How do we find the Taylor series centered at c for the function f(x) = 1 / (x – c + 1)?

To find the Taylor series centered at c for the function f(x) = 1 / (x – c + 1), we need to start by using the definition of a Taylor series.

The Taylor series of a function f(x) centered at c is given by:

T(x) = f(c) + f'(c)(x – c) + f”(c)(x – c)^2 / 2! + f”'(c)(x – c)^3 / 3! + …

First, we’ll calculate f(c):

f(c) = 1 / (c – c + 1) = 1 / 1 = 1

Next, we need to find the derivatives of f(x) at c. The first derivative, f'(x), can be computed using the quotient rule:

f'(x) = -1 / (x – c + 1)^2

Now, evaluate f'(c):

f'(c) = -1

Continuing this for higher derivatives, we find:

f”(x) = 2 / (x – c + 1)^3 and so on.

To generalize, the n-th derivative can be found using the pattern that each derivative introduces an additional factor of (-1)^{n}(n!)/(x-c+1)^{n+1} evaluated at c.

In the end, the Taylor series becomes:

T(x) = 1 – (x – c) + (x – c)^2 – (x – c)^3 + …

This continues indefinitely, giving us a power series centered at c. Ultimately, we get:

The final Taylor series is:

T(x) = 1 – (x – c) + (x – c)^2 – (x – c)^3 + … which converges for x near c.

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