power series

**Kat-M** · April 17th 2010, 06:14 PM

Let $\Sigma_{k=0} ^\infty a_k x^k$ and $\Sigma _{k=0} ^\infty b_k x^k$ be real power series which converge for $|x|<1$ . If $\Sigma_{k=0} ^\infty a_k x^k$ = $\Sigma _{k=0} ^\infty b_k x^k$ for $x=1/2,1/3,1/4,...$ , then prove that $a_k=b_k$ for all $k$

**Failure** · April 17th 2010, 08:56 PM

Originally Posted by Kat-M

Let $\Sigma_{k=0} ^\infty a_k x^k$ and $\Sigma _{k=0} ^\infty b_k x^k$ be real power series which converge for $|x|<1$ . If $\Sigma_{k=0} ^\infty a_k x^k$ = $\Sigma _{k=0} ^\infty b_k x^k$ for $x=1/2,1/3,1/4,...$ , then prove that $a_k=b_k$ for all $k$

Indirect proof: Suppose there existed a first summation index $k_0$ for which $a_{k_0}\neq b_{k_0}$ , then you have for all $x_n := \frac{1}{n}$ , that

$\sum_{k=0} ^\infty a_k x_n^k-\sum_{k=0} ^\infty b_k x_n^k = x_n^{k_0}\cdot\sum_{k=0}^\infty \left(a_{k_0+k}-b_{k_0+k}\right)x_n^k=0$ .
But because the function $f(x) := \sum_{k=0}^\infty \left(a_{k_0+k}-b_{k_0+k}\right)x^k$ is continuous at $x=0$ , we find that $f(0)=\lim_{n\to\infty}f(x_n)=0$ , from which it follows that $a_{k_0}-b_{k_0}=0$ , thus contradicting our assumption $a_{k_0}\neq b_{k_0}$ .

**Kat-M** · April 18th 2010, 02:14 AM

i am not getting how u get the equation. can u explain it please?

**Failure** · April 18th 2010, 03:16 AM

Originally Posted by Kat-M

i am not getting how u get the equation. can u explain it please?

How am I to figure out what "equation" you are talking about. Why can't you, please, be more specific?

From the givens, I conclude that $\sum_{k=0}^\infty a_k x^k-\sum_{k=0}^\infty b_k x^k{\color{red}=0}$ , for all $x=x_n := \frac{1}{n}, n\in \mathbb{N}, n \geq 2$ .

Next, if we assume (contrary to what is the case), that $a_k=b_k$ is not true for all indices $k=0,1,\ldots$ , then there would have to exist a specific smallest index $k_0$ , such that $a_{k_0}\neq b_{k_0}$ . Since, therefore, for all indices $k$ smaller than $k_0$ we have $a_k=b_k$ , and thus $a_k-b_k=0$ , it follows that

$\sum_{k=0}^\infty a_k x^k-\sum_{k=0}^\infty b_k x^k=\sum_{k=k_0}^\infty (a_k-b_k)x^k=x^{k_0}\cdot \sum_{k=0}^\infty(a_{k_0+k}-b_{k_0+k})x^k{\color{red}=0}$ , for all $x=x_n := 1/n$ .
From $x^{k_0}=(1/n)^{k_0}\neq 0$ it follows for the continuous[!] function $f(x) {\color{blue}:=} \sum_{k=0}(a_{k_0+k}-b_{k_0+k})x^k$ that $f(0)=\lim_{x\rightarrow 0}f(x)=\lim_{n\rightarrow\infty} f(x_n){\color{red}=}\lim_{n\to\infty}0{\color{gree n}=0}$ .
But this means that $f(0){\color{blue}=}a_{k_0}-b_{k_0}{\color{green}=0}$ , and therefore $a_{k_0}=b_{k_0}$ , contrary to our assumption that $k_0$ is the smallest summation index for which that is not the case. Thus no such smallest index $k_0$ can exist, and this means that for all indices $k$ , we have $a_k=b_k$ , what was to be shown.

**Kat-M** · April 18th 2010, 11:32 AM

thank u

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