\[\ \text{A monic polynomial f(x) with integer coefficients has the property:}\] \[\ \large f(x) * f''(x) - (f')^2 = -6x^{10} + 600x \] \[\ \text {Find f(2)}.\]
\[ f(x) = \sum_{k=0}^na_kx^k \]
Consider the largest that \(n\) could be.
\(f\) is degree \(n\) \(f'\) is degree \(n-1\) \(f''\) is degree \(n-2\)
So \(ff''\) is degree \(n+n-2 = 2n-2\) While \((f')^2\) is degree \(2(n-1) = 2n-2\)
it might be easier to just use differential equations.
This is actually making sense to me.
hm, how would that work?
Well \[ yy''-y'y' = -6x^{10}+600x \]
I'm not completely sure how to sole this diff eq other than guessing and checking.
Anyway \(2n-2\leq 10\implies n\leq 6\)
Supposing:\[ f(x) = \sum_{k=0}^{n}a_kx^k \]Then \[ f'(x) = \sum_{k=1}^{n}ka_kx^{k-1} \]And \[ f''(x) = \sum_{k=2}^{n}(k)(k-1)a_kx^{k-2} \]
We know that \((f')^2\) is even, so it can't contribute to the \(600x^1\) term
By even what I really should say is that all powers will be even.
Hmmm actually, maybe the best place to start is all the odd powers...
I may not be correct on the even powers thing.
I found something interesting though: \[ y^{(0)}y^{(2)} - y^{(1)}y^{(1)}=−6x^{10}+600x \]Differentiate: \[ y^{(1)}y^{(2)}+y^{(0)}y^{(3)} - y^{(2)}y^{(1)}-y^{(1)}y^{(2)} = y^{(0)}y^{(3)} - y^{(2)}y^{(1)} \]So \[ y^{(0)}y^{(3)} - y^{(2)}y^{(1)} = -60x^{9}+600 \]Differentiate again:\[ y^{(1)}y^{(3)} +y^{(0)}y^{(4)} - y^{(3)}y^{(1)}- y^{(2)}y^{(2)} = y^{(0)}y^{(4)} - y^{(2)}y^{(2)} \]So \[ y^{(0)}y^{(4)} - y^{(2)}y^{(2)} = -540x^{8} \]
From the first differentiation we get \[ a_0\times 3!a_3-2!a_2\times a_1=600 \]
Before that, we would have gotten \[ a_0\times 2!a_2-a_1\times a_1=0 \]
So \[ 2a_0a_2=(a_1)^2 \]
\[ 6a_0a_3-2a_2a_1=600 \]
Sham try 154
wio did you get 154 as a final answer? Jw
no. how did you get it?
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