       Re: iterative method?

• To: mathgroup at smc.vnet.net
• Subject: [mg95735] Re: [mg95699] iterative method?
• From: DrMajorBob <btreat1 at austin.rr.com>
• Date: Sun, 25 Jan 2009 21:50:07 -0500 (EST)
• References: <200901251151.GAA00605@smc.vnet.net>
• Reply-to: drmajorbob at longhorns.com

```If you've written Mathematica code and are trying to explain it, why not
copy/paste it into e-mail so that we have it right???

Highlight the code, select Edit>Copy As>Plain Text

and paste into e-mail. (This works for almost everything.)

Asspming I've read your post properly, here's a possible solution for
{a,b,c,d}={.1,.2,.3,.4}:

Clear[f]
f[a_, b_, c_, d_][{x_, y_}] := {(b + a x)/(c x), d x y}
pairs = NestList[f[.1, .2, .3, .4], {2, 5}, 50];

Now this plots the x values:

ListPlot@pairs[[All, 1]]

This plots the y values:

ListPlot@pairs[[All, 2]]

and this plots both at once:

ListPlot@Transpose@pairs

Warning: if a, b, c, d are undefined or exact (but not integers), the x
and y values get complicated.

For instance, this is nestng only TEN deep, not 50, and I've added
Simplify to each step:

Clear[f]
f[a_, b_, c_, d_][{x_, y_}] := Simplify /@ {(b + a x)/(c x), d x y}
pairs = NestList[f[Pi, E, Log, Sqrt], {2, 5}, 10];
Last@pairs

{(2 \[Pi]^10 + E^5 Log^4 (5 \[Pi] + Log) +
E \[Pi]^8 (\[Pi] + Log) +
E^4 \[Pi]^2 Log[
10]^3 (20 \[Pi] + Log) +
E^3 \[Pi]^4 Log[
10]^2 (21 \[Pi] + 4 Log + 2 Log +
Log) +
E^2 \[Pi]^6 Log[
10] (8 \[Pi] +
Log)\
)/(Log (2 \[Pi]^9 + E^5 Log^4 +
E^3 \[Pi]^3 Log[
10]^2 (15 \[Pi] + 4 Log + 2 Log) +
E^4 \[Pi] Log^3 (10 \[Pi] + Log) +
E \[Pi]^7 (\[Pi] + Log) +
E^2 \[Pi]^5 Log[
10] (7 \[Pi] +
Log))), (1/(
Log^8 Log))
31250 (2 \[Pi]^9 + E^5 Log^4 +
E^3 \[Pi]^3 Log[
10]^2 (15 \[Pi] + 4 Log + 2 Log) +
E^4 \[Pi] Log^3 (10 \[Pi] + Log) +
E \[Pi]^7 (\[Pi] + Log) +
E^2 \[Pi]^5 Log[
10] (7 \[Pi] +
Log))}

And here's another example:

Clear[f]
f[a_, b_, c_, d_][{x_, y_}] := FullSimplify /@ {(b + a x)/(c x), d x y}
pairs = NestList[f[a, b, c, d], {2, 5}, 20];
Last@pairs

{(a^19 (2 a + b) + 2 a^17 b (19 a + 9 b) c +
34 a^15 b^2 (9 a + 4 b) c^2 + 80 a^13 b^3 (17 a + 7 b) c^3 +
455 a^11 b^4 (8 a + 3 b) c^4 + 2002 a^9 b^5 (3 a + b) c^5 +
858 a^7 b^6 (7 a + 2 b) c^6 + 264 a^5 b^7 (13 a + 3 b) c^7 +
165 a^3 b^8 (6 a + b) c^8 + 10 a b^9 (11 a + b) c^9 +
2 b^10 c^10)/(c (a^18 (2 a + b) + a^16 b (36 a + 17 b) c +
8 a^14 b^2 (34 a + 15 b) c^2 + 35 a^12 b^3 (32 a + 13 b) c^3 +
91 a^10 b^4 (30 a + 11 b) c^4 + 143 a^8 b^5 (28 a + 9 b) c^5 +
132 a^6 b^6 (26 a + 7 b) c^6 + 66 a^4 b^7 (24 a + 5 b) c^7 +
15 a^2 b^8 (22 a + 3 b) c^8 + b^9 (20 a + b) c^9)), (1/(c^19))
5 (a^18 (2 a + b) + a^16 b (36 a + 17 b) c +
8 a^14 b^2 (34 a + 15 b) c^2 + 35 a^12 b^3 (32 a + 13 b) c^3 +
91 a^10 b^4 (30 a + 11 b) c^4 + 143 a^8 b^5 (28 a + 9 b) c^5 +
132 a^6 b^6 (26 a + 7 b) c^6 + 66 a^4 b^7 (24 a + 5 b) c^7 +
15 a^2 b^8 (22 a + 3 b) c^8 + b^9 (20 a + b) c^9) d^20}

Bobby

On Sun, 25 Jan 2009 05:51:27 -0600, matt <trex1704 at yahoo.com> wrote:

> How to find x_(n+1)=(a*x_n+b*y_n)/c*x_n , y_(n+1)=d*x_n*y_n using
> mathematica?
> I want to find a list of x and y when n=1,2,...,50
> so I defined the function as f[{x_,y_}]:={a*x_n+b*y_n)/c*x_n,d*x_n*y_...
> }
> and then use the function NestList[f,{2,5},50] with initial value of x
> is 2 and y is 5
> with this, I got the list of 50 values of (x,y).
> The problem is I want to plot graphs of x against n and y against n, n
> from 1 to 50 because I want to see what happened if n goes to infinity.
> But if use the method I just mentioned, I can only plot x against y.
> So do anyone have a clever method??
> Btw...a,b,c and d are constants and given and x_n is supposed to be x
> (subscript n)...same case with x_(n+1) and the others
>

--
DrMajorBob at longhorns.com

```

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