tukeywilliam Posted March 19, 2007 Share Posted March 19, 2007 I know that a set of vector functions [math]{\vec{v_{1}}(t), \vec{v_{2}}(t)+...+\vec{v_{n}}(t)}[/math] in a vector space [math]\mathbb{V}[/math] if [math]c_{i}= 0 [/math] for the following equation: [math]c_{1}\vec{v_{1}}(t)+c_{2}\vec{v_{2}}(t)+...+c_{n}\vec{v_{n}}(t) \equiv \vec{0}[/math] Where does the Wronskian come into play? Is it basically a determinant with functions and derivatives? Thanks Link to comment Share on other sites More sharing options...
CPL.Luke Posted March 19, 2007 Share Posted March 19, 2007 thats only if the functions are linearly independant. by taking the determinant (wronskian) you can find out whether or not they are linearly independant. Link to comment Share on other sites More sharing options...
tukeywilliam Posted March 19, 2007 Author Share Posted March 19, 2007 But why does the determinant involve taking derivatives? Link to comment Share on other sites More sharing options...
CPL.Luke Posted March 21, 2007 Share Posted March 21, 2007 is just a way of filling up the matrix, just think about what conditions you would have to impose for the derivatives of your functions to be linear combinations of the other derivatives of your function. ie. for one of the rows of the wronskian to be a linear combination of the others. Link to comment Share on other sites More sharing options...
Bignose Posted March 21, 2007 Share Posted March 21, 2007 You probably want to look up "Why Does the Wronskian Work?" by Mark Krusemeyer. The American Mathematical Monthly Vol 95, pp. 46-49, 1988. Link to comment Share on other sites More sharing options...
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