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Linear independence of $e^{at}$ has been answered multiple times. My favorite one is by Marc van Leeuwen in this one: Proof of linear independence of $e^{at}$. The answer uses the property that $e^{at}$ are eigenfunctions of the differentation operation. Now if we instead have an exponential function in two variables, say $e^{ax + by^2}$, it seems to me the linear independence of these functions can too be proved with the same technique, except now using the partial derivative operator:

Proof by induction as in the link except here $e^{a_1x + b_1y^2}$, $e^{a_2x + b_2y^2}$, $...$,$e^{a_{n-1}x + b_{n-1}y^2}$ are assumed linearly independent. As in Marc's proof, we then assume that $e^{a_1x + b_1y^2}$, $e^{a_2x + b_2y^2}$, $...$,$e^{a_{n}x + b_{n}y^2}$ are in turn dependent and thus have:

$e^{a_{n}x + b_{n}y^2}= c_1e^{a_1x + b_1y^2} + c_2e^{a_2x + b_2y^2} + ... + c_{n-1}e^{a_{n-1}x + b_{n-1}y^2}$. Applying operator $\frac{\partial}{\partial x} - a_nI$ will give $0= c_1(a_1-a_n)e^{a_1x + b_1y^2} + c_2(a_2-a_n)e^{a_2x + b_2y^2} + ... + c_{n-1}(a_{n-1}-a_n)e^{a_{n-1}x + b_{n-1}y^2}$ which would thus require all c to be zero, which essentially completes the induction proof (other argumentation as per link).

My question is: why does $y^2$ not seem to have any effect on the linear independence. Where does this stem from?

edit: assume all $a_k$ and $b_k$ are distinct.

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  • $\begingroup$ You have to take in consideration that the linear independence you present, is towards variable $x$, judging from the operator you are using. A 2 variable function $f(x,y)$ would need a 2-dimension operator. So what your assumption is that $e^{ax+bx^2}$ is linear towards $x$ which is true (its like considering $bx^2$ a constant) $\endgroup$ – Pookaros May 30 at 16:12
  • $\begingroup$ I suspected something like this (assume you meant $by^2$). I guess $\frac{\partial }{\partial x \partial y} $is then what youd sugggest $\endgroup$ – Lulu May 30 at 16:16
  • $\begingroup$ also, despite seeing your point on intuitive level, I fail to see the logic fully since I dont see which part of my argumentation fails in my question. After all $a_1f_1 + a_2f_2 + ... + a_nf_n = 0$ implies linear independence if all a_i are 0, no matter what the f_1 are (no matter how many variables e.g. second answer here mathhelpforum.com/calculus/…) $\endgroup$ – Lulu May 30 at 16:22
  • $\begingroup$ Note that your functions are products: $e^{a_jx} e^{b_jy^2}$. If a linear combination $\sum c_j e^{a_jx} e^{b_jy^2}$ is zero then by fixing $y$ it follows that a linear combination of the $e^{a_jx}$ is zero. $\endgroup$ – Martin R May 30 at 16:28
  • $\begingroup$ Good point. Am I right saying that then it would suffice to prove separately for each variable i.e. using partial differentiation wrt x and y each at turn $\endgroup$ – Lulu May 30 at 16:42
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Your proof essentially repeats the proof that the functions $e^{a_j x}$ are linearly independent. The “$y$-terms” have “no effect” because they occur as (non-zero) factors $e^{b_j y^2}$ which are constant with respect to $x$.

What you observed is this: If $g_1, \ldots, g_n : A \to \Bbb R$ and $h_1, \ldots, h_n : B \to \Bbb R$ are functions such that

  • $g_1, \ldots, g_n$ are linearly independent, and
  • there is a $y_0 \in B$ such that $h_j(y_0) \ne 0$ for all $j$,

then the functions $f_j : A \times B \to \Bbb R$, $f_j(x, y) = g_j(x) h_j(y)$, $j=1, \ldots, n$, are linearly independent.

The proof is straight forward: If $c_1, \ldots, c_n \in \Bbb R$ with $$ \sum_{j=1}^n c_j g_j(x) h_j(y) = 0 \text{ for all } (x, y) \in A \times B $$ then in particular $$ \sum_{j=1}^n \bigl( c_j h_j(y_0) \bigr) g_j(x) = 0 \text{ for all } x \in A $$ Since the $g_j$ are linearly independent it follows that $$ c_j h_j(y_0) = 0\text{ for } j = 1, \ldots, n \implies c_j = 0 \text{ for } j = 1, \ldots, n \,. $$

In your case $g_j(x) = e^{a_j x}$ and $h_j(y) = e^{b_j y^2}$.

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  • $\begingroup$ Ah amazing! So actually my proof was then "correct" (which I thought it wasnt based on Pookaros' comments above). I suppose his/her comment would be valid if x and y had a more complex relationship? $\endgroup$ – Lulu May 30 at 19:40

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