0
$\begingroup$

I have to prove the linear independence of $\{\sin^k(x)\}$, $ k\in \mathbb{N_0}$.

I haven't ideas for proving it. Linearly independent means that we cannot find solution of this : $\sin^{k_1}(x) = \sum_{i\ne k_1} l_i \cdot \sin^{i}(x)$ where $l_i \in \mathbb{R}$.

  • 1
    Usually Van der Monde determinant is helpful. – 2017-02-20
  • 1
    or you can consider to multiply by $sin(kx)$ and integrate in $[-\pi,\pi]$ – 2017-02-20
  • 0
    @Exodd: i think you see $sin(nx)$ instead $sin(x)^{n}$ – 2017-02-20
  • 2
    @Basti It's the same: $sin(x)^n$ can be written as sum of $sin(kx)$ for $k$sin(nx)$ appears only in $sin(x)^m$ with $m\ge n$ – 2017-02-20
  • 1
    Is $\sin^k(x)$ the $k$th iterate or the $k$th power of $\sin(x)$? – 2017-02-20

5 Answers 5

4

Suppose you have $$ \sum_{n=0}^N a_n\sin(x)^n = 0 $$ then the polynomial $$ p(y) = \sum_{n=0}^N a_n y^n $$ is of degree at most $N$, but $y=\sin(x)$ is a root for every $x$, so it has infinite roots. This means the polynomial is zero, or also said as $a_n=0$ for every $n$.

  • 0
    Thank you, but what means that $y = sin(x)$ is a root for every x. I mean root, why this polynom = 0 if $y = sin(x)$ – 2017-02-21
  • 0
    because of the first relations of the sines. If you substitute $y=sin(x)$ you obtain it. – 2017-02-21
1

Hint: If $\{f_k\circ g\}_{k\in\Bbb N}$ are linearly independent, then so are $\{f_k\}_{k\in\Bbb N}$.

Try $g=\arcsin$

  • 1
    Any additional condition on $g$? – 2017-02-20
  • 1
    @Clclstdnt No. (Note however that I use the $\{\}_{n\in\Bbb N}$ for *families*, not for *sets*). Assume $\sum a_k f_k=0$. Then $\sum a_k (f_k\circ g) = (\sum a_k f_k)\circ g=0\circ g=0$. Hence $a_k=0$ for all $k$. – 2017-02-21
0

If $\mathbb{sin}^{l}(x)=\sum_{k=1,k\neq l}^{n}l_{k}\mathbb{sin}^{k}(x)$ for all $x\in\mathbb{R}$, then $x^{l}=\sum_{k=1,k\neq l}^{n}l_{k}x^{k}$ for all $x\in[-1,1]$. So, the last equality holds for every $x\in\mathbb{C}$ (by an analytic continuation). We deduce $$x^{-1}=\sum_{k=1,k\neq l}^{n}l_{k}x^{k-l} $$ Integrating over the unitary circle we have $\frac{1}{2\pi i}=0$, because $z^{m}$ has a primitive for $m\neq-1$.

0

If $j,k\in \mathbb N_0$ with $j

Let $j_1,...,j_n$ be $n$ different members of $\mathbb N_0$ with $j_1=\min \{j_i:1\leq i\leq n\}.$

Let $g(x)=\sum_{i=1}^n A_i\sin^{(j_i)}x$ with each $A_i\ne 0.$

Then $\lim_{x\to 0}g(x)x^{-j_1}=A_1\ne 0$ so we cannot have $g(x)=0$ for all $x$.

0

Assume that $A$ is a finite subset of $\mathbb{N}^*$ and $$ f_A(x)=\sum_{a\in A} c_a \sin^a(x) = 0,\qquad C=c_{\max A}\neq 0.\tag{1} $$ $f_A(x)$ can be written as a Fourier series, and if $M=\max A$ the coefficient of $e^{iMx}$ in such Fourier series just depends on $C$, being $\pm\frac{C}{2^M}$. By Parseval's theorem $0=\int_{-\pi}^{\pi}f_A(x)^2\,dx $ is a fixed multiple of the sum of the squared coefficients of the previous Fourier series. In particular $C=0$, contradiction.