Proving that a fifth-degree polynomial has a root using just the IVT

In summary, the conversation discusses the proof of a case involving a polynomial function with different coefficients. The proof involves showing that the function approaches positive infinity for positive values of x and negative infinity for negative values of x. By defining a maximum value M based on the coefficients, it is shown that there exists an x in the interval [-5M, 5M] such that the function equals 0, proving the existence of a root for the polynomial function.
  • #1
Eclair_de_XII
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Homework Statement
Let ##f(x)=a_0+a_1x+a_2x^2+a_3x^3+a_4x^4+x^5##, where ##a_i\in \mathbb{R}##. Show that there is an ##x\in \mathbb{R}## such that ##f(x)=0##, without citing the Fundamental Theorem of Algebra.
Relevant Equations
Intermediate Value Theorem: Let ##g## be a real-valued function whose domain is contained in the field of real numbers. Suppose that for some ##a,b##, ##a<b##, in the domain, ##g(a)<0## and ##g(b)>0##. Then there is a ##c\in (a,b)## such that ##g(c)=0##.
I consider three cases, based on the sign of ##a_0##.

if ##a_0 == 0##:

Set ##x=0##.

\begin{align*}
f(0)&=&a_0+a_1\cdot 0+a_2\cdot 0^2+a_3\cdot0^3+a_4\cdot0^4+0^5\\
&=&a_0+0\\
&=&0+0\\
&=&0
\end{align*}

elif ##a_0<0##:

Define ##M=\max\{|a_i|:1\leq a_i\leq 5\}## and set ##x=5(M+1)\neq 0##.

Note: ##M+1>1##, so ##\frac{1}{M+1}<1##.

So for any summand in the bottom-most equation, the following must hold:

\begin{align*}

\frac{a_k}{(5(M+1))^k}&\leq&\frac{|a_k|}{5(M+1)}\cdot\frac{1}{(5(M+1))^{k-1}}\\

&\leq&\frac{1}{5}\cdot\frac{1}{(5(M+1))^{k-1})}\\

&<&\frac{1}{(5(M+1))^{k-1}}\\

&\leq&1

\end{align*}

Therefore, ##f(x)>0## for this choice of ##x##.

\begin{align*}

f(x)&=&x^{5}\left(\frac{a_0}{x^5}+\frac{a_1}{x^4}+\frac{a_2}{x^3}+\frac{a_3}{x^2}+\frac{a_4}{x}+1\right)\\

%

&=&5^5(M+1)^5 \left(1-\sum_{i=0}^4 \frac{|a_i|}{(5(M+1))^{5-i})}\right)\\

%

&\geq&5^5(M+1)^5 \left(1-\sum_{i=0}^4 \frac{1}{(5)^{5-i}}\right)\\

%

&>&5^5(M+1)^5 \left(1-\sum_{i=0}^4 \frac{1}{5}\right)\\

%

&=&5^5(M+1)^5 \left(1-5\cdot\frac{1}{5}\right)\\

%

&=&5^5(M+1)^5 \left(1-1\right)\\

%

&=&0

\end{align*}

I want someone to double-check my work before I post my solution for the case where ##a_0>0##.

But I feel like I could just take define some function ##h(x)=-f(x)## and the result would follow.
 
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  • #2
No need for proof by cases. It is enough to show that [itex]f(x) \to \infty[/itex] as [itex]x \to \infty[/itex] and [itex]f(x) \to -\infty[/itex] as [itex]x \to -\infty[/itex].

From that you can deduce the existence of [itex]R_1 \in \mathbb{R}[/itex] and [itex]R_2 \in \mathbb{R}[/itex] such that [itex]R_1 < R_2[/itex] and [itex]f(R_1) < 0 < f(R_2)[/itex].
 
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  • #3
Your proof has taken a wrong turn by dividing it into cases depending on ##a_0##. Other than that, I can see that your proof so far can not be correct since your definition of M only depends on the coefficients between 1 and 5. There may not be any coefficients like that, so M would not be defined. Try M = max({1,|##a_i##|, i=0,4}). Then if x>5*M, what can you say about f(x)? What about if x < -5*M?
 
  • #4
FactChecker said:
There may not be any coefficients like that

Oh. I had meant to put ##1\leq i\leq 4##, not ##1\leq a_i\leq 5##. In any case, I can see that the inequality immediately after that fails if all ##a_i=0##. Anyway:

FactChecker said:
Try ##M = \max({1,|a_i|, i=0,4})##. Then if x>5*M, what can you say about f(x)? What about if x < -5*M?

if ##x >5M##:

##x>5|a_i|## or alternatively, ##\frac{1}{5}>\frac{|a_i|}{x}## for all ##i##
##x>1## and ##\frac{1}{x}<1##

\begin{align}

f\left(x\right)&=&x^5\left(1+\sum_{k=0}^4 \frac{a_k}{x^{5-k}}\right)\\

&\geq&x^5\left(1-\sum_{k=0}^4 \frac{|a_k|}{x^{5-k}}\right)\\

&=&x^5\left(1-\sum_{k=0}^4 \frac{|a_k|}{x}\cdot\frac{1}{x^{5-k-1}}\right)\\

&>&x^5\left(1-\frac{1}{5}\sum_{k=0}^4 \frac{1}{x^{5-k-1}}\right)\\

&>&x^5\left(1-\frac{1}{5}\sum_{k=0}^4 1\right)\\

&=&x^5\left(1-\frac{1}{5}\cdot5\right)\\

&=&x^5\left(1-1\right)\\

&=&0
\end{align}

elif ##x<-5M##:

##x<0## since ##x<-5M## and ##-5M < 0##

\begin{align}
f\left(x\right)&=&x^5\left(1+\sum_{k=0}^4 \frac{a_k}{x^{5-k}}\right)\\
&=&x^5\left(1-\frac{a_3}{x^2}-\frac{a_1}{x^4}+\sum_{k=0}^2 \frac{a_{2k}}{x^{2k}}\right)\\
&\leq&x^5\left(1-\frac{2}{5}+\sum_{k=0}^2 \frac{a_{2k}}{|x^{2k}|}\right)\\
&\leq&x^5\left(1-\frac{2}{5}-\sum_{k=0}^2 \frac{|a_{2k}|}{|x^{2k}|}\right)\\
&\leq&x^5\left(1-\frac{2}{5}-\frac{1}{5}\sum_{k=0}^2 1\right)\\
&=&x^5\left(1-\frac{2}{5}-\frac{3}{5}\right)\\
&=&0
\end{align}

And so there is an ##x## in ##[-5M,5M]## such that ##f(x)=0##, by the intermediate value theorem.
 
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Related to Proving that a fifth-degree polynomial has a root using just the IVT

1. How do you use the Intermediate Value Theorem (IVT) to prove that a fifth-degree polynomial has a root?

The IVT states that if a function is continuous on a closed interval [a,b] and takes on values f(a) and f(b) at the endpoints, then for any value c between f(a) and f(b), there exists at least one value x in the interval [a,b] such that f(x) = c. To use the IVT to prove that a fifth-degree polynomial has a root, we need to show that the polynomial is continuous on a closed interval and takes on values with opposite signs at the endpoints. This will guarantee the existence of at least one root in the interval.

2. What is a fifth-degree polynomial?

A fifth-degree polynomial is a polynomial function with the highest degree term being x^5. It can be written in the form f(x) = ax^5 + bx^4 + cx^3 + dx^2 + ex + f, where a, b, c, d, e, and f are constants.

3. Can any fifth-degree polynomial be proven to have a root using the IVT?

No, not all fifth-degree polynomials have a root. The IVT can only be used to prove the existence of a root if the polynomial is continuous and takes on values with opposite signs at the endpoints of a closed interval.

4. How do you determine which interval to use for the IVT when proving a fifth-degree polynomial has a root?

To determine the interval, we need to find the values of the polynomial at the endpoints and see if they have opposite signs. If they do, then we can use that interval to apply the IVT. If not, we may need to try a different interval or use a different method to prove the existence of a root.

5. Can the IVT be used to find the exact value of the root of a fifth-degree polynomial?

No, the IVT only guarantees the existence of a root, it does not provide the exact value. To find the exact value, we may need to use other methods such as factoring or the quadratic formula.

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