Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Stewart wants me to prove stuff but I have no idea how to.

a) Show that a polynomial of degree 3 has at most three real roots.

b) Show that a polynomial of degree n has at most n real roots.

share|improve this question
    
$n$-odd? ${}{}$ –  user21436 Apr 6 '12 at 23:21
    
This is the same idea as your other question. The Mean Value Theorem implies that between any two roots of a polynomial, there has to be a root of the derivative of the polynomial (between any two $0$s, there has to be a critical point). –  Arturo Magidin Apr 7 '12 at 1:49
    
@Arturo I am confused, I thought it wasn't specfically roots unless it is Rolle's Theorem. –  user138246 Apr 7 '12 at 1:56
    
It's the same idea: between any two roots of the function, there has to be a critical point/root of the polynomial. Both the previous problem and this one revolve around the same idea: if you know how many roots the derivative can have, then you know how many roots the function can have. For example: since a degree 1 polynomial can have at most one root, then a degree 2 polynomial (whose derivative is degree 1) can have at most two roots (between any two roots of the degree 2 polynomial, there has to be one of the derivative). –  Arturo Magidin Apr 7 '12 at 1:58
    
I understand that the derivative determines if the function can change direction so does that mean that the zeros can never exceed the degree? –  user138246 Apr 7 '12 at 2:04
show 1 more comment

3 Answers

up vote 7 down vote accepted

I'll do some lower dimensional cases. You can build them up.

A degree-one polynomial has at most 1 real root. This, I take for granted.

Now, we would like to say that a quadratic polynomial as at most 2 real roots. Suppose it has three. Then by the mean value theorem (in fact, just Rolle's Theorem), the derivative of our quadratic has at least 2 real roots. But this is a contradiction, as above I mentioned that a linear polynomial has at most 1 real root. So a quadratic has at most 2 real roots.

Do you see how this continues?

share|improve this answer
    
I don't follow. –  user138246 Apr 6 '12 at 23:35
    
I know Rolle's Theorem, being applied here it just means that there is a derivative that is equal to zero. I don't understand how that means there can only be 3. –  user138246 Apr 6 '12 at 23:39
    
@Jordan: Ah, so we know the derivative is equal to zero between the first and second root, and between the second and third root. Thus the derivative is equal to zero twice. But the derivative is a linear, one degree polynomial. And those can't have 2 roots. –  mixedmath Apr 6 '12 at 23:41
    
How do we know there are 2 roots and not just one? –  user138246 Apr 6 '12 at 23:42
    
@Jordan: suppose the 3 assumed roots were at $a, b, c$ with $a < b < c$. Then Rolle guarantees a root between $a$ and $b$, and another between $b$ and $c$. –  mixedmath Apr 6 '12 at 23:44
show 22 more comments

According to Rolles theorem and the polynomial theorem any third degree polynomial sports at most 1 to 3 roots. Therefore, any polynomial of degree n possesses at least one root, whether real or complex. Moreover, a polynomial with a degree n bears at most n roots. To back this argument with Rolles theorem assume that there are three roots such that a>b>c. According to Rolles theorem there must be a number m such that f'(m) = 0 between a and b. Likewise there must be a value n such that f'(n) = 0 between b and c. This implies that m and n are minimums or maximums. Since there are two solutions for f'(x) because a third degree polynomial's derivative is a degree two function. Therefore the third degree polynomial acquires at most three real roots. If the function is twice differentiated there must be such a "c" between m and n such that f"(c) = 0 thus revealing that indeed that a third degree function bears at least one root. Moreover if you took a quadratic function and analyze between x- intercepts there is either a max or min value in between the two thus asserting the fact that when the derivative equals to zero there is indeed two zeros per critical point. As a result, there must be two c.p s to have three real roots.

share|improve this answer
add comment

Part (a)

Let $f(x)=K_3x^3+K_2x^2+K_1x+K_0$ Since $f(x)$ is a polynomial, it is continuous and differentiable everywhere. Suppose $f(x)$ has four roots $a_1, a_2, a_3$ such that $f(a_1)=f(a_2)=f(a_3)=f(a_4)=0$. By Rolle's Theroem there exists; $b_1$ in $(a_1, a_2)$, $b_2$ in $(a_2, a_3)$, $b_3$ in $(a_3, a_4)$ such that $f'(b_1)=f'(b_2)=f'(b_3)=0$. So, $f'(x)=3K_3x^2+2K_2x+K_1$ is also a polynomial and thus continuous and differentiable everywhere, by Rolle's Theorem again there exists; $c_1$ in $(b_1, b_2)$ and $c_2$ in $(b_2, b_3)$ such that $f''(c_1)=f''(c_2)=0$. So, $f''(x)=6K_3x+K_2$ which is still continuous and differentiable everywhere, by Rolle's Theorem there should exist; $d_1$ in $(c_1, c_2)$ such that $f'''(x)=0$ however $f'''(x)=6K_3$ which is not zero. So $f'''(x)$ never equals zero and thus; $f''(x)$ can only have at most one root, $f'(x)$ can only have at most two roots, and $f(x)$ can only have at most three roots.

Part (b)

First, all polynomials are continuous everywhere and differentiable everywhere.

Let $f(x)$ be a polynomial of degree $n$, $f(x)=K_nx^n+K_{n-1}x^{n-1}+...+K_2x^2+K_1x+K_0$. Suppose that $f(x)$ has $n+1$ roots, $a_{n+1}, a_n, a_{n-1}, ... a_2, a_1$, such that $f(a_n)=f(a_{n-1})=...=f(a_2)=f(a_1)=0$. Then by Rolle's Theorem there exists some $b_n$ in $(a_{n+1}, a_n)$, $b_{n-1}$ in $(a_n, a_{n-1})$,..., $b_2$ in $(a_3, a_2)$, $b_1$ in $(a_2, a_1)$, such that $f'(b_n)=f'(b_{n-1})=...=f'(b_2)=f'(b_1)=0$. By applying Rolle's Theorem again there exists some $c_{n-1}$ in $(b_n, b_{n-1})$, $c_{n-2}$ in $(b_{n-1}, b_{n-2})$,..., $c_2$ in $(b_3, b_2)$, $c_1$ in $(b_2, b_1)$ such that $f''(c_{n-1})=f''(c_{n-2})=...=f''(c_2)=f''(c_1)=0$. If we keep applying Rolle's Theorem we find that eventually there exists two roots $z_2$ in $(y_3, y_2)$ and $z_1$ in $(y_2, y_1)$ such that $f^n(a_2)=f^n(a_1)=0$. Where $y_3, y_2$ and $y_1$ are the roots of $f^{n-1}(x)$. But $f^n(x)=n!K_n$ and so $f^n(x)>0$ for all $x$ and so the $(n-1)$th derivative of $f(x)$ can only have one root. Therefore the $(n-2)$th derivative can have at most 2 roots, the $(n-3)$th derivative can have at most 3 roots,..., the 3rd derivative can have at most $(n-3)$ roots, the 2nd derivative can have at most $(n-2)$ roots, the 1st derivative can have at most $(n-1)$ roots and hence $f(x)$, a polynomial of degree $n$, can have at most $n$ roots.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.