Problems of functional series

1. Study the point convergence and the uniform convergence of the succession of functions with general term f_n\left(x\right)=\sin^n\left(x\right)\cos\left(x\right)

We first study the pointwise limit for n\rightarrow\infty. It must be taken into account when calculate the limit that the variable x can take any value in the domain of the function. Recalling that the function \sin\left(x\right) is bounded in the interval [- 1,1], and also for x\neq k\mathrm\pi,\;k=0,1,2,\dots is bounded in the open interval \lim_{n\rightarrow\infty}k^n=0\; whenever k\in\left(-1,1\right)\; we get:

\lim_{n\rightarrow\infty}f_n\left(x\right)=\lim_{n\rightarrow\infty}\sin^n\left(x\right)\cos\left(x\right)=\left\{\begin{array}{l}0\;\text{si }x\neq k\mathrm\pi,\;k=0,1,2,\dots\\0\;\text{si }x=k\mathrm\pi,\;k=0,1,2,\dots\end{array}\right.

Since when\sin\left(x\right)=1\Leftrightarrow\cos\left(x\right)=0\;. In the following illustration we see the graphs of three terms: f_1\left(x\right)=\sin\left(x\right)\cos\left(x\right),\;f_3\left(x\right)=\sin^3\left(x\right)\cos\left(x\right),\;\;f_{20}\left(x\right)=\sin^{20}\left(x\right)\cos\left(x\right), in black, red and green, respectively, as the index n increases the function f_n tends to take smaller values:

Algunos términos de la sucesión funcional del problema

Some terms of the functional series

Therefore f_n (x) is point-wise convergent to the null function f (x) = 0. Let's now look at the uniform convergence.

Criterion of uniform convergence of a sequence f_n(x)
f_n(x) Converges uniformly to the point boundary f (x) whenever it is fulfilled:
\lim_{n\rightarrow\infty}{\text{Sup}}_x\left|f_n\left(x\right)-f\left(x\right)\right|\text{=0}.

We apply it: {\text{Sup}}_x\;\left|f_n\left(x\right)-f(x)\right|={\text{Sup}}_x\left|\sin^n\left(x\right)\cos\left(x\right)-0\right|{\text{=Sup}}_x\left|\sin^n\left(x\right)\cos\left(x\right)\right|. To find the maximum we derive and equate to zero, the absolute value does not concern us because the functions f_n (x) are symmetric respect to the abscissa axis (see previous figure):

\begin{array}{l}D_x\left(\sin^n\left(x\right)\cos\left(x\right)\right)=n\sin^{n-1}\left(x\right)\cdot\cos\left(x\right)\cdot\cos\left(x\right)-\sin^n\left(x\right)\cdot\sin\left(x\right)=0\Leftrightarrow\\\sin^{n-1}\left(x\right)\left[n\cdot\cos^2\left(x\right)-\sin^2\left(x\right)\right]=0\Leftrightarrow\\\left\{\begin{array}{l}\sin^{n-1}\left(x\right)=0\Leftrightarrow x=k\mathrm\pi,\;\mathrm k=0,1,2,\dots\\n\cdot\cos^2\left(x\right)-\sin^2\left(x\right)=0\Leftrightarrow\sin^2\left(x\right)=n\cdot\cos^2\left(x\right)\end{array}\right.\end{array}

The first case is discarded because it provides a minimum not a maximum, since \sin^{n-1}\left(x\right)=0\Leftrightarrow\sin\left(x\right)=0\Leftrightarrow f_n(x)=0. For the second case, using the equality \sin^2\left(x\right)+\cos^2\left(x\right)=1 and calling x_0 to the point at which we have a maximum:

\begin{array}{l}\sin^2\left(x_0\right)=n\cdot\cos^2\left(x_0\right)\Leftrightarrow1-\cos^2\left(x_0\right)=n\cdot\cos^2\left(x_0\right)\\\Leftrightarrow\sqrt{\frac1{\left(n+1\right)}}=\cos\left(x_0\right)\Leftrightarrow\sin\left(x_0\right)=\sqrt{1-\frac1{\left(n+1\right)}}=\sqrt{\frac1{n+1}};\\f_n\left(x_0\right)=\sin^n\left(x_0\right)\cos\left(x_0\right)=\left(\frac1{n+1}\right)^\frac n2\sqrt{\frac1{\left(n+1\right)}}.\end{array}

We apply the limit:

\lim_{n\rightarrow\infty}{\text{Sup }}_x\left|f_n\left(x\right)\right|=\lim_{n\rightarrow\infty}\left(\frac1{n+1}\right)^\frac n2\sqrt{\frac1{\left(n+1\right)}}=\lim_{n\rightarrow\infty}\frac1{\left(n+1\right)^{\left(n+1\right)/2}}=0.

Therefore the sequence converges uniformly.

separador2

2. Study the pointwise convergence and the uniform convergence of the sequence of functions with general term f_n\left(x\right)=\left\{\begin{array}{l}\text{inf }\left(n,\frac1x\right)\text{ si }x>0\\0\;\text{si }x=0\end{array}\right. Also study the uniform convergence in the constrained interval (0,1].


For small values of x the fraction 1 / x takes large values, and the \text{inf}\left(n,\frac1x\right)=n, whereas for large values the opposite happens, \text{inf}\left(n,\frac1x\right)=1/x. In the following image we have the graph of f_1 (x) (in the point x = 0 the image is not shown, which is zero):

exercici2_succ_funcions

The transition point between the two values of \text {inf} \left (n, \frac1x\right) can be easily found: n = 1 / x\Leftrightarrow x = 1 / n; At point x_0 = 1 / n we have \text {inf} \left (n, \frac1x\right) = n = \frac1x, and we can then redefine the function definition:

f_n\left(x\right)=\left\{\begin{array}{l}\frac1x\;\text{si }x>\frac1n\\n\;\text{si }0<x\leq\frac1n\\0\;\text{si }x=0\end{array}\right.

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