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Following the question I asked here and is:

Let $P(\lambda)=(\lambda-\lambda_{0})^{r}$where $r$ is a positive integer. Prove that the equation $P(\frac{d}{dt})x(t)=0$ has solutions $t^{i}e^{\lambda_{0}t},i=0,1,\ldots,r-1$

I now wish to prove the solutions are linearly independent.

I have two questions regarding this:

  1. I learned to prove such independence with the Wronskian, but I am having trouble calculating it in (I calculated the derivatives of $e^{\lambda_{0}t},te^{\lambda_{0}t}$ but its getting too hard when it is a greater power of $t$ since I am getting longer and longer expressions). How can I calculate the Wronskian ?

  2. If I think of the vector space that is the smooth real valued functions then it seems that this set (if I take the power of $t$ to be as big as I want, but finite) is linearly independent. did I deduce right ?

I would appriciate any help!

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    I was unsure what other tags to add, feel free to add tags if you think it is related.2012-06-16

2 Answers 2

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To show that $t^i e^{\lambda_0t}$ are linearly independent, it suffices to show that $t^i$ are linearly independent.

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    @Begli if there is a point where the function which is a factor of all the terms you are working with is zero and you are trying to do a proof of linear independence, then work away from that point. In practice this has never caused me a problem with the proofs. The thing to remember is that functions are different if they differ at a single point. As you say, this is not a problem here.2012-06-16
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Since you have some doubts, I'll try to give you a longer answers and maybe clear them.

The assertion that the $n$ functions $f_k(t)=t^k e^{\lambda_0 t} \text{ ; }{k=0,1,\dots,n-1}$ are linearly independent is that if

$\sum_{k=0}^{n-1} c_k f_k(t)= e^{\lambda_0 t}(c_0+c_1 t +c_2 t^2+\cdots+c_{n-1} t^{n-1})=0$

then

$c_0=c_1=\cdots=c_{n-1}=0$

Since $e^{\lambda_0 t}\neq 0$ for any $t$, it suffices to prove that if

$c_0+c_1 t +c_2 t^2+\cdots+c_{n-1} t^{n-1}=0$

then $c_0=c_1=\cdots=c_{n-1}=0$ or that the Wronskian determinant of the $n$ functions, $p_k(t)=t^k \text{ ; }{k=0,1,\dots,n-1}$ is never zero $0$.

For example, for the case $n=3$, we have the functions

$y_0(t)=c_0$ $y_1(t)=c_1 t$ $y_2(t)=c_2 t^2$

The Wronskian determinant is

$W(y_1,y_3,y_3)=\begin{vmatrix} {1}& { t} &{ t^2} \\ {0}& {1} &{2 t} \\ {0}& {0} &{2 } \end{vmatrix}=2 \cdot 1 \cdot 1 = 2! 1!$

since all other combinations will occurr with a $0$.

You can try and prove the Wronskian determinant

$\begin{vmatrix} 1 & t & \cdots & {{t^{n - 2}}} & {{t^{n - 1}}} \\ 0 & 1 & \cdots & {\left( {n - 2} \right){t^{n - 3}}} & {\left( {n - 1} \right)} \\ \vdots & \vdots & \vdots & \vdots & \vdots \\ 0 & 0 & 0 & {\left( {n - 2} \right)!} & {\left( {n - 1} \right)!t} \\ 0 & 0 & 0 & 0 & {\left( {n - 1} \right)!} \end{vmatrix}$

will be equal to $1! 2! 3! \cdots (n-1)!$ so it cannot be zero.

Alternatively, one can prove the result by induction. You can assume the result is proven for $1,2,\dots, n-2$ and show the result is true for $n-1$. I think it is much easier to use the Wronskian determinant.

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    @Artem Sure. The OP explicitly asked for the use of the Wronskian, so I used that.2012-06-16