MathGroup Archive 2000

[Date Index] [Thread Index] [Author Index]

Search the Archive

Re: sum of recursive fn: solving for n

  • To: mathgroup at smc.vnet.net
  • Subject: [mg22123] Re: sum of recursive fn: solving for n
  • From: "Robert Nowak" <robert.nowak at ims.co.at>
  • Date: Wed, 16 Feb 2000 02:34:39 -0500 (EST)
  • Organization: telecom.at (Vienna, Austria)
  • References: <888afd$c7p@smc.vnet.net>
  • Sender: owner-wri-mathgroup at wolfram.com

your function f[x_] seems to be 2^x so the following will solve your problem

In[1]:=
f[x_] = 2^x;

In[2]:=
g[n] = Sum[f[x], {x, 1, n}]

Out[2]=
2*(-1 + 2^n)

In[3]:=
Solve[g[n] == 62, n]

Solve::"ifun" : "Inverse functions are being used by \!\(Solve\), so some \
solutions may not be found."

Out[3]=
{{n -> 5}}


hope that helps
Robert

--
---
Robert Nowak (robert.nowak at ims.co.at)
Ionen Mikrofabrikations Systeme GmbH
A-1020 Wien, Schreygasse 3, Austria
Phone: (+43 1)2144894-32, Fax: (+43 1)2144894-99

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
>




  • Prev by Date: Re: sum of recursive fn: solving for n
  • Next by Date: Graphics and animation
  • Previous by thread: Re: sum of recursive fn: solving for n
  • Next by thread: Re: sum of recursive fn: solving for n