Derivatives of Bessel functions?
1940L...@gmail.com
find in print or on the internet. Now, I've seen many times that J' 0
(x) {that is, the first derivative of the Bessel function of the first
kind of order zero) is equal to -J 1 (x). Am I correct that:
J' 0 (ax) = -aJ 1(ax),
where a is a constant?
Thanks.
David Moran
<1940L...@gmail.com> wrote in message
news:1150132059.3...@j55g2000cwa.googlegroups.com...
I am pretty sure that this is right, but I am not 100% sure. I haven't seen
Bessel functions since I got my degree in math.
Dave
G. A. Edgar
<1940L...@gmail.com> wrote:
Go to http://functions.wolfram.com/ and follow your nose... I got this:
<http://functions.wolfram.com/BesselAiryStruveFunctions/BesselJ/20/01/02
/>
and yours is the 4th equation listed.
For the version with constant a, you need the chain rule.
[Someone writing that table assumed the reader would know the chain
rule, so he did not put in the more general version...]
--
G. A. Edgar http://www.math.ohio-state.edu/~edgar/
Ronald Bruck
If, for the LHS, you meant d/dx J_0(ax), then of course; that's the
chain rule.
But technically, J_0'(ax) is -J_1(ax). Assuming J_0' = -J_1, which
I'm pretty sure I remember, too.
--Ron Bruck
Dave L. Renfro
I tried some obvious searches, such as
http://www.google.com/search?q=derivatives+Bessel-functions
http://books.google.com/books?q=derivatives-of-Bessel-functions
and came up with quite a bit, in particular the derivative
formula you asked about. For example, see
"Special Functions and Their Applications
by N N Lebedev", Dover Publications, 1972,
p. 100, eq. 5.2.9. [Use "Search within this
book" = "Derivatives of Bessel functions"]
http://books.google.com/books?vid=ISBN0486606244
Dave L. Renfro
Dave L. Renfro
>> I'd like to verify something I believe is true but
>> haven't been able to find in print or on the internet.
>> Now, I've seen many times that J' 0 (x) {that is, the
>> first derivative of the Bessel function of the first
>> kind of order zero) is equal to -J 1 (x). Am I correct that:
>>
>> J' 0 (ax) = -aJ 1(ax),
>>
>> where a is a constant?
Dave L. Renfro wrote:
> I tried some obvious searches, such as
>
> http://www.google.com/search?q=derivatives+Bessel-functions>
>
> http://books.google.com/books?q=derivatives-of-Bessel-functions
>
> and came up with quite a bit, in particular the derivative
> formula you asked about. For example, see
>
> "Special Functions and Their Applications
> by N N Lebedev", Dover Publications, 1972,
> p. 100, eq. 5.2.9. [Use "Search within this
> book" = "Derivatives of Bessel functions"]
> http://books.google.com/books?vid=ISBN0486606244
Re-reading your post, it appears that you're asking how
to get from J'0(x) = -J1(x) to J'0(ax) = -a*J1(ax).
Maybe this isn't what you intended to write (because
anyone studying Bessel functions should know this
from a first semester calculus course), but in case
this is really what you were asking, use the chain rule.
Dave L. Renfro
eugene
J'0 (ax) = - J1 (ax), and if it were true it must have been -aJ1(ax) = -J1 (ax) and a = 1. So the only case when you this identity is true is a = 1.
David C. Ullrich
What you meant to ask is true - what you
wrote is wrong.
The derivative of the function J0(ax) is
indeed -aJ1(ax). But what you wrote was
not the derivative of the function J0(ax),
it was the derivative of J0, evaluated at ax.
The value of _that_ is -J1(ax).
Anyway, find a calculus book and read up on the chain rule.
>Thanks.
************************
David C. Ullrich
1940L...@gmail.com
functions as it does to more commonly-encountered functions.
Gene Ward Smith
1940L...@gmail.com wrote:
> Very good: I wasn't clear whether the chain rule applied to Bessel
> functions as it does to more commonly-encountered functions.
That's why they have these general theorems instead of just having a
separate rule for each function.
Lee Rudolph
And that is why mathematicians are such lusers. Clearly, if each
function had to obtain its own license for, e.g., chain-rule
applicability, *someone* would be raking it in. But, oh no,
it's come-one-come-all, this is Liberty Hall. Bah, humbug.
Lee Rudolph
David C. Ullrich
>Very good: I wasn't clear whether the chain rule applied to Bessel
>functions as it does to more commonly-encountered functions.
??? What does the chain rule say?
I don't mean to ask what the formula is. What are the
_hypotheses_ in the chain rule?
>David C. Ullrich wrote:
>> On 12 Jun 2006 10:07:39 -0700, 1940L...@gmail.com wrote:
>>
>> >I'd like to verify something I believe is true but haven't been able to
>> >find in print or on the internet. Now, I've seen many times that J' 0
>> >(x) {that is, the first derivative of the Bessel function of the first
>> >kind of order zero) is equal to -J 1 (x). Am I correct that:
>> >
>> >J' 0 (ax) = -aJ 1(ax),
>> >
>> >where a is a constant?
>>
>> What you meant to ask is true - what you
>> wrote is wrong.
>>
>> The derivative of the function J0(ax) is
>> indeed -aJ1(ax). But what you wrote was
>> not the derivative of the function J0(ax),
>> it was the derivative of J0, evaluated at ax.
>> The value of _that_ is -J1(ax).
>>
>> Anyway, find a calculus book and read up on the chain rule.
>>
>> >Thanks.
>>
>>
>> ************************
>>
>> David C. Ullrich
************************
David C. Ullrich
Jorge Pais
The only approach I found is the derivative of J1(x) which is J0(x)-1/x*J1(x)
Thanks
Jorge
Alois Steindl
> Based on the post where the derivative of J0(ax) is -aJ1(ax), I am
> asking the derivative of J1(ax).
Was there an answer?
>
> The only approach I found is the derivative of J1(x) which is J0(x)-1/x*J1(x)
>
> Thanks
> Jorge
Ever learned the chain rule?
Alois
Jorge Pais
> asking the derivative of J1(ax).
Was there an answer?
Yes, there is an answer.