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

show steps please

3 Answers

Relevance
  • 1 decade ago
    Favorite 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

  • cubit
    Lv 4
    4 years ago

    Find The Taylor Series

  • 4 years ago

    1/x-(1/x^2)(x-2) - (1/x^3)(x-2)^2

    Source(s): Chemical Engineer SK
Still have questions? Get your answers by asking now.