lim_{n\rightarrow \infty}\left (\sin (x)-\frac{nx}{1+n^2}\right )=\sin(x)$$ So $f_n(x)$ converges pointwise to$f^{\star}=\sin(x)$.
We have that $$\left |f_n(x)-f^{\star}(x)\right |=\left |\sin (x)-\frac{nx}{1+n^2}-\sin(x)\right |=\left |-\frac{nx}{1+n^2}\right |$$ We have to calculate first the supremum for $x\in \mathbb{R}$ and then the limit for $n\rightarrow \infty$.
Isn't the supremum $x\in \mathbb{R}$ the infinity? :unsure:
 
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mathmari said:
Isn't the supremum $x\in \mathbb{R}$ the infinity?
Hey mathmari!

Yes, it is. (Nod)
 
Klaas van Aarsen said:
Yes, it is. (Nod)

So $f_n$ doesn't converge uniformly to$f^{\star}$, right? :unsure:
 
mathmari said:
So $f_n$ doesn't converge uniformly to$f^{\star}$, right?

Indeed. :geek:
 
Comment on Grammer: "uniformly" is an adverb and so modifies to verbs, adjectives, and other adverbs. Here "converge" is a noun and so requires the adjective "uniform".

One can ask "Does this converge uniformly?" or "Is this convergence uniform?" but not "Is this convergence uniformly".

(Yes, I realize this was probably just a typo but I couldn't help myself!)
 
We all know the definition of n-dimensional topological manifold uses open sets and homeomorphisms onto the image as open set in ##\mathbb R^n##. It should be possible to reformulate the definition of n-dimensional topological manifold using closed sets on the manifold's topology and on ##\mathbb R^n## ? I'm positive for this. Perhaps the definition of smooth manifold would be problematic, though.

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