have an index that runs through all values of nu found in the
expression, including perhaps symbolic ones.
for each value n,
divide the expression by b(n-1,z)+b(n+1,z), and get quotient Q and
remainder R.
change the expression to
2*n/z*b(n,z)*Q +R.
This is equivalent, by the identity below.
Go on to the next value of n.
This could also be done by using ratsubst, but you would still have
to sequence through all the expressions b(n-1,x)+b(n+1,x)..
for all n (and all x).
You would have to think about how to make sure this terminates if
you produce new values of nu and want to reduce those, too.
RJF
Billinghurst, David (CALCRTS) wrote:
>>> bessel_j(nu-1,z)+bessel_j(nu+1,z) = (2*nu/z) * bessel_j(nu,z)
>>>
>>>
>>...
>>
>>
>>>After substituting a solution into a differential equation
>>>I have the following expression that should equal zero....
>>>
>>>
>>I don't know of any straightforward way to do this using patterns, or
>>any other way to do it automatically (automagically as we say), but
>>here's the approach I would take. For simplicity, I will use b(n) as a
>>shorthand for bessel_j(n,x^2/2). By the way, in my version of Maxima
>>(5.9.0), the arguments to bessel_j are in the other order! anyway....
>>
>>
>
>Thanks for the ideas. I will investigate further.
>
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