# Prove that $e^x, xe^x,$ and $x^2e^x$ are linearly independent over $\mathbb{R}$

Question: Prove that $e^x, xe^x,$ and $x^2e^x$ are linearly independent over $\mathbb{R}$.

Generally we proceed by setting up the equation $$a_1e^x + a_2xe^x+a_3x^2e^x=0_f,$$ which simplifies to $$e^x(a_1+a_2x+a_3x^2)=0_f,$$ and furthermore to $$a_1+a_2x+a_3x^2=0_f.$$

From here I think it's obvious that the only choice to make the sum the zero function is to let each scalar equal 0, but this is very weak reasoning.

As an undergraduate we learned to test for independence by determining whether the Wronskian is not identically equal to 0. But I can only use this method if the functions are solutions to the same linear homogeneous differential equation of order 3. In other words, I cannot use this method for an arbitrary set of functions. I was not given a differential equation, so I determined it on my own and got that they satisfy $$y'''-3y''+3y'-y = 0.$$

I found the Wronskian, $2e^{3x}\neq0$ for any real number. Thus the set is linearly independent. But it took me some time to find the differential equation and even longer finding the Wronskian so I'm wondering if there is a stronger way to prove this without using the Wronskian Test for Independence.

• The Wronskian works for your functions as long as the functions are $n–1$ times differentiable. It is not necessary to construct the differential equation. – Zuriel Dec 30 '16 at 5:42
• @Zuriel: My text does not say that. So is your statement equivalent to saying that the functions satisfy a linear ODE of order $n$? – SOULed_Outt Dec 30 '16 at 5:52
• Not exactly. Each one of them must be $n-1$ times differentiable (if there are $n$ functions). This is not the same as saying that they satisfy some linear ODE of order $n$ (For example, one function may be missing in the ODE). – Zuriel Dec 30 '16 at 6:01
• @Zuriel: I'm sorry. I should have been more detailed in my question. In order to use Wronskian test for independence, the functions are solutions to the same linear homogeneous differential equation of order n. But to find the Wronskian the functions must be $n–1$ times differentiable – SOULed_Outt Dec 30 '16 at 6:06
• You are right. For general functions, "a common misconception is that $W = 0$ everywhere implies linear dependence." – Zuriel Dec 30 '16 at 6:09

Setting $x = 0$ in the equation $a_1 + a_2x + a_3x^2 = 0$ results in $a_1 = 0$. Then $a_2x + a_3x^2 = 0$ for all $x\in \Bbb R$. Setting $x = 1$ gives $a_2 + a_3 = 0$, and setting $x = -1$ gives $-a_2 + a_3 = 0$. Solving the system of equations will yield $a_2 = a_3 = 0$.

• Did you even read the question? – Maffred Dec 30 '16 at 5:46
• @Maffred in the prompt, it states "...I'm wondering if there is a stronger way to prove this without using the Wronskian." I have provided such a way. – kobe Dec 30 '16 at 5:58

Suppose $a_1e^x + a_2xe^x+a_3x^2e^x=0$ for all $x$.

Setting $x=0$ shows that $a_1 = 0$. Now note that $a_2xe^x+a_3x^2e^x=0$ for all $x$ and hence $a_2e^x+a_3xe^x=0$ for all $x \neq 0$. Taking limits as $x \to 0$ shows that $a_2 = 0$, and setting $x=1$ shows that $a_3 = 0$.

• Did you even read the question? – Maffred Dec 30 '16 at 5:45
• @Maffred: I did. What is your point? The OP was looking for a way to prove this without using the Wronskian. Did you read the question? – copper.hat Dec 30 '16 at 5:47

Though it is overkill and not really in the spirit of the problem (see kobe's answer for the most elementary approach), you can use the fundamental theorem of algebra (or the weaker result that a real polynomial of degree $n>1$ has at most $n$ roots) on your reduced equation to conclude that $a_3$ and $a_2$ must be zero (otherwise you'd have a positive degree polynomial with infinitely many roots). Of course $a_1$ must be zero, too: this follows just from setting $x=0$.