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Find $f:\mathbb{R}\to\mathbb{R}$ such that $f(x)+f(x-2) = x^2-2x+1$ for all $x\in\mathbb{R}$, given that $f$ is is an even function.

I have tried putting $$x = x+2$$ and also $$x=-x$$but that gives the same thing, I don't know I am unable to make any progress. I also tried differentiating the function, because we were taught sometimes it could be solved like that, and I also know differentiation of an even function is an odd function, but it's also not helping.

Paprika7191
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Kutubkhan Bhatiya
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1 Answers1

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Suppose $p$ is a real polynomial that satisfies $p(x)+p(x-2)=x^2-2x+1$. Note that the leading terms of $p(x)$ and $p(x-2)$ do not cancel out when they sum together. Thus, $p$ must be a quadratic polynomial. By letting $p(x)=ax^2+bx+c$, we can find that $p(x)=\frac{x^2-1}{2}$ as in the comments.

Now, suppose $f$ is an even function that satisfies $$ f(x)+f(x-2)=x^2-2x+1. \tag{1} $$ We can write $f=p+g$, where $g$ is another function. Substituting back into $(1)$, we have $$ g(x)=-g(x-2) \tag{2} $$ for all $x\in\mathbb{R}$. Since $f$ and $p$ are even functions, $g$ is also even, so $$ g(x)=g(-x) \tag{3} $$ for all $x\in\mathbb{R}$. Note that equations $(2)$ and $(3)$ are also sufficient conditions for $p+g$ being a solution.

Equations $(2)$ and $(3)$ show that $g$ is completely determined by its image on $[0,1)$, and $g(x)=0$ for all $x\in 1+2\mathbb{Z}$. This is not hard to show. First, by equation $(2)$, $g$ is completely determined by its image on $[0,2)$. Now, for all $x\in (1,2)$, we have $g(x)=-g(x-2)=-g(2-x)$. Since $2-x\in (0,1)$ for all $x\in (1,2)$, the image of $g$ on $(1,2)$ is determined by the image of $g$ on $(0,1)$. Also, $g(1)=-g(2-1)=-g(1)$, so $g(1)=0$ and hence $g(x)=0$ for all $x\in 1+2\mathbb{Z}$ by equation $(2)$.

Since equations $(2)$ and $(3)$ are sufficient conditions for $p+g$ being a solution, any arbitrary function $h:[0,1)\to \mathbb{R}$ induces a solution as shown above. In other words,

An even function $f:\mathbb{R}\to\mathbb{R}$ satisfies $f(x)+f(x-2)=x^2-2x+1$ for all $x\in\mathbb{R}$ if and only if there exists a function $h:[0,1)\to\mathbb{R}$ such that $f(x)=\frac{1}{2}x^2-\frac{1}{2}+g(x)$ for all $x\in\mathbb{R}$, where $$g(x)= \begin{cases} h(x-\lfloor x \rfloor) & x\in (0,1)+4\mathbb{Z} \\ -h(1-(x-\lfloor x \rfloor)) & x\in (1,2)+4\mathbb{Z} \\ -h(x-\lfloor x \rfloor) & x\in (2,3)+4\mathbb{Z} \\ h(1-(x-\lfloor x \rfloor)) & x\in (3,4)+4\mathbb{Z} \\ h(0) & x\in 4\mathbb{Z} \\ -h(0) & x\in 2+4\mathbb{Z} \\ 0 & x\in 1+2\mathbb{Z} \end{cases}. $$

Paprika7191
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  • Why not just algebraically manipulate the given equation, or find various values to significantly narrow down the guesswork? – Derek Luna May 19 '23 at 14:50
  • @DerekLuna Where is the guesswork? I did not "guess" the answer. The idea behind my approach is seeing that there is a unique polynomial solution, and by adding a function $g$ we can obtain all other solutions. Finding various values does not help in this case. Also, I did algebraically manipulate the given equation to obtain equation $(2)$. – Paprika7191 May 19 '23 at 14:53
  • Actually, since the answer is not unique, you cannot obtain a solution by solving linear equations in $f(x), f(x+2), f(x-2)$, which is likely what you meant by "algebraically manipulating the given equation". – Paprika7191 May 19 '23 at 14:56
  • I was just confused by the way the question was asked, and the "work" shown by the asker. It seemed that one must simply find a function, which could have been done much more easily as you know. I did not know if the asker meant to ask for the general solution. – Derek Luna May 19 '23 at 14:58
  • @DerekLuna actually this was a question in my test, So I was only required to find the polynomial that he found out easily, but I was not aware that we could say directly it would be a polynomial which was quite obvious, but yeah that's the story. – Kutubkhan Bhatiya May 21 '23 at 13:31