1. ## Produce a taylor series please....

for fx=(1/x) centered at c=1

2. Originally Posted by tiga killa
for fx=(1/x) centered at c=1

$f(x)=\frac{1}{x}=\frac{1}{1- [-(x-1)]}$

This is a geometric series with $r=-(x-1)$

so

$f(x)=\sum_{n=0}^{\infty}(-1)^n(x-1)^n$

3. Originally Posted by tiga killa
for fx=(1/x) centered at c=1
Assume that $f(x)$ can be expressed as a polynomial.

Since it is centred at $c = 1$, it will be of the form

$x^{-1} = a + b(x - 1) + c(x - 1)^2 + d(x - 1)^3 + e(x - 1)^4 + \dots$.

At $x = 1$ we find $a = 1$.

Take the derivative of both sides

$-x^{-2} = b + 2c(x - 1) + 3d(x - 1)^2 + 4e(x - 1)^3 + 5f(x - 1)^4 + \dots$

At $x = 1$ we find $b = -1$.

Take the derivative of both sides

$2x^{-3} = 2c + 6d(x - 1) + 12e(x - 1)^2 + 20f(x - 1)^3 + \dots$

At $x = 1$ we find $c = 1$.

Take the derivative of both sides

$-6x^{-4} = 6d + 24e(x - 1) + 60f(x - 1)^2 + \dots$

At $x = 1$ we find $d = -1$.

Take the derivative of both sides

$24x^{-5} = 24e + 120f(x - 1) + \dots$

At $x = 1$ we find $e = 1$.

Take the derivative of both sides

$-120x^{-6} = 120f + \dots$.

At $x = 1$ we find $f = -1$.

Looking at our polynomial we have

$\frac{1}{x} = 1 - x + x^2 - x^3 + x^4 - x^5 + \dots$

$\frac{1}{x} = \sum_{n = 0}^{\infty}{(-1)^n(x-1)^n}$.

4. Originally Posted by TheEmptySet
$f(x)=\frac{1}{x}=\frac{1}{1- [-(x-1)]}$

This is a geometric series with $r=-(x-1)$

so

$f(x)=\sum_{n=0}^{\infty}(-1)^n(x-1)^n$
Of course, we're assuming that

$|r| < 1$.

So $|-(x - 1)| < 1$

$|x - 1| < 1$

$-1 < x - 1 < 1$

$0 < x < 2$.

5. Originally Posted by Prove It
Of course, we're assuming that

$|r| < 1$.

So $|-(x - 1)| < 1$

$|x - 1| < 1$

$-1 < x - 1 < 1$

$0 < x < 2$.
Of course it has the same radius of convergence as the series you came up with above.