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MathGroup Archive 2001

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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>
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