If $$\sum_{k=0}^{\infty}a_kt^k = \sum_{k=0}^{\infty}b_kt^k \quad \forall t \in \mathbb{R} \implies (a_k=b_k \ \forall k \in \mathbb{N_{0}} ) $$ Prove or make counterexample.I think it's true but don't know how to proceed.
I start $$ \lim_{n\to\infty}\sum_{k=0}^{n}a_kt^k =\lim_{m\to\infty}\sum_{k=0}^{m}b_kt^k $$ I can only get that $a_0 = b_0$ by setting $t=0$.But then i tried same trick to do for $a_1 = b_1 $ but that didn't help me. Maybe it is something trivial but I can't see it.
On
This is a property of power series if it is assumed that the series converge in $(-r,r)$ for some $r>0$. The coefficients of a convergent power series $\sum a_n t^{n}$ are simply $\frac {f^{(n)} (0)} {n!}$ where $f$ is the sum of the series.
Reference: Any book on Complex Analysis starts with this this theorem. But you can simply replace complex numbers by real numbers in the proof and it works fine for real series also.
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Let $f(t) = \sum_{k = 0}^\infty a_kt^k$, $g(t) = \sum_{k = 0}^\infty b_kt^k$. By hypothesis, $f(t) = g(t)$ $ \forall t \in \mathbb{R}$. Then $$ f'(0) = \lim_{h \to 0} \frac{f(h) - f(0)}{h} = a_1$$ and $$ g'(0) = \lim_{h \to 0} \frac{g(h) - f(0)}{h} = b_1.$$ Hence $a_1 = b_1$. By continuing this way, you prove equality $a_k = b_k$ for every $k \in \mathbb{N}$.
Notice that you don't actually need equality of the two series for every $t \in \mathbb{R}$, but only for every $t$ in a fixed neighborhood of zero.
The property is equivalent to
$$\forall t:d(t):=\sum_{k=0}^\infty d_kt^k=0\implies \forall k:d_k=0.$$
Clearly, $d(0)=0\implies d_0=0$.
Then as $d(t)$ converges, so does $e(t):=\dfrac{d(t)}t=\displaystyle\sum_{k=1}^\infty d_kt^{k-1}$, and $\lim_{t\to0}e(t)=0=d_1$. You can iterate.