B Properties of roots of polynomials

[Rishabh Narula]

i have some doubts from chapter 1 of the book Mathematical methods for physics and engineering.
i have attached 2 screenshots to exactly point my doubts.
in the first screenshot...could you tell me why exactly the 3 values of f(x) are equal.
the first is the very definition of polynomials...but why is f(x) also equal to product of those factors(i.e the other two equations of f(x))?
And in the second screenshot...how exactly are properties 1.12,1.13,1.14 derived from those 3 equations of f(x)?
Also how do you use property 1.14...how do you evaluate that double summation?

 

[fresh_42]

The book is a bit sloppy as it doesn't explain the roots very well, at least on what you copied. I assume it is done beforehand.

We start with $f(x)=a_nx^n+a_{n-1}x^{n-1}+\ldots +a_1x+a_0$.
$f(x)$ has at most $n$ zeros $\alpha_1,\ldots ,\alpha_n$, that is $f(\alpha_i)=0$.

Now by the Euclidean algorithm (division with remainder) we can see that $(x-\alpha_i)$ divides $f(x)$.
If we then multiply all terms $(x-\alpha_i)$ we get a polynomial of degree $n$ which divides $f(x)$.
Because they have the same degree, they can only differ by a constant factor. This factor has to be $a_n$, because we have
\begin{align*}
f(x)&=a_nx^n+a_{n-1}x^{n-1}+\ldots +a_1x+a_0\\ &=a_n\cdot \left( x^n+\dfrac{a_{n-1}}{a_n}x^{n-1}+\ldots +\dfrac{a_1}{a_n}x+\dfrac{a_0}{a_n} \right) \\ &= a_n \cdot (x-\alpha_1)\cdot \ldots \cdot (x-\alpha_n)
\end{align*}
Formula $(1.12)$ is what we get by comparison of the coefficients at the term $x^0=1$, i.e. the absolute term of the two polynomial expressions after we multiplied it out.
Formula $(1.13)$ is what we get by comparison of the coefficients at the term $x^{n-1}$, i.e. the second term of the two polynomial expressions after we multiplied it out.
Formula $(1.14)$ is what we get by comparison of the coefficients at the term $x^{n-2}$, i.e. the third term of the two polynomial expressions after we multiplied it out.

What might be a bit confusing is the second line of our polynomial expressions. It reads
$$
f(x) = a_n\cdot (x-\alpha_1)^{m_1} \cdot \ldots \cdot (x-\alpha_r)^{m_r}
$$
This is because until now we haven't considered, that some of the $\alpha_i$ might occur multiple times. So as sets we have $\{\,\alpha_1,\ldots,\alpha_n\,\}=\{\,\alpha_1,\ldots,\alpha_r\,\}$, that is we have only $r$ distinct roots. Each of the $\alpha_i$ occurs $m_i$ times. So the second row in your copy is simply achieved by gathering all identical $\alpha_i$. At the end we get $n=m_1+\ldots +m_r$ and the $\{\,\alpha_1,\ldots,\alpha_r\,\}$ are pairwise distinct, whereas the $\{\,\alpha_1,\ldots,\alpha_n\,\}$ are not.

 

[Rishabh Narula]

The book is a bit sloppy as it doesn't explain the roots very well, at least on what you copied. I assume it is done beforehand.

We start with $f(x)=a_nx^n+a_{n-1}x^{n-1}+\ldots +a_1x+a_0$.
$f(x)$ has at most $n$ zeros $\alpha_1,\ldots ,\alpha_n$, that is $f(\alpha_i)=0$.

Now by the Euclidean algorithm (division with remainder) we can see that $(x-\alpha_i)$ divides $f(x)$.
If we then multiply all terms $(x-\alpha_i)$ we get a polynomial of degree $n$ which divides $f(x)$.
Because they have the same degree, they can only differ by a constant factor. This factor has to be $a_n$, because we have
\begin{align*}
f(x)&=a_nx^n+a_{n-1}x^{n-1}+\ldots +a_1x+a_0\\ &=a_n\cdot \left( x^n+\dfrac{a_{n-1}}{a_n}x^{n-1}+\ldots +\dfrac{a_1}{a_n}x+\dfrac{a_0}{a_n} \right) \\ &= a_n \cdot (x-\alpha_1)\cdot \ldots \cdot (x-\alpha_n)
\end{align*}
Formula $(1.12)$ is what we get by comparison of the coefficients at the term $x^0=1$, i.e. the absolute term of the two polynomial expressions after we multiplied it out.
Formula $(1.13)$ is what we get by comparison of the coefficients at the term $x^{n-1}$, i.e. the second term of the two polynomial expressions after we multiplied it out.
Formula $(1.14)$ is what we get by comparison of the coefficients at the term $x^{n-2}$, i.e. the third term of the two polynomial expressions after we multiplied it out.

What might be a bit confusing is the second line of our polynomial expressions. It reads
$$
f(x) = a_n\cdot (x-\alpha_1)^{m_1} \cdot \ldots \cdot (x-\alpha_r)^{m_r}
$$
This is because until now we haven't considered, that some of the $\alpha_i$ might occur multiple times. So as sets we have $\{\,\alpha_1,\ldots,\alpha_n\,\}=\{\,\alpha_1,\ldots,\alpha_r\,\}$, that is we have only $r$ distinct roots. Each of the $\alpha_i$ occurs $m_i$ times. So the second row in your copy is simply achieved by gathering all identical $\alpha_i$. At the end we get $n=m_1+\ldots +m_r$ and the $\{\,\alpha_1,\ldots,\alpha_r\,\}$ are pairwise distinct, whereas the $\{\,\alpha_1,\ldots,\alpha_n\,\}$ are not.

okay.thanks.i also tried expanding the formulas for n=2 and 3 and on comparing could see why the identities are true..thanks for explaing though.helpful.