UniformConvergence

• Visualize uniform convergence for the sequence $f_n(x)=\frac{\sin(nx)}{n}\qquad (x\in[0,2\pi]),$ and its limit function $f(x)=0.$

• Use the slider $n$ to choose the current index $n$ (the solid blue curve is $f_n$).

• Use the slider $\varepsilon$ to set the tolerance $\varepsilon>0$.

• Drag the point $x$ on the $x$-axis to choose the evaluation point $x\in[0,2\pi]$.

• The solid blue curve is the current function $f_n(x)=\sin(nx)/n$.

• The faint dashed blue curves are earlier functions $f_k(x)=\sin(kx)/k$ (for $k<n$), to show how the whole family shrinks as $n$ increases.

• The red horizontal line is the limit function $f(x)=0$.

• The dashed red lines are the $\varepsilon$-tube around the limit: $y=\pm\varepsilon$.

• The vertical gray guide marks the chosen $x$.

• The highlighted point $P_n=(x,f_n(x))$ shows the value at the chosen $x$.

• The dashed segment from $(x,0)$ to $(x,f_n(x))$ visualizes the error $|f_n(x)-0|$.

  • choose an $\varepsilon>0$,
  • increase $n$ until the entire blue curve stays inside the $\varepsilon$-tube $y=\pm\varepsilon$,
  • now drag $x$ across the whole interval $[0,2\pi]$ and notice that $P_n$ stays green everywhere.
  
  

Description

Link to code.

Reference: Lecture Notes Calculus 1 ( May22,2022 ) Definition8.1 page 144

This applet visualizes uniform convergence using the sequence of functions

The blue curve is $f_n$ and the red line is the limit function $f(x)=0$.

Two sliders are used: n selects the current index $n$ in $f_n(x)=\frac{\sin(nx)}{n}$, and ε selects the tolerance $\varepsilon>0$. A draggable point x chooses the evaluation point $x\in[0,2\pi]$.

The point x as well as the limit and the ε tube turn green when:

const ok = Math.abs(fnx - fx) < eps;

By selecting an n and fixing an ε large enough, the student can observe that $|f_n(x) - f(x)| < \varepsilon$ holds for all x by dragging the point x, so that n depends only on ε.