# Why can linear independence of $e^{ax +by^2}$ be proven without considering $y$?

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.

• 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) – Pookaros May 30 at 16:12
• I suspected something like this (assume you meant $by^2$). I guess $\frac{\partial }{\partial x \partial y}$is then what youd sugggest – Lulu May 30 at 16:16
• 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/…) – Lulu May 30 at 16:22
• 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. – Martin R May 30 at 16:28
• 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 – Lulu May 30 at 16:42

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 $$h_j(y) = e^{b_j y^2}$$.

• 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? – Lulu May 30 at 19:40