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I'm confused about how this related to the roots of an equation. Here's the definition:

We say that ##f## has a zero of order ##m## at ##z_0## if

##f^{(k)}(z_0)=0## for ##k=0,\dots , m-1##, but ##f^{(m)}(z_0)\ne 0##

or equivalently that

##f(z) = \sum_{k=m}^\infty a_k(z-z_0)^k##, ##a_m \ne 0##.

For example the function ##(z^2+z-2)^3## has zeros of order ##3## for ##z=2## and ##z=-1## which is the multiple of the roots so the easiest way to determine the order seems to just note this instead of differentiating ##4## times.

However for a function like ##z^2(1-\cos z)## this doesn't work anymore. If you differentiate this you actually find a zero of order ##4## at ##z=0## and zeros of order ##2## at ##z = 2\pi n, \; = \pm 1, \pm 2, \dots## which isn't what I would expect just looking for roots to the equation.

So when determining the order of a zero is the approach always to differentiate? Since for simple functions like polynomials they seem to be the same as the roots.

We say that ##f## has a zero of order ##m## at ##z_0## if

##f^{(k)}(z_0)=0## for ##k=0,\dots , m-1##, but ##f^{(m)}(z_0)\ne 0##

or equivalently that

##f(z) = \sum_{k=m}^\infty a_k(z-z_0)^k##, ##a_m \ne 0##.

For example the function ##(z^2+z-2)^3## has zeros of order ##3## for ##z=2## and ##z=-1## which is the multiple of the roots so the easiest way to determine the order seems to just note this instead of differentiating ##4## times.

However for a function like ##z^2(1-\cos z)## this doesn't work anymore. If you differentiate this you actually find a zero of order ##4## at ##z=0## and zeros of order ##2## at ##z = 2\pi n, \; = \pm 1, \pm 2, \dots## which isn't what I would expect just looking for roots to the equation.

So when determining the order of a zero is the approach always to differentiate? Since for simple functions like polynomials they seem to be the same as the roots.

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