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Here is equation :

$$2f(x+2)+3f(x-2)=x+5$$

How to find $f(1)$ ?

Umberto
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qwr
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  • I don't think there is any way to solve for $f(1)$. – Thomas Andrews Apr 24 '14 at 12:49
  • Is there any other condition on this problem, like maybe that $f$ is a linear function? – Thomas Andrews Apr 24 '14 at 12:56
  • @ThomasAndrews actually I think that I have to proof that this function can be only linear. With linearity it is solved easily. But indeed on question there is not such condition. Anyway I gave to see if there is other optoins to solve this problem – qwr Apr 24 '14 at 13:01
  • No, it definitely can be non-linear. – Thomas Andrews Apr 24 '14 at 13:04
  • Does your first part of answer proving that without aditional condition it can not be solved? Like my collegues sugested me to use recursion or system equation – qwr Apr 24 '14 at 13:05
  • Yes. It's true that if $f(x)$ is a polynomial, it must be linear, but being a polynomial is not a condition of the problem. – Thomas Andrews Apr 24 '14 at 13:06

3 Answers3

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If $x=y-2$ then $2f(y+4)+3f(y)=y+3$, or $$f(y+4)=\frac{y+3-3f(y)}{2}$$

So you can choose any set of values for $f$ on $[0,4)$ and complete the function to the real line.

This means that $f(1)$ is not determined by this equation alone.

Perhaps there was another condition on $f$? Was it supposed to be linear?

Thomas Andrews
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You could make some hypothesis. If you suppose that $f(x)$ is a polynomial, then you could write $f(x)=ax+b$ and find $a$ and $b$ and therefore find $f(1)$. But you need start somewhere...

Umberto
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If we assume that $f(x)$ is linear, that is to say that $f(x)=a x+b$, we have $$2f(x+2)+3f(x-2)=2[a(x+2)+b]+3[a(x-2)+b]=5ax-2a+5b$$ If this has to be equal to $x+5$, we have two equations and two unknowns $$5a=1$$ $$5b-2a=5$$ from which $a=\frac{1}{5}$ and $b=\frac{27}{25}$. Then, $f(1)=\frac {28} {25}$.

If you assume that $f(x)$ is quadratic, that is to say that $f(x)=a x^2+b x+c$, expanding $2f(x+2)+3f(x-2)$ will lead to the fact that $a=0$; then, if polynomial, $f(x)$ is linear.

In fact, you could easily prove that, if $f(x)$ is a polynomial, it must be of the same degree as the rhs.

Claude Leibovici
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