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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.

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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