If $f(x+1)+f(x-1)=4x^2-2x+10$ then what is $f(x)$ 
If $$f(x+1) +f(x-1)= 4x^2 -2x +10$$  then what is $f(x)$ 

What is strategy of solving this kind of  problems ?
Thank you for help
 A: *

*Find one solution


Try to find it as $f_0(x) = ax^2 +bx + c$
$f_0(x+1) + f_0(x-1) = 2ax^2 + 2a + 2bx +2c$ => $f_0(x)=2x^2-x + 3$


*$f(x) = f_0(x) + g(x)$ => $g(x-1) + g(x+1) =0$


Find all $g(x)$
A: First, you can see that 
$$f(x)=-f(x-2) + 4(x-1)^2-2(x-1)+10$$
$$f(x-2) = -f(x-4) + 4(x-3)^2-2(x-3)+10$$
$$f(x-4) = -f(x-6) + 4(x-5)^2 - 2(x-5)+10$$
$$\dots$$
You can see that you need a basis or some additional constraint to express $f(x)$, because now a solution is not unique. Once you find a solution $f_0(x)$, the general form of the answer would be $f(x) = f_0(x) + g(x)$, where $g(x)$ is a function satisfying $$g(x) = -g(x-2)$$
For example for $g(x) = \sin\left(\dfrac{\pi x}{2}\right)$ and $f_0(x) = 2x^2-x+3$ we can obtain a new solution
$$f_1(x) = 2x^2-x+3 + \sin\left(\dfrac{\pi x}{2}\right)$$
A: Using Taylor's formula for polynomials:


*

*$f(x+1)=f(x)+f'(x)\cdot 1+f''(x)\cdot\dfrac 12$,

*$f(x-1)=f(x)-f'(x)\cdot 1+f''(x)\cdot\dfrac 12$.


Therefore $\;f(x+1)+f(x-1)=2f(x)+f''(x)=4x^2-2x+10$. If we set $f(x)=ax^2+bx+c$, this equation simplifies to
$$ax^2+bx+c+a=2x^2-x+5,\enspace\text{whence}\quad a=2, b=-1, c=3.$$
A: Hint...let $f(x)=ax^2+bx+c$ and equate coefficients
A: $$ 4x^2 -2x +10 = 2(x+1)^2-(x+1)+3+2(x-1)^2-(x-1)+3$$
which means $f(x)=2x^2-x+3$
