matematicar73
New member
- Joined
- Dec 16, 2015
- Messages
- 8
Hi there,
I'm stuck with an example that probably is very simple but I can't find the way around it.
f(z,t) = sum_(-∞)^∞ [ t^(n) * J_(n) (z)] where J_(n) (z) is a Bessel function of order n.
I need to obtain a first-order differential function relating df/dz and f(z,t) and I have to deduce that f(z,t) = exp [z/2 * (t - 1/t)].
so far I know that df/dz = sum_(-∞)^∞ [t^(n) * J'_(n) (z)] and with the Bessel function property 2J'_(n) (z) = J_(n-1) (z) - J_(n+1) (z)
there is:
df/dz = sum_(-∞)^∞ [1/2(t^n * J_(n-1) (z) - t^n * J_(n+1) (z))
any help going [FONT=Helvetica Neue, Helvetica, Arial, sans-serif]further?
more given properties of Bessel functions are:
J_(n) (-z) = J_(-n) (z) = (-1)^n * J_(n) (z)
J_(n) (0) = 0 for all n greater than 0
J_(0) (0) = 1[/FONT]
I'm stuck with an example that probably is very simple but I can't find the way around it.
f(z,t) = sum_(-∞)^∞ [ t^(n) * J_(n) (z)] where J_(n) (z) is a Bessel function of order n.
I need to obtain a first-order differential function relating df/dz and f(z,t) and I have to deduce that f(z,t) = exp [z/2 * (t - 1/t)].
so far I know that df/dz = sum_(-∞)^∞ [t^(n) * J'_(n) (z)] and with the Bessel function property 2J'_(n) (z) = J_(n-1) (z) - J_(n+1) (z)
there is:
df/dz = sum_(-∞)^∞ [1/2(t^n * J_(n-1) (z) - t^n * J_(n+1) (z))
any help going [FONT=Helvetica Neue, Helvetica, Arial, sans-serif]further?
more given properties of Bessel functions are:
J_(n) (-z) = J_(-n) (z) = (-1)^n * J_(n) (z)
J_(n) (0) = 0 for all n greater than 0
J_(0) (0) = 1[/FONT]