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From: barefoot gigantor on 24 Nov 2009 06:03
Dear Math Group:-

Suppose we have:-

x = a * e + (2*b - 3*a^2) * e^2

Now let us find such functions:

F(x) = 1 + a * e + (2 * b - a^2) * e^2 + ......

we are just interested in the first three terms.

Now two such functions can be:

F1(x) = 1 + x + 2 * x^2

and

F2(x) = (1-x)/(1-2*x)

How can we find all such functions F(x)?

From: dh on 25 Nov 2009 02:29


barefoot gigantor wrote:

> Dear Math Group:-

>

> Suppose we have:-

>

> x = a * e + (2*b - 3*a^2) * e^2

>

> Now let us find such functions:

>

> F(x) = 1 + a * e + (2 * b - a^2) * e^2 + ......

>

> we are just interested in the first three terms.

>

> Now two such functions can be:

>

> F1(x) = 1 + x + 2 * x^2

>

> and

>

> F2(x) = (1-x)/(1-2*x)

>

> How can we find all such functions F(x)?

>

Hi,

all such function can be written as:

F[x]== 1 + a * e + (2 * b - a^2) * e^2 + e^3 PS[e]

where PS[e] is a power series (terminating or not, converging in a

circle) in e.

To get an expression in x we may solve x for for e:

sol=Solve[x==a * e + (2*b - 3*a^2) * e^2, e]

we may use these solutions above to get F[x] in terms of x. PS[e] can

then be replaced by any analytic function of x.

E.g. one possibility:

F[x]=1 + (a (a - Sqrt[a^2 - 12 a^2 x + 8 b x]))/(

2 (3 a^2 - 2 b)) + ((-a^2 + 2 b) (a - Sqrt[

a^2 - 12 a^2 x + 8 b x])^2)/(4 (3 a^2 - 2 b)^2) + (

fun (a - Sqrt[a^2 - 12 a^2 x + 8 b x])^3)/(8 (3 a^2 - 2 b)^3)

where fun is a function of x.



Daniel



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