challenge for programmable calculators

[Han]

(12-25-2013 06:15 AM)Thomas Klemm Wrote:  

(12-25-2013 01:27 AM)Han Wrote:  Once you've solved a problem completely using mathematics, is there really any need for a program? I always held the opinion that programs were useful when we reached a mathematical stumbling block.

You'd still have to verify that the determinant \((1+a^2-9a)^2-4a(11a+1)\) is only positive for a = 1. So we're down at a loop of 7. Thus a program is still helpful.
Either that or you solve \((1+a^2-9a)^2-4a(11a+1)=0\) and figure that out by the roots.

Cheers
Thomas

This requires nothing more than calculus. Let \( f(x) = (x^2-9x+1)^2-4x(11x+1) \). Note that \( f(1) = 1 \) and \( f(2) = -15 \). Verify that \( f'(x) = 4x^3-54x^2+78x-22 \) and \( f ' ' (x) = 12x^2-108x+78 \). Using the quadratic formula, we see that \( f ' ' (x) = 0 \) when \( x = \frac{9}{2} \pm \frac{\sqrt{55}}{2} \). Since \( 7 < \sqrt{55} < 8 \),
\[ 0 < \frac{9}{2} - \frac{\sqrt{55}}{2} < 1 \]
\[ 8 < \frac{9}{2} + \frac{\sqrt{55}}{2} < 9 \]
We can easily check that \( f ' ' (x) < 0 \) on \( [ 1, 8 ] \) -- it's a parabola that opens upward and has two real zeros. Since \( f ' ' (x) < 0 \) on \( [ 1, 8 ] \), the function \( f'(x) \) is decreasing on the interval \( [1,8] \). Since \( f'(2) = -50 \), and \( f'(x) \) is decreasing, then \( f'(x) < 0 \) on the interval \( [2,8] \). This implies \( f(x) \) is decreasing on \( [ 2, 8 ] \). Since \( f(2) = -15 \), it follows that \( f(x) < 0 \) on \( [2,8] \). Given that the parameter \( a \in [ 1, 7 ] \cap \mathbb{Z} \), it follows that \( a=1 \) is the only value for which \( f(a) > 0 \).