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Assume that $f(x) = \sum_{n=0}^{\infty} c_n x^n$ is convergent in $(-1,1)$ and that $f(1/k) = 0$ for all $k\in \mathbb{N}$. Show that $c_n = 0$ for all $n\geq 0$.

I'm not sure what I can use to prove this. I've tried creating various alternating series, none of which sum to 0, but I'm sure there's a theorem of some sort that would be more useful than attempting to prove this by hand.

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  • $\begingroup$ Note that, if it exists (which it does by Taylor's Theorem), the derivative of $f(x)$ at $0$ is equal to $\lim_{k\to\infty}\frac{f(0)+f(0+1/k)}{1/k}$ $\endgroup$ – Grant B. Apr 6 '17 at 4:25
  • $\begingroup$ @GrantB. hmm...and since $f(1/k) = 0$, $f'(0) = lim_{k->\infty} k f(0)$. Is this the right direction? $\endgroup$ – mizichael Apr 6 '17 at 4:30
  • $\begingroup$ And since we know this limit exists (since the power series exists)... $\endgroup$ – Grant B. Apr 6 '17 at 4:37
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Hint: Use mean value theorem (to be more precisely, it should be Rolle's Theorem) to show that there exists a sequence $a_n \rightarrow 0$ such that $f'(a_n) = 0$.

Inductively, you could prove $f^{(n)}(0) = 0$ for all $n\geq 0$.

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Let $f$ be represented by the series

$$f(x)=\sum_{n=0}^\infty c_nx^n$$

for $|x|<1$. Furthermore, we are given that $f(1/k)=0$ for all $k\in \mathbb{N}$. Clearly we have

$$0=f(1/k)=\sum_{n=0}^\infty c_n\left(\frac1k\right)^n\tag 1$$


Letting $k\to \infty$ in $(1)$ reveals that $c_0=0$. Then, we have

$$f(1/k)=\sum_{n=1}^\infty c_n\left(\frac1k\right)^n \tag 2$$

whereupon multiplying both sides of $(2)$ by $k$ yields

$$kf(1/k)=\sum_{n=1}^\infty c_n\left(\frac1k\right)^{n-1}\tag 3$$

Noting that the left-hand side of $(3)$ is $0$, we take the limit as $k\to \infty$ of the right-hand side of $(3)$ to find that $c_1=0$.


Continuing recursively, we have for all $m\ge 1$

$$k^mf(1/k)=\sum_{n=m}^\infty c_n\left(\frac1k\right)^{n-m}\tag 4$$

whence letting $k\to \infty$ reveals that $C_m=0$.

And we are done!

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