# find taylor series centered at c=1 of f(x)= 1/x?

show steps please

### 3 Answers

- 1 decade agoFavorite Answer
The taylor series expansion of 1/x centered at c = 1 can be written as:

f(c)/0! + f'(c)(x - c)/1! + f''(c)(x - c)^2 / 2! + f'''(c)(x - c)^3 / 3! + ... + f^n(c)(x - c)^n / n!

Under sigma notation, this is represented as:

Sum(from n = 0 → ∞) [ f^n(c) (x - c)^n / n! ]

Where f^n(c) denotes the nth derivative of f at c.

First we find a general expression for the nth derivative of f, and then use that to find the nth coefficient in the series.

f(x) = 1/x → f(1)/0! = 1

f'(x) = -1/x^2 → f'(1)/1! = -1

f''(x) = 2/x^3 → f''(1)/2! = 1

f'''(x) = -6/x^4 → f'''(1)/3! = -1

At this point we can see that f^n(x) = (-1)^n n! / x^(n+1) and f^n(c) / n! = (-1)^n * n! / n! = (-1)^n

So (-1)^n is the nth coefficient in the series.

With this information, we can write the taylor series for 1 / x centered at 1.

Sum(from n = 0 → ∞) [ f^n(c) (x - c)^n / n! ]

= Sum(from n = 0 → ∞) [ (-1)^n * (x - 1)^n ]

I will now use the ratio test to find the radius of convergence of the series.

By the ratio test, the series will converge absolutely whenever

lim (n → ∞)| (x - 1)^(n+1) / (x - 1)^n | = L and L < 1.

Note that I've taken the absolute value of the nth and nth + 1 term of the series.

lim (n → ∞)| (x - 1)^(n+1) / (x - 1)^n |

= lim (n → ∞) | x - 1 | = | x - 1 |

The series converges if | x - 1 | < 1

-1 < x - 1 < 1

0 < x < 2

The series converges if 0 < x < 2

But we must still check the endpoints.

Let x = 0

Then the series is

Sum(from n = 0 → ∞) [ (-1)^n * (0 - 1)^n ] = 1 + 1 + 1 + 1....

This series diverges...

Let x = 2

Then the series is

Sum(from n = 0 → ∞) [ (-1)^n * (2 - 1)^n ] = 1 + -1 + 1 + -1 + ...

This oscillating series is divergent as well.

So the series :

Sum(from n = 0 → ∞) [ (-1)^n * (x - 1)^n ]

converges to 1 / x on the interval 0 < x < 2.

PS. If you take x = 1 don't forget that 0^0 = 1 so that the sum does equal 1 / x at that point.

The sum for x = 1 would be

0^0 + 0^1 + 0^2 + 0^3 + ... + 0^n = 1