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Suppose, we have a quadratic equation $Q(x)$ and, we shift the input by $\alpha$ i.e: the function is now $Q(x -\alpha)$, such that $Q(x) = Q(x -\alpha)$

Suppose,

$$ Q(x) = ax^2 + bx +c$$

and,

$$ Q(x-\alpha) = a(x-\alpha)^2 + b(x-\alpha) +c$$

Since, $$Q(x-\alpha)= Q(x)$$ for all $x$ in $\mathbb{R}$

The left hand side,

$$ Q(\alpha) = 0 + 0 + c$$

And, the right hand side,

$$ Q(\alpha) = a\alpha^2 + b \alpha +c$$

and since,

$$ Q(x) = Q(x-\alpha)$$

$$ a \alpha^2 + b\alpha +c = c$$

and, hence,

$$ \alpha = \{ 0, -\frac{b}{a} \}$$

This means that,

$$ Q(x- 0 ) = Q( x + \frac{b}{a} ) = Q(x)$$

For all x!

I can't grasp the result which I have got, why is there a unique shift such the function does not change at all? thinking of larger polynomials suggests that for cubics there must be two such shifts such that the cubic function is invariant and more generally for a 'n' degree polynomial, n-1 such shifts

Edit: I solved most part of the problem myself, if anyone can give some geometric insight, then I"ll accept that answer :)

tryst with freedom
  • 11,538

2 Answers2

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It is not possible for a quadratic $f(x)$ to be invariant under a linear shift of $x$. Let $f(x)=Ax^2+Bx+C$, if we transform $x \to y-p$, the we get $$f(y-p)=A(y-p)^2+B(y-p)+C = Ay^2+(B-2Ap)y+Ap^2-Bp+c$$, now if it has to be invariant then $(B-2Ap)=B, Ap^2-Bp+C=C \implies Ap=0, \& Bp=0,$ which implies that the assume quadratic does not exist.

But $f(x)$ can be invariant under reflection about $x=p/2$ when $f(x)=f(p-x).$ We can cobstruct such a quadratic as $$f(x)=Ax^2+A(p-x)^2+Bx(p-x)+C= (2A-B)x^2-(B-2Ap)x+Ap^2+C.$$ This quadratic for any arbitrary values of $A,B,C,p$ is invariant under $x\to p-x$.

Z Ahmed
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$$a(x-\alpha)^2+b(x-\alpha)+c=ax^2+(b-2a\alpha)x+a\alpha^2-b\alpha+c=ax^2+bx+c$$

can only be an identity when

$$b=b-2a\alpha$$

or $$\alpha=0.$$

More generally, no polynomial is invariant to an horizontal shift. $P(x)-P(x-\alpha)$ is a polynomial of one degree less and cannot be identically zero.