Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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!

share|improve this question
    
I was unsure what other tags to add, feel free to add tags if you think it is related. –  Belgi Jun 16 '12 at 12:45

2 Answers 2

up vote 2 down vote accepted

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

share|improve this answer
    
Why, can you please add an explanation ? –  Belgi Jun 16 '12 at 12:53
    
Write the definition of "linearly independent". If you don't know it, look it up. Compare the two cases I mention. You may use the fact that $e^{\lambda_0 t} \ne 0$. –  GEdgar Jun 16 '12 at 13:00
    
We need to add that the vecotr space of all smooth real valued function have no zero divisors for this ? Edit: do you need it not to be the zero function, or that for all $t$ it doesn't equal to zero ? –  Belgi Jun 16 '12 at 13:02
    
"zero divisor" is not a concept for vector space. –  GEdgar Jun 16 '12 at 13:07
    
Can we say that such a conclusion is right even if there was $t$ such that $e^{\lambda_{0}t}=0$ ? maybe we can say that this is a ring with the pointwise multiplication and then add the argument about zero divisors ? –  Belgi Jun 16 '12 at 13:09

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.

share|improve this answer
    
The Wronskian is not required here actually, the linear independence follows just from the fact that the polynomial has no more than $n-1$ roots. –  Artem Jun 16 '12 at 16:45
    
@Artem What do you mean by "required"? The Wronskian is a fair argument, as valid as you argument. I think saying "just from the fact" is rather disrespectful to the Fundamental Theorem of Algebra. Now, if you can adjoin a proof of that, we'll be cool. =D –  Pedro Tamaroff Jun 16 '12 at 16:49
    
What I meant was that the notion of the Wronskian is far less "obvious" (known, etc) than the "common wisdom" fact about the roots of a polynomial. –  Artem Jun 16 '12 at 16:56
    
@Artem Sure. The OP explicitly asked for the use of the Wronskian, so I used that. –  Pedro Tamaroff Jun 16 '12 at 17:03

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.