Using the power series expansions of e^x, sin x, cos x

Using the power series expansions of $e^x$, $\sin x$, $\cos x$, $\ln(1+x)$, $(1+x)^a$, expand $\ln(1+3x+2x^2)$ in a power series in $x$ and indicate when the expansion is valid.

Step 1 · Factor the argument

The roots of $2x^2+3x+1=0$ are $x=\dfrac{-3\pm1}4$, i.e. $x=-\frac12,-1$. So $$ 1+3x+2x^2 = (1+x)(1+2x). $$

Step 2 · Split and expand each logarithm

$$ \ln(1+3x+2x^2) = \ln(1+x)+\ln(1+2x). $$ Using $\ln(1+t)=\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}n t^n$: $$ \ln(1+x)=\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}nx^n, \qquad \ln(1+2x)=\sum_{n=1}^{\infty}\frac{(-1)^{n-1}2^n}nx^n. $$

Step 3 · Combine and determine validity

$$ \ln(1+3x+2x^2) = \sum_{n=1}^{\infty}\frac{(-1)^{n-1}(1+2^n)}nx^n = 3x-\frac{5x^2}2+3x^3-\frac{17x^4}4+\frac{33x^5}5-\cdots $$ The series for $\ln(1+x)$ converges on $-1

Summary Table

Additional Notes

• The radius of convergence, $\frac12$, is set by the nearer singularity of $\ln(1+2x)$ at $x=-\frac12$, exactly the same "closer factor wins" principle used in the rational-function problems of Problems 246–260.
• The left endpoint is excluded (logarithmic divergence at $x=-\frac12$) while the right endpoint is included, mirroring the one-sided endpoint behavior of the basic $\ln(1+x)$ series itself.
• This factor-then-sum-logarithms technique generalizes immediately to any polynomial argument with real roots, always producing a coefficient of the form $\sum(\pm r_i)^n/n$ summed over the roots' reciprocals.