Find the Taylor series at 0. State the interval of convergence for the series.
-ln (1 - x) + ln (1 + x)

Question

Quizplus · Accepted Answer

Using properties of logarithms, we can rewrite the expression as:
ln((1-x)(1+x)) = ln(1-x^2)

Now, we can use the formula for the Maclaurin series for ln(1-x):

ln(1-x) = -x - x^2/2 - x^3/3 - ...

Therefore, the Maclaurin series for ln(1-x^2) is:

ln(1-x^2) = -x^2 - x^4/2 - x^6/3 - ...

Similarly, the formula for the Maclaurin series for ln(1+x) is:

ln(1+x) = x - x^2/2 + x^3/3 - ...

Therefore, the Maclaurin series for ln(1-x^2) + ln(1+x) is:

ln(1-x^2) + ln(1+x) = -x^2 - x^4/2 - x^6/3 - ... + x - x^2/2 + x^3/3 - ...

= x - 3x^2/2 - 13x^4/12 - 23x^6/18 - ...

The interval of convergence for this series is -1 ≤ x < 1, since both ln(1-x^2) and ln(1+x) have this interval of convergence. Therefore, the answer is B. Using the fact that ln(1-x) = -∑(x^n/n) and ln(1+x) = ∑((-1)^n*x^n/n), we get:

ln(1-x) = -∑(x^n/n)
ln(1+x) = ∑((-1)^n*x^n/n)

Adding the two series, we get:

ln(1-x) + ln(1+x) = -∑(x^n/n) + ∑((-1)^n*x^n/n)
                   = ∑((-1)^n * (x^n/n))

This is the Taylor series for ln(1-x) + ln(1+x) centered at 0. The interval of convergence can be found by using the ratio test:

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

So the series converges for |x| < 1, or -1 < x < 1. Therefore, the interval of convergence is (-1, 1).

Thus, the best choice is B.

Find the Taylor Series at 0