The polynomial \(\p(x)\) is given by
\[
\ds \p(x)= x^n +\sum\limits_{r=0}^{n-1}a_rx^r\,,
\]
where \(a_0\), \(a_1\), \(\ldots\) , \(a_{n-1}\) are fixed real
numbers and \(n\ge1\). Let \(M\) be the greatest value of \(\big\vert \p(x) \big\vert\) for $\vert x \vert\le
1\(. Then Chebyshev's theorem states that \)M\ge 2^{1-n}$.
Prove Chebyshev's theorem in the case \(n=1\) and verify that Chebyshev's theorem holds in the following cases:
\( \p(x) = x^2 - \frac12\,\);
\(\p(x) = x^3 - x \,\).
Use Chebyshev's theorem to show that the curve
$ \ y= 64x^5+25x^4-66x^3-24x^2+3x+1
\ $
has at least one turning point in the interval \(-1\le x \le 1\).
Solution:
If \(n = 1\) the theorem is \(\max_{x \in [-1,1]} \left ( |x + a_0 |\right) \geq 1\), but clearly \(\max(1+a_0, |a_0 - 1|) \geq 1\) (taking according to the sign of \(a_0\))
\( \p(x) = x^2 - \frac12\,\) - take \(x = 0\) then \(|p(0)| = \frac12 \geq 2^{1-2} = \frac12\)
\(\p(x) = x^3 - x \,\). take \(x = \frac1{\sqrt{2}}\), then \(|p\left ( \frac1{\sqrt{2}}\right)| = |\frac12 \frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}}| = \frac{1}{2\sqrt{2}} > \frac14 = 2^{1-3} \)
Consider \(p(x) = \frac{1}{64} \left ( 64x^5+25x^4-66x^3-24x^2+3x+1\right)\), then \(p\) satisfies the conditions of the theorem, therefore \(\max |p(x)| \geq 2^{1-5} = \frac1{16} = \frac{4}{64}\).
However, \(p(-1) = \frac{1}{64}\) and \(p(1) = \frac{3}{64}\), so it cannot be strictly increasing or decreasing and there must be at turning point to achieve \(\frac{4}{64}\)
Let \[\mathrm{f}(t)=\frac{\ln t}t\quad\text{ for }t>0.\]
Sketch the graph of \(\mathrm{f}(t)\) and find its maximum
value. How many positive values of \(t\) correspond to a
given value of \(\mathrm f(t)\)?
Find how many positive values of \(y\) satisfy
\(x^y=y^x\) for a given positive value of \(x\). Sketch the
set of points \((x,y)\) which satisfy \(x^y=y^x\) with \(x,y>0\).