Re: Series Expansion of 1/x ?

On 28 oct, 03:32, Ray Vickson <C6...@xxxxxxx> wrote:
On Oct 27, 8:47 am, John Schutkeker <jschutke...@xxxxxxxxxxxxxxxxxxxx>
wrote:

Is there a series expansion for 1/x in positive powers of x, or have I made
an obvious mistake. I tried using Taylor's formula, but that's for a small
perturbation a from x. If I set x=0, then every single term in the
expansion goes to zero, and I'm left with nothing. TIA.'

I'm surprised. By setting x = 0 in the expansion f(y) = sum [f^(n)(x)/
n!](y-x)^n, I would get each term equal to + or - infinity, not even
close to zero.

R.G. Vickson

Bon Dimanche,

I'm not sure it helps you...
but within the neighbourhood of c we may write :
1/x = 1/(c -(c-x)) = 1/c *1/(1 -(1-x/c))
one term approx. 1/x <> 2/c -x/c^2
two terms 3/c -3x/c^2+x^2/c^3
three terms 4/c-6x/c^2+4x^2/c^3-x^3/c^4
n terms corresponding to
1/x <> 1/c*(1 -(1-x/c)^(n+1))/(1 - (1-x/c))

Alain

.

Relevant Pages

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