Htykuut

Can someone please explain to me why, intuitively, does Descartes rule of signs work?

I realize there is a previous answer for this, that "Basically, at different values of $x$ different terms in the polynomial "dominate." So every the sign switches, there will be a change in the direction of the curve. Either

1. This will result in crossing the $x$-axis and a root or


2. There will have to be another change, meaning "losing roots" will always happen in pairs.

So the roots are equal to, or less than
by an even number, the number of sign changes." But perhaps because my
math understanding is not good enough, I still fail to see why this
ensures that the Descartes rule of signs works.

Why would different values of $x$ dominate in different areas?

And why would this result in crossing the $x$-axis or losing roots?

Can you please explain to me why the rule of signs can find the number of real zeroes for any polynomial? Please keep the explanation simple.

But perhaps because my math understanding is not good enough, I still fail to see why this ensures that the Descartes rule of signs works.

In math, intuition never ensures that something works; the only thing that ensures that it works is a proof.

It is important to remember that Descartes' rule of signs says that a polynomial has at least as many sign changes as it has positive real roots. Let's say we have a polynomial $p(x)$ with one positive real root factored out: $p(x)=(x-a)q(x)$ where $q(x)$ is a polynomial. Let's say the last term of $q(x)$ is $b_nx^n$. When we multiply this term by $(x-a)$ we get $(b_nx^{n+1}-b_nax^n)$ which has a sign change. If the second to last term in $q(x)$ is $b_{n-1}x^{n+1}$ then when multiplying the last 2 terms by $(x-a)$ we get $b_{n-1}x^{n+2}+(b_n-b_{n-1}a)x^{n+1}-b_nax^n$. If $b_{n-1}a<b_n$ then the signs of these terms are $++-$ or $-++$ depending on whether b_n is positive or negative. If $b_{n-1}a>b_n$ then the signs of these terms are either $+--$ or $-++$ depending on whether $b_{n-1}$ is positive or negative. Either way, there is a sign change. Our intuition tells that we can repeat this process with as many terms of $q(x)$ as we need to show that there will always be an extra sign change after multiplying by $(x-a)$. This means "adding a root" to $q(x)$ adds a sign change.

If we want to produce the polynomial $p(x)$ we simply let $q(x)$ equal with the product of all its factors that don't produce positive real roots and multiply $q(x)$ by $x-r_n$ for all of its positive real roots. By the previous argument the result (which is $p(x)$) will have as many sign changes as $q(x)$ did plus however many positive real roots it had. This shows that any polynomial has at least as many positive real roots as it has sign changes, which is what Descartes rule of signs says.

To show that the difference between the number of positive real roots and the number of sign changes is even, we just have to show that the number of sign changes in $q(x)$ is even. Since $q(x)$ does not have any positive real roots, all of its roots must either be negative real numbers or complex numbers. We can ignore the negative roots because multiplying a polynomial by $(x+a)$ where $a$ is positive does not change any signs.

Now we must show that the product of the complex roots has an even number of sign changes. Since $p(x)$ has real coefficients, the complex conjugate root theorem tells us that the complex roots are in conjugate pairs $a+bi$ and $a-bi$. When we multiply such a pair of factors together, we get the quadratic $x^2-2ax+(a^2+b^2)$ which has 2 sign changes. Since the product of the other complex factors is a product of quadratics we must show that the product of quadratics whose sequence of terms is $+-+$ has an even number of sign changes. We could do this by noticing that when we multiply an even degree polynomial whose signs are $+-+-...-+$ by a quadratic whose signs are $+-+$ we get a polynomial where the signs of the even exponents are $+$ and the signs of the odd exponents are $-$, which is again of the form $+-+-...-+$. Intuition again tells us we can repeat this until we have multiplied are the complex factors together and we will get a polynomial with an even number of sign changes, which is what we needed to show.

It might be interesting to note that we have shown that the number of sign changes minus the number of real roots is precisely the number of complex roots.