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

[Eclair_de_XII]

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)\\

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

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

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

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

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

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&=&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.

 

[pasmith]

No need for proof by cases. It is enough to show that $f(x) \to \infty$ as $x \to \infty$ and $f(x) \to -\infty$ as $x \to -\infty$.

From that you can deduce the existence of $R_1 \in \mathbb{R}$ and $R_2 \in \mathbb{R}$ such that $R_1 < R_2$ and $f(R_1) < 0 < f(R_2)$.

 

[FactChecker]

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?

 

[Eclair_de_XII]

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:

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.