how to find a function
barefoot gigantor
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)?
dh
barefoot gigantor wrote:
> Dear Math Group:-
>
> Suppose we have:-
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>
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>
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> and
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> F2(x) = (1-x)/(1-2*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