Logarithmic Series - Alpha and Beta are the roots of a given equation

Question

If $\alpha$ and $\beta$ are the roots of the equation $x^2 - px +q= 0$, then the value of

Is,

a) $\log(1-px+qx^2)$

b) $\log(1+px-qx^2)$

c) $\log(1+px+qx^2)$

d) None of these

For this I split it into $\log(1+\alpha x)$ and $\log(1+\beta x)$ and then added both the equations. I think that there is misprinting in the question and the Denominator of the last term should be three. And if I consider that and solve like mentioned, I get the answer as c) $\log(1+px+qx^2)$

But that is much different from the answer and solution given,

I am having trouble understanding that am I making huge mistakes or am I correct? I don't understand the solution and I believe that they did a mistake in solution from step 2.

Am I wrong? What's the correct answer and why?

I am new here, so pardon me if the question is having problems and let me know for any suggestions

Answer 1

The book's answer is incorrect. First, you are correct that the series representation of $\log(1+\alpha x)$ is

$$\log(1+\alpha x)=\sum_{n=1}^\infty \frac{(-1)^{n-1}(\alpha x)^n}{n}$$

for $-1<\alpha x\le 1$.

Note that the series representation of $\log(1-\alpha x)$ is

$$ \log(1-\alpha x)=-\sum_{n=1}^\infty \frac{(\alpha x)^n}{n} $$

for $-1\le \alpha x<1$.

Hence, the given series represents $\log((1+\alpha x)\,(1+\beta x))$.

And what is $(1+\alpha x)\,(1+\beta x)$ when $\alpha$ and $\beta$ are the roots of the given quadratic equation? It is $1+px+qx^2$.

The answer is $\log(1+px+qx^2)$.

Answer 2

In two different places the book has a $2$ where a $3$ should appear in the denominator. The first is where the numerator is $\alpha^2+\beta^3.$ The second is where the fraction is multiplied by $(\alpha x)^3.$

The basic series should be $$ \log(1+x) = x - \frac{x^2} 2 + \frac{x^3} 3 - \frac{x^4} 4 + \cdots. $$ Therefore \begin{align} (\alpha+\beta)x - \left( \frac{\alpha^2+\beta^2} 2 \right) x^2 + \left( \frac{\alpha^2+\beta^3} 3 \right) x^3 - \cdots & = \log(1+\alpha x) + \log(1+\beta x) \\[10pt] & = \log(1+(\alpha+\beta) x + \alpha\beta x^2) \\[10pt] & = \log(1 + px + qx^2). \end{align}