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I was working on a problem and faced a this system of equations ($y_i$ and $x_i$ are givens)

$$ y_i = \sum_{j=0}^{n} c_j e^{jx_i} \quad0 \le i \le n$$

is there a way to determine this system is solvable or not?

Glyphack
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    Presumably, $y_i$ and $c_j$ are givens, and you are solving for $x_i?$ – Thomas Andrews Jun 12 '21 at 19:32
  • No $y_i$ and $x_i$ are givens, thank you for mentioning that. – Glyphack Jun 13 '21 at 05:42
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    Are the $x_i$ real? –  Jun 13 '21 at 06:06
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    If the $z_i=e^{x_i}$ are distinct (which is different from $x_i$ distinct, in the case of complex numbers.) This is because the Vanermonde matrix is invertible unless the $z_i$ repeat. https://en.wikipedia.org/wiki/Vandermonde_matrix – Thomas Andrews Jun 13 '21 at 06:06
  • @Gae.S. x_i are real, I was reading your comment. so I see the first $p(z)$ you mentioned does exist and it's the interpolation polynomial but about the second part why is it true? – Glyphack Jun 13 '21 at 06:07
  • @ThomasAndrews Thanks but aren't the elements in Vandermonde matrix the form of $x^i$ ? here I have $e^{x_i}$ I could not see how to write it in that form. – Glyphack Jun 13 '21 at 06:11
  • @Glyphack The comment I've deleted was just wrong. –  Jun 13 '21 at 06:16
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    @Gae.S. it is the Vandermonde matrix of $z_i=e^{x_i}$. – Thomas Andrews Jun 13 '21 at 06:17

2 Answers2

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This is just a Vandermonde matrix (https://en.wikipedia.org/wiki/Vandermonde_matrix). It is invertible if the x_i are distinct.

Bronzong
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You get exactly one solution if you have $y_i,x_i$ for $i=0,\dots,n$ (rather than $1,\dots,n$) and the values $z_i=e^{x_i}$ are distinct. (If the $x_i$ are real, this is equivalent to the $x_i$ being distinct.)

If you do have $n+1$ equations consider the $(n+1)\times(n+1)$ Vadermonde matrix:

$$V=(v_{ij})_{i,j=0}^n=(z_i^j)=(e^{jx_i})$$

Then you are seeking the solution to:

$$V\begin{pmatrix}c_0\\c_1\\\vdots\\c_n\end{pmatrix}=\begin{pmatrix}y_0\\y_1\\\vdots\\y_n\end{pmatrix}$$

But if the $z_i$ are distinct, then $V$ is invertible, and you get:

$$\begin{pmatrix}c_0\\c_1\\\vdots\\c_n\end{pmatrix}=V^{-1}\begin{pmatrix}y_0\\y_1\\\vdots\\y_n\end{pmatrix}$$

So, as your question is stated, given $y_i,x_i$ for $i=1,\dots,n$ with $e^{x_i}$ distinct, then you can get infinitely many solutions by picking any pair $x_0,y_0$ so that the $e^{x_0}$ is distinct from the other $e^{x_i}.$

Or, alternatively, you can find a unique solution where $c_n=0$ using $i=1,\dots,n.$

You can get a solution when the $z_i$ are not distinct, but only if $y_i=y_j$ whenever $z_i=z_j.$

Thomas Andrews
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