The Derivative of Bessel Function of the Second Kind

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  • #1
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Hello,

What is [tex]\frac{d}{dx}K_v\left(f(x)\right)=?[/tex]

Thanks in advance
 

Answers and Replies

  • #2
Mute
Homework Helper
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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).
 
  • #3
1,367
61
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).

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
 
  • #4
3
0
Hello,

What is [tex]\frac{d}{dx}K_v\left(f(x)\right)=?[/tex]

Thanks in advance

Do you get the answer?
 
  • #5
1,367
61
Do you get the answer?

let [tex]y=f(x)[/tex] and then use the chain rule.
 
  • #6
3
0
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

i think this is right
 
  • #7
3
0
let [tex]y=f(x)[/tex] and then use the chain rule.

thanks a lot
 

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