Since a infinite series for Phi (f) must equate to its polynomial solution value of  [Sqrt(5)+1]/2, we can try to expand this via a Taylor series.

The general form of a Taylor series is described by EQ. 1:

So by letting f = [f(x) + 1]/2                                   (EQ. 2)

we can get a power series representation for Phi by using  f(x) =  =  x^1/2  when f(x) is developed into a Taylor series that is evaluated at x=5. My approach expands this Taylor series for f(x) about a=4 (this is a trick that later will conveniently make the (x-a)ⁿ terms simplify to unity when evaluating at x=5).

To get the first part of the numerator of the series for f(x), which is fⁿ(a),  we look at the value of successive (as n increases) derivatives of x^1/2 evaluated at a=4.

This gets a little messy but after some inspection there can be found a recursive pattern describing  fⁿ(a) :

for n > 2  (at a=4) we get;

for n=0 (at a=4) we get  f⁰(a) = f(a) = = 2

and for n=1 (at a=4) we get  f¹(a) = 1/2 (4)^-1/2 = (1/2)(1/2) = 1/4

Overall then (for a=4)    =  f(x)  expands via Eq. 1 to be

Equation 2 then becomes, when evaluating f(x) at  x=5 in Eq. 3;

combining terms and adjusting the index of the series to start at 0 (by replacing n by n+2 throughout) this equals:

Separately, one can manipulate a binomial series, as an additional way to check this Taylor series.

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