1. ## power series

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

2. Originally Posted by Kat-M
Let $\displaystyle \Sigma_{k=0} ^\infty a_k x^k$ and $\displaystyle \Sigma _{k=0} ^\infty b_k x^k$ be real power series which converge for $\displaystyle |x|<1$. If $\displaystyle \Sigma_{k=0} ^\infty a_k x^k$= $\displaystyle \Sigma _{k=0} ^\infty b_k x^k$ for $\displaystyle x=1/2,1/3,1/4,...$, then prove that $\displaystyle a_k=b_k$ for all $\displaystyle k$
Indirect proof: Suppose there existed a first summation index $\displaystyle k_0$ for which $\displaystyle a_{k_0}\neq b_{k_0}$, then you have for all $\displaystyle x_n := \frac{1}{n}$, that

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

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

4. 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 $\displaystyle \sum_{k=0}^\infty a_k x^k-\sum_{k=0}^\infty b_k x^k{\color{red}=0}$, for all $\displaystyle x=x_n := \frac{1}{n}, n\in \mathbb{N}, n \geq 2$.

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

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

5. thank u