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Re: sum of recursive fn: solving for n
*To*: mathgroup at smc.vnet.net
*Subject*: [mg22169] Re: sum of recursive fn: solving for n
*From*: "Allan Hayes" <hay at haystack.demon.co.uk>
*Date*: Thu, 17 Feb 2000 01:23:58 -0500 (EST)
*References*: <888afd$c7p@smc.vnet.net>
*Sender*: owner-wri-mathgroup at wolfram.com
"fiona" <reply at newsgroup.please> wrote in message
news:888afd$c7p at smc.vnet.net...
>
> what am i doing wrong here?
>
> f[x_] := (f[x-1])*2
> f[1] =2
> Solve[Sum[f[x], {x, 1,n}] ==62, n]
>
> tia,
> fiona
>
Fiona:
Your doing nothing wrong, its just the way Mathematica works.
First, if we try to evaluate f[x], when x is not a posive integer, then we
never reach f[2] and we go on until stopped by the recursion linit.
f[x]
$RecursionLimit::"reclim": "Recursion depth of \!\(256\) exceeded."
5789604461865809771178549250434395392663499233282028201972879200395656481996
8 \
Hold[f[(-254 + x) - 1]]
Now, why do we get
Sum[f[x], {x, 1, n}]
$RecursionLimit::"reclim": "Recursion depth of \!\(256\) exceeded."
...........
\!\(\[Sum]\+\(x = 1\)\%n f[x]\)
This happens since {x,1,n} does not allow Mathematica to construct an
explicit sequence of values (not knowing what n is).
In these circumstances it evaluates f[x] , to val say, and tries to find a
formula for Sum[val,{x,1,n}]
So it tries find a formula for
Sum[578960446186580977117854925043439539266349923328202820197287920039565648
19\
968 Hold[f[(-254 + x) - 1]], {x, 1, n}]
It fails to find one, and so returns the original input.
Finally , Solve can't deal with this.
Her are three little experiments
1)
i = 1; While[Sum[f[x], {x, 1, i}] < 62, ++i]; i
5
Sum[f[x], {x, 1, 5}]
62
Clear[f, i]
2)
We can see that f[n] = 2^n (and we could prove it by induction)
Or we could use
<< DiscreteMath`RSolve`
RSolve[{f[i] == 2 f[i - 1], f[1] == 2}, f[i], i]
\!\({{f[i] -> 2\^i}}\)
Sum[2^y, {y, 1, n}] // InputForm
2*(-1 + 2^n)
And then use Solve
Solve[% == 62]
Solve::"ifun": "Inverse functions are being used by \!\(Solve\), so some \
solutions may not be found."
{{n -> 5}}
Clear[f, i]
3) Use
RSolve[{S[n + 1] == S[n] + f[n + 1], S[1] == f[1], f[n] == 2 f[n - 1],
f[1] == 2}, {f[n], S[n]}, n] // InputForm
{{f[n] -> 2^n, S[n] -> 2*(-1 + 2^n)}}
Solve[(S[n] /. %) == {62}, n]
Solve::"ifun": "Inverse functions are being used by \!\(Solve\), so some \
solutions may not be found."
{{n -> 5}}
Allan
---------------------
Allan Hayes
Mathematica Training and Consulting
Leicester UK
www.haystack.demon.co.uk
hay at haystack.demon.co.uk
Voice: +44 (0)116 271 4198
Fax: +44 (0)870 164 0565
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