Understanding the definition of a multiple (double) root
According to the definition on Wolfram, a multiple root is a root with multiplicity $n\geq 2$. For example, in the equation $(x-1)^2 = 0$, $1$ it the multiple (double) root.
Question: Why is $x = 1$ the multiple root of $(x-1)^2 = 0$? What does multiplicity mean in this case? According to Wolfram multiplicity is the number of values for which a condition holds, or the number of times for which a polynomial equation has a root at a given point. In what way does the polynomial $x^2 - 2x + 1 = 0$ have multiple roots?
$\endgroup$4 Answers
$\begingroup$An intuitive explanation: if you consider the polynomial $(x-1)(x-1-\varepsilon)$ $(\varepsilon\ne0)$, it has two roots, $1$ and $1+\varepsilon$. When $\varepsilon\to0$, the second root tends to $1$, so we consider that, in the equation $(x-1)^2=0$, the root $1$ counts for two, whence the multiplicity $2$.
$\endgroup$ $\begingroup$The equation $(x-1)^2+\epsilon =0$ (for $\epsilon >0 $ and small) will have two distinct roots that are very close to $x=1$. As $\epsilon$ tends to zero these two roots will coincide with one another. We say there is a root at $x=1$ with a multiplicity of $2$.
$\endgroup$ $\begingroup$The polynomial $x^2-2x+1$ has a double root (which is $1$). This is so because $x^2-2x+1$ is a multiple of $(x-1)^2$ but not a multiple of $(x-1)^3$.
$\endgroup$ 2 $\begingroup$You seem to be having a problem with the definition, and the use of the word multiplicity. If you go back to the Wolfram definition of multiplicity you linked you will see that it refers to a power series example.
Well you can regard $f(x)=(x-1)^2$ as a power-series expansion about $x=1$, equivalent to: $$0\cdot (x-1)^0+0\cdot (x-1)^1+ 1\cdot (x-1)^2 + 0\cdot (x-1)^3+0\cdot (x-1)^4 \dots$$ and the example transfers over, with $$f(1)=(1-1)^2=0; f'(1)=2(1-1)=0; f''(1)=2\neq 0$$
It has multiplicity $2$ at $x=1$ in this sense because the first two derivatives are zero, but not the third.
But really the power series example is best regarded as a generalisation of the simple case of polynomials, and the natural definition of multiplicity in the case of polynomials is normally that the multiplicity $p(x)$ at $a$ is the highest power of $x-a$ which is a factor of $p(x)$. This is like saying that $2$ is a double factor of $12=2^2\times 3$.
It is also true that $p(x)$ has multiplicity at least $r$ at $x=a$ if the first $r$ derivatives (starting with $f^{(0)}=f(x)$) evaluate to zero at $x=a$. If $f^{(r)}(a)\neq 0$ the multiplicity is exactly $r$. But this is not so natural a definition in elementary work with polynomials.
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