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From: Sam on 5 Jan 2010 01:47
I am using mathematica 7, and am trying to solve a recurrence relation
using the code below:

RSolve[{f[x] == Sum[f[i]*f[x - i], {i, 1, x - 1}], f[1] == 1}, f[x],
x]

but it gives me the RSolve::piarg error. I have found that the
solution for the above problem is in fact

f[x_] = Product[(4*i - 6)/i, {i, 2, x}]

so it is theoretically solveable, but doing this sort of thing by hand
is extremely error prone. Is there any way of solving this type of
problem with mathematica or is it necessary to do it by hand?

From: dh on 6 Jan 2010 06:00


Hi Sam,

in fact, Mathematica can solve your equation. Straight from the manual:



Nonlinear convolution equation:

RSolve[{a[n + 1] == Sum[a[m] a[n - m], {m, 0, n}], a[0] == 1}, a[n],

n]



Note that counting starts at zero here.

However, Mathematica fails if you start counting from 1 instead of 0:



RSolve[{a[n + 1] == Sum[a[m] a[n - m + 1], {m, 1, n}], a[1] == 1},

a[n], n]



this produces your error message.



I think this is a bug in inpout section of RSolve. Please report this to

Wolfram.



Daniel



Sam wrote:

> I am using mathematica 7, and am trying to solve a recurrence relation

> using the code below:

>

> RSolve[{f[x] == Sum[f[i]*f[x - i], {i, 1, x - 1}], f[1] == 1}, f[x],

> x]

>

> but it gives me the RSolve::piarg error. I have found that the

> solution for the above problem is in fact

>

> f[x_] = Product[(4*i - 6)/i, {i, 2, x}]

>

> so it is theoretically solveable, but doing this sort of thing by hand

> is extremely error prone. Is there any way of solving this type of

> problem with mathematica or is it necessary to do it by hand?

>



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