Obtain the Taylor series in powers of x + 1 for f(x) = x/(2 + x), giving
the general term and the interval of convergence.
To obtain the Taylor series in powers of x + 1 for the given function f(x) = x/(2 + x), we will need to follow a few steps:
Step 1: Find the maclaurin series of f(x)
Since the given function is not centered around x = 0, we will first shift the function by substituting u = x + 1. This will center the function around u = 0.
Let u = x + 1. We can express f(x) in terms of u as:
f(u - 1) = (u - 1)/(2 + u - 1) = (u - 1)/(u + 1)
Step 2: Expand f(u - 1) in a Maclaurin series
To expand the function f(u - 1) in a Maclaurin series, we will need to express it as a power series.
Using the formula for the geometric series, we can write f(u - 1) as:
f(u - 1) = (u - 1)/(u + 1) = [(u - 1)/(1 - (-u))] (1/(u + 1))
Now, we can express 1/(1 - (-u)) as a power series:
1/(1 - (-u)) = 1 + (-u) + (-u)^2 + (-u)^3 + ...
Substituting this into the previous expression, we get:
f(u - 1) = [(u - 1)/(1 - (-u))] (1/(u + 1)) = (u - 1)(1 + (-u) + (-u)^2 + (-u)^3 + ...) (1/(u + 1))
Step 3: Simplify the expression
We can now simplify the expression by multiplying out the terms and collecting like powers of u. Since we are looking for the Taylor series in powers of x + 1, we will substitute u = x + 1 back into the series.
f(u - 1) = (u - 1)(1 - u + u^2 - u^3 + ...) (1/(u + 1))
= (x + 1 - 1)(1 - (x + 1) + (x + 1)^2 - (x + 1)^3 + ...) (1/(x + 1 + 1))
= x(1 - x + (x + 1)^2 - (x + 1)^3 + ...) (1/(x + 2))
Step 4: Determine the general term and interval of convergence
From the simplified expression, we can see that the general term of the Taylor series is given by:
a_n = (-1)^n * (x + 1)^(n - 1)
To determine the interval of convergence, we need to consider the convergence of the series. The series will converge as long as the absolute value of (x + 1) is less than 1. So the interval of convergence is -2 < x < 0.
Therefore, the Taylor series in powers of x + 1 for f(x) = x/(2 + x) is:
f(x) = x(1 - x + (x + 1)^2 - (x + 1)^3 + ...) (1/(x + 2))
= ∑[(-1)^n * (x + 1)^(n - 1)] (from n = 0 to infinity)