The discussion revolves around the derivative of the Bessel function of the second kind, specifically the expression \(\frac{d}{dx}K_v(f(x))\). Participants explore the application of the chain rule and related identities in the context of Bessel functions, addressing both theoretical and practical aspects of differentiation with more complex arguments.
Participants express varying levels of understanding and confidence regarding the derivative of the Bessel function with complex arguments. There is no clear consensus on the final expression or method, and some participants remain uncertain about the correctness of their approaches.
Participants reference identities and methods that may depend on specific definitions or assumptions about the functions involved. The discussion includes unresolved mathematical steps and varying interpretations of how to apply the chain rule in this context.
Mute said:For [itex]\nu[/itex] not necessarily an integer, [itex]C_{\nu}(y) = e^{\nu \pi i}K_{\nu}(y)[/itex] satisfies the identity
[tex]2\frac{dC_\nu}{dy} = C_{\nu-1}(y) + C_{\nu + 1}(y)[/tex]
Then let [itex]y = f(x)[/itex] and use the chain rule.
Identities like this can usually be found on wikipedia, and for something as studied and used as frequently as the Bessel Functions, are generally correct.
http://en.wikipedia.org/wiki/Bessel_function
For a more 'official' reference, see something like http://www.math.sfu.ca/~cbm/aands/page_437.htm (scans of a reference book).
S_David said:Hello,
What is [tex]\frac{d}{dx}K_v\left(f(x)\right)=?[/tex]
Thanks in advance
skyspeed said:Do you get the answer?
S_David said:I know that
[tex]z\frac{d}{dz}K_v(z)+vK_v(z)=-zK_{v-1}(z)[/tex]
but I was confused when we have more complicated arguments such as
[tex]z=\sqrt{x^2+x}[/tex].
But after your posting, I have now a simple method to move from simple to more complicated arguments. So, I can say the following:
[tex]\frac{f(z)}{f'(z)}\frac{d}{dz}K_v(f(z))+vK_v(f(z))=-f(z)K_{v-1}(f(z))[/tex]
Am I right?
Best regards
S_David said:let [tex]y=f(x)[/tex] and then use the chain rule.