Minimum of $f(x,y)=\sum_{n=0}^{+\infty}\frac{(n^2−nx−y)^2}{2^n}$

Question

Show that $$f(x,y)=\sum_{n=0}^{+\infty}\frac{(n^2−nx−y)^2}{2^n}$$ is defined on $\Bbb{R}^2$, it has a minimum and find for which couple $(x, y)$ the minimum is reached.

The first point is okay, for the second one I tried to expand the square to havea quadratic form whose coefficients are series : but this is not a quadratic form.

I do not know how to continue ? I think it's about distance ?

Thanks in advance for your time.

Answer 1

Jimmy's response shows how to sum the series to find the polynomial $f(x,y)$, and suggests using standard derivative methods to minimize the function. Here's a hint for a faster way of finding the minimum of $f(x,y)$ without calculus:

$$\begin{align} f(x,y)&=150-52x+6x^2-12y+4xy+2y^2\\ &=\frac23(3x+y-13)^2+\frac43(y^2+4y+28)\\ &=\frac23(3x+y-13)^2+\frac43(y+2)^2+32\\ &\ge32. \end{align}$$

$f(x,y)$ will only attain this absolute minimum value of $32$ when the square terms are zero, i.e., when

$$\begin{cases} 3x+y-13=0,\\ y+2=0. \end{cases}$$

Now solve the system of equations.

Answer 2

Expanding gives us $f(x,y) = \displaystyle\sum_{n = 0}^{\infty}2^{-n}\left(n^4-2 n^3 x+n^2 x^2-2 n^2 y+2 n x y+y^2\right)$

We can show that $\displaystyle\sum_{n = 0}^{\infty}2^{-n} = 2$, $\displaystyle\sum_{n = 0}^{\infty}n2^{-n} = 2$, $\displaystyle\sum_{n = 0}^{\infty}n^22^{-n} = 6$, $\displaystyle\sum_{n = 0}^{\infty}n^32^{-n} = 26$, $\displaystyle\sum_{n = 0}^{\infty}n^42^{-n} = 150$.

Thus, $f(x,y) = 150-52x+6x^2-12y+4xy+2y^2$.

Now, just take the derivative w.r.t. $x$ and $y$, set them equal to zero, and solve.

$\dfrac{\partial f}{\partial x} = 12x + 4y - 52 = 0$, $\dfrac{\partial f}{\partial y} = 4x + 4y - 12 = 0$

$(x,y) = (5,-2)$.