Re: Equation

*To*: mathgroup at smc.vnet.net*Subject*: [mg26634] Re: [mg26614] Equation*From*: Tomas Garza <tgarza01 at prodigy.net.mx>*Date*: Sat, 13 Jan 2001 22:36:13 -0500 (EST)*References*: <200101111539.KAA07047@smc.vnet.net>*Sender*: owner-wri-mathgroup at wolfram.com

I assume you want a numerical solution, since I don't think you'll find a neat closed formula for any arbitrary value of i. For a given value of i, say, i = 0.03, In[1]:= FindRoot[0.03 == r - r((1 + .05)^10 - 1)/((1 + r)^10 - 1), {r, 0.01, 1}] Out[1]= {r -> 0.074549} (remember not to use capital letters to name your variables). By the way, a visualization of r as a function of i looks very much linear to me, at least in the range i = 0.01 to i = 0.25. Look at In[2]:= ListPlot[Table[{i, FindRoot[i == r - r((1 + .05)^10 - 1)/((1 + r)^10 - 1), {r, 0.01, 1}][[1, 2]]}, {i, 0.01, 0.25, 0.01}], PlotJoined -> True, AxesOrigin -> {0, 0}, PlotRange -> {{0, 0.25}, {0, 0.25}}] If that range of values of i is acceptable for your purposes, you might fit a straight line and use that as an approximation: In[3]:= Fit[Table[{i, FindRoot[i == r - r((1 + .05)^10 - 1)/((1 + r)^10 - 1), {r, 0.01, 1}][[1, 2]]}, {i, 0.01, 0.25, 0.01}], {1, i}, i] Out[3]= 0.0473053 + 0.872371 i I.e., in conventional notation, r( i ) = 0.0473053 + 0.872371 i (in the range referred to above). Tomas Garza Mexico City ----- Original Message ----- From: "Jose Lasso" <jml at accessinter.net> To: mathgroup at smc.vnet.net Subject: [mg26634] [mg26614] Equation > Hi, > > I have little experience with Mathematica, I change to Mathematica > few months ago, I want to solve: i = R - R((1 + .05)^10-1)/((1 + > R)^10-1) for R, how can I do this task in Mathematica??. Thx in > advance. Regards. > > Jose M Lasso > > PS:Happy new year!! >

**References**:**Equation***From:*Jose Lasso <jml@accessinter.net>