Question

If $\alpha $ and $\beta $ are roots of the quadratic equation $a{{x}^{2}}+bx+c$ = 0, then show that:
$\log \left( a-bx+c{{x}^{2}} \right)=\log a+\left( \alpha +\beta \right)x-\left( \dfrac{{{\alpha }^{2}}+{{\beta }^{2}}}{2} \right){{x}^{2}}+\left( \dfrac{{{\alpha }^{3}}+{{\beta }^{3}}}{3} \right){{x}^{3}}-.......$

Answer 1

Hint: First we will find the value of $\alpha $+$\beta $, and then after that we will approach to prove this from the left hand side of the equation. We will apply tailor expansion on log and then we will try to convert it in the given form on the right hand side by using all the information that is given.

Complete step-by-step solution -
First let’s look at the formula of taylor expansion:
$f\left( x \right)=f\left( {{x}_{0}} \right)+f'\left( {{x}_{0}} \right)\left( x-{{x}_{0}} \right)+f''\left( {{x}_{0}} \right)\left( \dfrac{{{(x-{{x}_{0}})}^{2}}}{2!} \right)+f'''\left( {{x}_{0}} \right)\left( \dfrac{{{(x-{{x}_{0}})}^{3}}}{3!} \right)+......$
As we have stated the formula for taylor expansion one thing to keep in mind is that here ${{x}_{0}}$ is any value for x .
In this question we will put ${{x}_{0}}$= 0, to prove the above statement and so that we can make it in the form that is given in the right hand side.
After putting ${{x}_{0}}$ = 0 we get,
$f\left( x \right)=f\left( 0 \right)+f'\left( 0 \right)\left( x \right)+f''\left( 0 \right)\left( \dfrac{{{(x)}^{2}}}{2!} \right)+f'''\left( 0 \right)\left( \dfrac{{{(x)}^{3}}}{3!} \right)+......$
Now as per the question,
$f\left( x \right)=\log \left( a-bx+c{{x}^{2}} \right)$
Now will find the respective derivatives:
\[\begin{align}
  & \dfrac{d\log x}{dx}=\dfrac{1}{x} \\
 & f'\left( x \right)=\dfrac{2cx-b}{a-bx+c{{x}^{2}}} \\
 & \\
\end{align}\]
Now if $f=\dfrac{u}{v}\text{ then }f'=\dfrac{u'v-v'u}{{{v}^{2}}}$
We will use this formula.
\[\begin{align}
  & f''\left( x \right)=\dfrac{2c\left( a-bx+c{{x}^{2}} \right)-\left( 2cx-b \right)\left( 2cx-b \right)}{{{(a-bx+c{{x}^{2}})}^{2}}} \\
 & f''\left( x \right)=\dfrac{2ca-2bcx+2{{c}^{2}}{{x}^{2}}-4{{c}^{2}}{{x}^{2}}+4bcx-{{b}^{2}}}{{{(a-bx+c{{x}^{2}})}^{2}}} \\
 & f''\left( x \right)=\dfrac{2ca+2bcx-2{{c}^{2}}{{x}^{2}}-{{b}^{2}}}{{{(a-bx+c{{x}^{2}})}^{2}}} \\
\end{align}\]
Again using the same formula the third derivative is:
\[f'''\left( x \right)=\dfrac{(2bc-4{{c}^{2}}x){{(a-bx+c{{x}^{2}})}^{2}}-\left( 2ca+2bcx-2{{c}^{2}}{{x}^{2}}-{{b}^{2}} \right)\left( 2(a-bx+c{{x}^{2}})\left( 2cx-b \right) \right)}{{{(a-bx+c{{x}^{2}})}^{4}}}\]
Now we will put x = 0, and then find the values of all the derivatives,
$\begin{align}
  & f'\left( 0 \right)=\dfrac{-b}{a} \\
 & f''\left( 0 \right)=\dfrac{2ca-{{b}^{2}}}{{{a}^{2}}} \\
 & f'''(0)=\dfrac{2bc{{a}^{2}}+2ab\left( 2ca-{{b}^{2}} \right)}{{{a}^{4}}} \\
\end{align}$
Now we use all these values in the tailor expansion and see what we get,
$f\left( x \right)=f\left( 0 \right)+f'\left( 0 \right)\left( x \right)+f''\left( 0 \right)\left( \dfrac{{{(x)}^{2}}}{2!} \right)+f'''\left( 0 \right)\left( \dfrac{{{(x)}^{3}}}{3!} \right)+......$
$f\left( x \right)=\log a+\left( \dfrac{-b}{a} \right)\left( x \right)+\left( \dfrac{2ca-{{b}^{2}}}{{{a}^{2}}} \right)\left( \dfrac{{{(x)}^{2}}}{2!} \right)+\left( \dfrac{2bc{{a}^{2}}+2ab\left( 2ca-{{b}^{2}} \right)}{{{a}^{4}}} \right)\left( \dfrac{{{(x)}^{3}}}{3!} \right)+......$
Now we will use $a{{x}^{2}}+bx+c$= 0, and we will write the relation between roots,
The formula is:
$\alpha $+$\beta $= $\dfrac{-b}{a}$ and ($\alpha $$\beta $) = $\dfrac{c}{a}$ .
So, we will use this formula and substitute this in the above expansion.
And some more formulas:
$\begin{align}
  & {{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab \\
 & {{a}^{3}}+{{b}^{3}}=\left( a+b \right)\left( {{a}^{2}}-ab+{{b}^{2}} \right) \\
\end{align}$
After substituting we get,
$\begin{align}
  & f\left( x \right)=\log a+\left( \alpha +\beta \right)\left( x \right)+\left( {{\alpha }^{2}}+{{\beta }^{2}} \right)\left( \dfrac{{{(x)}^{2}}}{2!} \right)+\left( \dfrac{2bc}{{{a}^{2}}}+\dfrac{2b\left( \dfrac{2c}{a}-\dfrac{{{b}^{2}}}{{{a}^{2}}} \right)}{a} \right)\left( \dfrac{{{(x)}^{3}}}{3!} \right)+...... \\
 & f\left( x \right)=\log a+\left( \alpha +\beta \right)\left( x \right)-\left( {{\alpha }^{2}}+{{\beta }^{2}} \right)\left( \dfrac{{{(x)}^{2}}}{2!} \right)+2\left( {{\alpha }^{3}}+{{\beta }^{3}} \right)\left( \dfrac{{{(x)}^{3}}}{3!} \right)+...... \\
\end{align}$
Hence, we have proved that LHS = RHS.

Note: Keep in mind that we have used the formula of taylor expansion and in that there is a very good chance of making mistakes while finding the derivatives and rearranging them in the form that is given in the question.