4

Consider the set $S=\left\{x^{n}\,\colon\, n\in\mathbb{N}\right\}$. (Note that $x\in\mathbb{R}$) Is this set linearly dependent?

Well thinking about it we want to find some non-trivial values $\lambda_{n}$ such that $$\lambda_{1}x+\lambda_{2}x^{2} + \ldots + \lambda_{n}x^{n} + \ldots = 0.$$ In other words we want to find if $$\sum_{k=1}^{\infty}\lambda_{n}x^{n} = 0$$ for any $\lambda_{n}\not=0$. However i'm not really sure how to proceed.

user2850514
  • 3,689
  • How many roots does a non-zero polynomial of degree $n$ have? You want finite sums, I think, by the way... – David Mitra Jan 27 '15 at 20:07
  • Note that $k$ doesn't eventually go till infinity, as only finite linear combinations are considered in the definition of (in-)dependence. – Berci Jan 27 '15 at 20:08
  • @Berci where does it stop then since we are considering all $n\in\mathbb{N}$ and David $n$ roots. – user2850514 Jan 27 '15 at 20:10
  • It stops, say, at index $K\in\Bbb N$. So, eventually what you will write up is nothing but a polynomial with coefficients $\lambda_n$. That makes it easier.. – Berci Jan 27 '15 at 20:11
  • Related: http://math.stackexchange.com/q/152133/ – Jonas Meyer Jan 27 '15 at 21:22

4 Answers4

5

An approach that doesn't involve differentiation: it suffices to show that $$ f(x) = \lambda_{1}x+\lambda_{2}x^{2} + \cdots + \lambda_{n}x^{n} = 0 \implies \lambda_i = 0 $$ Note that if any of the $\lambda_i$ is non-zero, $\lim_{x \to \infty} f(x)$ is either $\infty$ or $-\infty$, depending on the sign of the coefficient of the non-zero term of greatest degree. If $|\lim_{x \to \infty} f(x)| = \infty$, then $f(x) \neq 0$.

Thus, if any of the $\lambda_i$ are non-zero, then $f(x) \neq 0$. By contrapositive, the conclusion follows.

Ben Grossmann
  • 225,327
  • How do you know if $\lim_{x\to\infty} f(x)|=\infty$ then $f(x)\not=0$? For example $f(0)=0$.. – user2850514 Jan 27 '15 at 21:21
  • 1
    $f(x) = 0$ means that $f(a)$ = 0 for every $a \in \Bbb R$. If $f(x) = 0$, then $\lim_{x \to \infty} f(x) = 0$. Thus, if $\lim_{x \to \infty} f(x) \neq 0$, then $f(x) \neq 0$. – Ben Grossmann Jan 27 '15 at 21:27
4

You are probably considering the set as a subset of vector space of real functions on $\mathbb{R}$ (or continuous functions, perhaps, but it's the same).

A set is linearly independent if and only if any finite subset is linearly independent, and so we can simply show that $$ \{1,x,x^2,\dots,x^n\} $$ is linearly independent. Suppose $\lambda_0+\lambda_1x+\dots+\lambda_nx^n=0$ (that is, the constant zero function); this is a polynomial that has infinitely many roots, so it's the zero polynomial, which means $\lambda_0=\lambda_1=\dots=\lambda_n=0$.

egreg
  • 238,574
2

I presume you are talking about the functions $b_n(x) = x^n$.

Suppose you have $\alpha_k$ ($k=1,..,N$) such that $f(x)=\sum_k \alpha_k b_k(x) = 0$ for all $x$. Note that $f^{(k)}(0) = 0$ for all $k \in \mathbb{N}$.

It is not hard to show that $f^{(k)}(0) = k! \alpha_k $, from which it follows that $\alpha_k = 0$. Hence the $b_n$ are linearly independent.

copper.hat
  • 172,524
  • Hi, I'm not really sure how you have noted that $f^{(k)}(0) = 0$ for all $k\in\mathbb{N}$. Take $k=1$ for example, at this point $f(x) = \alpha_{1}x$, then $f'(x)=f'(0)=\alpha_{1}\not=0$. Where is my fault? – user2850514 Jan 27 '15 at 21:15
  • Since $f(x) = 0$ for all $x$, all derivatives are zero. You are trying to show that if there are $\alpha_k$s such that $f(x) = 0$ everywhere then the $\alpha_k = 0$. – copper.hat Jan 27 '15 at 21:17
  • Ah okay that makes sense, so in effect we have an equality $f(x)=0$ and we have differentiated both sides $k$ times. – user2850514 Jan 27 '15 at 21:22
2

Let $\alpha_1<\ldots<\alpha_k\in\Bbb N$ and let $\lambda_1,\ldots,\lambda_k\in \Bbb R$ such that

$$P(x)=\lambda_1 x^{\alpha_1}+\cdots+\lambda_k x^{\alpha_k}=0$$ Now we have successively $P^{(\alpha_i)}(x)=(\alpha_i)!\lambda_i=0$ for $i=\alpha_k,\ldots,\alpha_1$. Conclude.