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Prove that A Real Root Exists in [-1, 1]

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anemone

MHB POTW Director
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Feb 14, 2012
3,755
Given \(\displaystyle f(x)=5tx^4+sx^3+3rx^2+qx+p\) for $f(x)\in R$. If $r+t=-p$, prove that there is a real root for $f(x)=0$ in $[-1,1]$.
 

Ackbach

Indicium Physicus
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Jan 26, 2012
4,197
Some ideas:

We rewrite the polynomial as
$$f(x)=5tx^4+sx^3+3rx^2+qx-r-t=0,$$
where we are setting it equal to zero. We evaluate $f(1)$ and $f(-1)$:
\begin{align*}
f(1)&=5t+s+3r+q-r-t=4t+2r+s+q\\
f(-1)&=5t-s+3r-q-r-t=4t+2r-s-q.
\end{align*}
Idea: if $f(1)\cdot f(-1)<0$, then by the Intermediate Value Theorem, we would have shown there is a root in $[-1,1]$. Now
$$f(1)\cdot f(-1)=(4t+2r)^{2}-(s+q)^{2}.$$
Not seeing where to go with this. There's nothing stopping $s=q=0$, with $4t+2r\not=0$, in which case I have not proved what I want to prove.
 

zzephod

Well-known member
Feb 3, 2013
134
Given \(\displaystyle f(x)=5tx^4+sx^3+3rx^2+qx+p\) for $f(x)\in R$. If $r+t=-p$, prove that there is a real root for $f(x)=0$ in $[-1,1]$.
Consider the polynomial \(\displaystyle p(x)=tx^5+(s/4)x^4+rx^3+(q/2)x^2+px-(s/4+q/2)\).

Then \(\displaystyle p(1)=p(-1)=0\) and so \(\displaystyle p\) has an extremum in \(\displaystyle (-1,1)\), so \(\displaystyle p'(x)\) has a root in \(\displaystyle (-1,1)\) ...

.
 
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anemone

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Feb 14, 2012
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Thanks to both, Ackbach and zzephod for participating...

Consider the polynomial \(\displaystyle p(x)=tx^5+(s/4)x^4+rx^3+(q/2)x^2+px-(s/4+q/2)\).

Then \(\displaystyle p(1)=p(-1)=0\) and so \(\displaystyle p\) has an extremum in \(\displaystyle [-1,1]\), so \(\displaystyle p'(x)\) has a root in \(\displaystyle [-1,1]\) ...

.
WoW!!! What an elegant way to approach this problem! Well done, zzephod!(Clapping)

And there is another quite straightforward and beautiful way to tackle it as well...therefore I'll wait for the inputs from other members for now...(Sun)
 
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anemone

MHB POTW Director
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Feb 14, 2012
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Another method proposed by other to solve this challenge problem is by using the integration method:

Hint:

\(\displaystyle \int_{-1}^1 p(x) dx=0\)