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Revision difference : Wronskian determinant
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Given functions $f_1, f_2, \dotsc, f_n$, then the \emph{Wronskian determinant} (or simply the Wronskian) $W(f_1, f_2, f_3, \dotsc, f_n)$ is the determinant of the square matrix Given functions $f_1, f_2, \dotsc, f_n$, then the \emph{Wronskian determinant} (or simply the Wronskian) $W(f_1, f_2, f_3, \dotsc, f_n)$ is the determinant of the square matrix
\[ \[
W(f_1, f_2, f_3, \dotsc, f_n) = \left\lvert\begin{array}{@{}ccccc@{}} W(f_1, f_2, f_3, \dotsc, f_n) = \left\lvert\begin{array}{@{}ccccc@{}}
f_1 & f_2 & f_3 & \cdots & f_n\\ f_1 & f_2 & f_3 & \cdots & f_n\\
f_1' & f_2' & f_3' & \cdots & f_n'\\ f_1' & f_2' & f_3' & \cdots & f_n'\\
f_1'' & f_2'' & f_3'' & \cdots & f_n''\\ f_1'' & f_2'' & f_3'' & \cdots & f_n''\\
\vdots & \vdots & \vdots & \ddots & \vdots\\ \vdots & \vdots & \vdots & \ddots & \vdots\\
f_1^{(n-1)} & f_2^{(n-1)} & f_3^{(n-1)} & \cdots & f_n^{(n-1)}\\ f_1^{(n-1)} & f_2^{(n-1)} & f_3^{(n-1)} & \cdots & f_n^{(n-1)}\\
\end{array}\right\rvert \end{array}\right\rvert
\] \]
where $f^{(k)}$ indicates the $k$th derivative of $f$ (not exponentiation). where $f^{(k)}$ indicates the $k$th derivative of $f$ (not exponentiation).
The Wronskian of a set of functions $F$ is another function, which is zero over any interval where $F$ is linearly dependent. Just as a set of vectors is said to be linearly dependent when there exists a non-trivial linear relation between them, a set of functions $\{f_1, f_2, f_3, \dotsc, f_n\}$ is also said to be dependent over an interval $I$ when there exists a non-trivial linear relation between them, i.e., The Wronskian of a set of functions $F$ is another function, which is zero over any interval where $F$ is linearly dependent. Just as a set of vectors is said to be linearly dependent when there exists a non-trivial linear relation between them, a set of functions $\{f_1, f_2, f_3, \dotsc, f_n\}$ is also said to be dependent over an interval $I$ when there exists a non-trivial linear relation between them, i.e.,
\[ \[
a_1 f_1(t) + a_2 f_2(t) + \dotsb + a_n f_n(t) = 0 a_1 f_1(t) + a_2 f_2(t) + \dotsb + a_n f_n(t) = 0
\] \]
for some $a_1, a_2, \dotsc, a_n$, not all zero, at any $t \in I$. for some $a_1, a_2, \dotsc, a_n$, not all zero, at any $t \in I$.
Therefore the Wronskian can be used to determine if functions are independent. This is useful in many situations. For example, if we wish to determine if two solutions of a second-order differential equation are independent, we may use the Wronskian. Therefore the Wronskian can be used to determine if functions are independent. This is useful in many situations. For example, if we wish to determine if two solutions of a second-order differential equation are independent, we may use the Wronskian.
\paragraph{Examples} \paragraph{Examples}
Consider the functions $x^2$, $x$, and $1$. Take the Wronskian: Consider the functions $x^2$, $x$, and $1$. Take the Wronskian:
\[ \[
W = \left\lvert\begin{array}{@{}ccc@{}} W = \left\lvert\begin{array}{@{}ccc@{}}
x^2 & x & 1\\ x^2 & x & 1\\
2x & 1 & 0\\ 2x & 1 & 0\\
2 & 0 & 0\\ 2 & 0 & 0\\
\end{array}\right\rvert = -2 \end{array}\right\rvert = -2
\] \]
Note that $W$ is always non-zero, so these functions are independent everywhere. Consider, however, $x^2$ and $x$: Note that $W$ is always non-zero, so these functions are independent everywhere. Consider, however, $x^2$ and $x$:
\[ \[
W = \left\lvert\begin{array}{@{}cc@{}} W = \left\lvert\begin{array}{@{}cc@{}}
x^2 & x\\ x^2 & x\\
2x & 1\\ 2x & 1\\
\end{array}\right\rvert = x^2 - 2x^2 = -x^2 \end{array}\right\rvert = x^2 - 2x^2 = -x^2
\] \]
Here $W = 0$ only when $x = 0$. Therefore $x^2$ and $x$ are independent except at $x = 0$. Here $W = 0$ only when $x = 0$. Therefore $x^2$ and $x$ are independent except at $x = 0$.
Consider $2x^2+3$, $x^2$, and $1$: Consider $2x^2+3$, $x^2$, and $1$:
\[ \[
W = \left\lvert\begin{array}{@{}ccc@{}} W = \left\lvert\begin{array}{@{}ccc@{}}
2x^2 + 3 & x^2 & 1\\ 2x^2 + 3 & x^2 & 1\\
4x & 2x & 0\\ 4x & 2x & 0\\
4 & 2 & 0\\ 4 & 2 & 0\\
\end{array}\right\rvert = 8x - 8x = 0 \end{array}\right\rvert = 8x - 8x = 0
\] \]
Here $W$ is always zero, so these functions are always dependent. This is intuitively obvious, of course, since Here $W$ is always zero, so these functions are always dependent. This is intuitively obvious, of course, since
\[ \[
2x^2 + 3 = 2(x^2) + 3(1) 2x^2 + 3 = 2(x^2) + 3(1)
\] \]
Given $n$ linearly independant functions $f_1, f_2, \dotsc, f_n$, we can use Given $n$ linearly independant functions $f_1, f_2, \dotsc, f_n$, we can use
the Wronskian to construct a linear differential equation whose solution space the Wronskian to construct a linear differential equation whose solution space
is exactly the span of these functions. Namely, if $g$ satisfies the equation is exactly the span of these functiosn. Namely, if $g$ satisfies the equation
\[ \[
W(f_1, f_2, f_3, \dotsc, f_n, g) = 0, W(f_1, f_2, f_3, \dotsc, f_n, g) = 0,
\] \]
then $g = a_1 f_1(t) + a_2 f_2(t) + \dotsb + a_n f_n(t)$ for some choice of then $g = a_1 f_1(t) + a_2 f_2(t) + \dotsb + a_n f_n(t)$ for some choice of
$a_1, a_2, \dotsc, a_n$. $a_1, a_2, \dotsc, a_n$.