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Given $p_1 = 40+ 7x + 51x^3$, $p_2=8+x+24x^2+7x^3$, $p_3 = 8x + 4x^3$, $p_4 = 24+3x+21x^3$, find the value of the wronskian of $\{p_1, p_2, p_3, p_4\}$ at $x=0$.

Isn't the Wronskian at $x=0$ just $0$?

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  • $\begingroup$ This doesn't make sense. Don't you need two functions at least to calculate $fg' - gf'$? $\endgroup$ – John Dec 9 '13 at 5:55
  • $\begingroup$ Eh? I thought the Wronskian was the matrix of f^(n-1) derivatives $\endgroup$ – user113027 Dec 9 '13 at 5:59
  • $\begingroup$ @user113027: See Wronskian. I get a constant $829440$. It is possible for it to be a function, in which case you would put $x=0$, but constant for these functions. $\endgroup$ – Amzoti Dec 9 '13 at 6:06
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We have as the Wronskian:

$\begin{vmatrix} f_1 & f_2 & f_3 & f_4 \\ f_1' & f_2' & f_3' & f_4' \\ f_1'' & f_2'' & f_3'' & f_4'' \\ f_1''' & f_2''' & f_3''' & f_4''' \end{vmatrix} = \begin{vmatrix} 40 + 7 x + 51 x^3 & 8 + x + 24 x^2 + 7 x^3 & 8 x + 4 x^3 & 24 + 3 x + 21 x^3\\ 7 + 153 x^2 & 1 + 48 x + 21 x^2 & 8 + 12 x^2 & 3 + 63 x^2 \\ 306 x & 48 + 42 x & 24 x & 126 x \\ 306 & 42 & 24 & 126 \end{vmatrix}$

The Wronskian is:

$$829440$$

In this case, it is just a constant, but can be a function of $x$ depending on the functions.

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