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Find and prove an upper bound on the number of times that two distinct polynomials of degree $d$ can intersect.

What if the polynomials' degrees differ?

My attempt:

let $p(x)$ and $q(x)$ be two distinct polynomials of degree $d$.

And they intersect if and only if $p(x_i)=q(x_i)$ for some $x_i$

Then the problem becomes to find the number of roots of $f(x) = p(x_i)-q(x_i) =0$

Since both of polynomials are degree of $d$, then $f(x)$ which is the difference between these two polynomials can only have degree of at most $d$, then the number of roots of $f(x)$ is $d$.

If the degrees of these two polynomial differ, then the number of roots of $f(x) =\max(\text{ degree of p}, \text{ degree of q}).$

I feel like I have done something wrong here since based on Paul's link, it should intersect at most $d^2$ times.

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How many variables are the polynomials functions of? –  Anthony Deluca Dec 8 '11 at 7:57
Your approach is exactly the right approach. It yields a correct upper bound for the number of roots; examples are easily found to show that the upper bound is sharp. –  Greg Martin Dec 8 '11 at 8:13
only one variable, namely, $x$ if you will –  geraldgreen Dec 8 '11 at 8:14
@GregMartin but what about the link Paul posted? –  geraldgreen Dec 8 '11 at 8:15
The "degree" there allows for $y$ to be raised to a power, as far as I can tell. –  David Mitra Dec 8 '11 at 8:17

2 Answers 2

up vote 2 down vote accepted

If the polynomials, $Q$ and $P$ have the same degree $d$, then the number of intersections points of their graphs is at most $d$, since these intersection points correspond to the zeroes of the polynomial $Q-P$. It is easy to see that this is a sharp estimate. For example, take $Q$ to be a polynomial with $d$ distinct real roots and set $P=-Q$.

If the degrees differ, the same argument shows that the number of intersection points of their graphs is at most $\max\{\text {deg}(P),\text{deg }(Q)\}$. This is a sharp estimate. For example take a polynomial with $r$ distinct real roots and write it as $$ \underbrace{a_r x^r + a_{r-1} x^{r-1}+\cdots+a_{q+1} x^{q+1}}_Q+ \underbrace{a_q x^q+\cdots+a_1 x +a_0}_{-P} $$

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Bézout's Theorem

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True, but wasteful for polynomials in only one variable. –  Greg Martin Dec 8 '11 at 8:14
While this link may answer the question, it is better to include the essential parts of the answer here and provide the link for reference. Link-only answers can become invalid if the linked page changes. –  tomasz Aug 17 '12 at 13:39

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