Is it possible that by using the following transformations: \[ f(x) \mapsto x^2 \cdot f \left(\frac{1}{x}+1 \right) \ \ \ \text{or} \ \ \ f(x) \mapsto (x-1)^2 \cdot f\left(\frac{1}{x-1} \right)\] the function $f(x)=x^2+5x+4$ is sent to the function $g(x)=x^2+10x+8$ ?
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Tags: function, quadratics, algebra proposed, algebra
FantasyLover
13.03.2011 20:42
We claim that it is impossible for the function $f(x)=x^2+5x+4$ to be sent to $g(x)=x^2+10x+8$ using any combination of the given transformations.
Without loss of generality, we will only consider the transformations of the quadratic function $ax^2+bx+c$.
If the transformation of the first kind is applied to $ax^2+bx+c$, we obtain $x^2\left(a\left(\frac{1}{x}+1\right)^2+b\left(\frac{1}{x}+1\right)+c\right)=(a+b+c)x^2+(2a+b)x+a$.
If the transformation of the second kind is applied to $ax^2+bx+c$, we obtain $(x-1)^2\left(\frac{a}{(x-1)^2}+\frac{b}{x-1}+c\right)=cx^2+(b-2c)x+(a-b+c)$.
Now, observe that both of these transformations preserve the determinant of the original function.
However, as the determinant of $f(x)$ is not equal to that of $g(x)$, it is impossible to transform $f(x)$ into $g(x)$, as desired. $\blacksquare$
mike93
21.03.2011 19:05
We can also observe that the parity of the coefficent of $x$ doesn't change during any kind of operation. Initially it was odd, so we can not obtain an even number.
littletush
30.10.2011 08:06
it suffices to prove that the discriminant remains unchanged. since the discriminants of the two are 9 and 68 respectively,the transformation can't be achieved.