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If for every $t$ $$\sum_{i=0}^{k_1}\left[\left(a_i^Tx-b_i\right)t^i\right]=0$$ where $a_i \in \mathbb{R}^{n \times 1}$, $x \in \mathbb{R}^{n \times 1}, \forall i \in \{0,\dots, k_1\}$, and $b_i \in \mathbb{R}$, then $$Ax=b$$ where the rows of $A$ are the vectors $a_i^T$ and the components of $b$ are the values $b_i$, $\forall i \in \{0,\dots, k_1\}$.

For the case $$\sum_{i=0}^{k_1}\left[\left(a_i^Tx-b_i\right)t^i\right]+t^\lambda\sum_{i=0}^{k_2}\left[\left(c_i^Tx-d_i\right)t^i\right]=0$$ where $\lambda$ is not an integer, can we form a similar system of equations? For example, $$\left[\begin{matrix} A \\ C \end{matrix}\right]x=\left[\begin{matrix} b \\ d \end{matrix}\right]$$ where the rows of $C$ are the vectors $c_i^T$ and the components of $d$ are the values $d_i$, $\forall i \in \{0,\dots, k_2\}$?

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    $\begingroup$ so a Piseux series with infinitely many coefficients 0 ? $\endgroup$ – Roddy MacPhee Apr 18 at 19:26
  • $\begingroup$ That's correct. $\endgroup$ – ApPs Apr 19 at 20:24

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