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While studying, I stumbled upon a statement that all the roots of $q(x) = x^{(p^n)} - x$, where p is a prime number are different, because otherwise $q$ would share some factors with its derivative.

This lead me to wonder, what happens to the roots when we derive, as my basic understanding would say that the multiplicity of all of the roots drops by one, but I cannot find some proof for this or counter-example.

Can anyone please describe what happens (preferably also with a link to the proof).

klemen kobau
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1 Answers1

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Let $p(x)=\prod_{k=1}^n (x-x_k)^{m_k}$ where the $m_k$ are the multiplicities of the roots.

Then by the logarithmic derivative,

$$\frac{p'(x)}{p(x)}=\sum_{i=1}^n\frac{m_i}{x-x_i}$$ and close to a root, let $x_i$, we have $\dfrac1{|x-x_i|}\gg\dfrac1{|x-x_k|}$ and

$$p'(x)\approx\frac{m_i\,p(x)}{x-x_i},$$ meaning that $p$ loses a factor $x-x_i$.

When a root is simple, it simply disappears.

In blue a polynomial with a simple, a triple and a double root, from left to right.

enter image description here

Another way to look at it is to see multiple roots as very close but distinct ones. By continuity they are separated by extrema, which correspond to roots of the derivative, one less.

enter image description here

  • I see that there are some new intersections with the y axis, so I take it that also new roots appear? – klemen kobau Jun 12 '18 at 19:39
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    Well observed. By the FTA, the sum of the multiplicities can be as large as the degree (in case all roots are real). Taking the derivative will decrease all multiplicities by one, so that the total multiplicity decreases by the number of distinct roots, let $r$. So there is room for $r-1$ new roots. –  Jun 12 '18 at 19:45
  • Oh I get it now, can I just ask what do you mean by FTA? I don't understand the english terms that well. – klemen kobau Jun 12 '18 at 19:50
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    @klemenkobau: fundamental theorem of algebra. –  Jun 12 '18 at 19:54