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Find $f(x)$ if $f(x)-2f(\frac{x}{2})+f(\frac{x}{4})=x^2$, where $x, f(x)\in (-\infty , \infty)$ and $f(x)$ is continuous.

Math Lover
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Samar Imam Zaidi
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  • what is $y$? There is no expression with $y$ in it. – Anurag A Aug 28 '18 at 21:08
  • I am using y in lieu of f(x) – Samar Imam Zaidi Aug 28 '18 at 21:09
  • Modified the question – Samar Imam Zaidi Aug 28 '18 at 21:10
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    Let $g(x) = f(x) - f(x/2)$ then $g(x) - g(x/2) = x^2$. Try to apply this many times: $g(x) = g(x/2) + x^2 = g(x/4) + x^2 + (x/2)^2 = \ldots$. Use continuity at $x=0$. – Winther Aug 28 '18 at 21:12
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    What have you tried? Worth noting: $f(x)$ is not uniquely determined. If $f(x)$ is one solution, then $f(x)+c$ is another. – lulu Aug 28 '18 at 21:13
  • Try $f(x)=ax^2$ and solve for $a$. – Paul Aug 28 '18 at 21:13
  • @lulu, that is not true. – Paul Aug 28 '18 at 21:20
  • how $f(x)+c$ is another solution din't get that – Deepesh Meena Aug 28 '18 at 21:20
  • @Paul. I believe it is. If $f(x)$ works then let $g(x)=f(x)+c$. Then $g(x)-2g\left( \frac x2 \right)+g\left(\frac x4\right)=f(x)+c-2f\left( \frac x2 \right)-2c+f\left(\frac x4\right)+c=x^2$ as the constants all cancel. – lulu Aug 28 '18 at 21:23
  • @lulu, ah I see yes that’s right. – Paul Aug 28 '18 at 21:25
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    And one can generalize the problem: let $\square f(x) = f(x) - f(x/2)$ then the equation $\square^n f(x) = p(x)$ has the solution $f(x) = f(0) + \sum_{k_1,k_2,\ldots,k_n=0}^\infty p\left(\frac{x}{2^{k_1+k_2+\ldots+k_n}}\right)$ which for case $p(x) = \sum a_m x^m$ simplifies to $f(x) = f(0) + \sum a_m x^m \left(\frac{2^m}{2^m-1}\right)^n$. The current problem is $n=2$ with $p(x) = x^2$. – Winther Aug 28 '18 at 21:50

3 Answers3

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OK, since the hint seems not to be enough, here is the solution.

First, Setting $g(x)=f(x)-f(\frac{x}{2})$ we have $$g(x)-g(\frac{x}{2})=x^2 \\ f(x)-f(\frac{x}{2})=g(x) \tag{*}$$

We solve the first equation.

By induction $$g(x)-g(\frac{x}{2^{n}})=x^2+\frac{x^2}{4}+\frac{x^2}{16}+...+\frac{x^2}{4^{n-1}}=x^2 \frac{1-\frac{1}{4^n}}{1-\frac{1}{4}}$$

Since $f$ is continuous at $x=0$, so is $g$. Taking the limit by $n$ you get $$g(x)-g(0)=\frac{4}{3} x^2$$

Moreover, $f(x)-f(\frac{x}{2})=g(x)$ implies $g(0)=0$.

Now, we need to solve $$f(x)-f(\frac{x}{2})=\frac{4}{3}x^2$$ which is exactly the equation above, multiplies by $\frac{4}{3}$. Therefore, solving exactly as above we get $$f(x)-f(0)=\frac{4}{3}\frac{4}{3}x^2=\frac{16}{9}x^2$$

This shows that $$f(x)=\frac{16}{9}x^2 +c$$ where $c=f(0)$ are all the solutions.

P.S. To make this more clear, the equation can be reduced via (*) to two equations of the type $$h(x)-h(\frac{x}{2})=r(x)$$ with $h, r$ continuous. This equation can be solved as above: By induction $$h(x)-h(\frac{x}{2^n})=\sum_{k=0}^{n-1}r(\frac{x}{2^n})$$ Using the continuity of $h$ at $x=0$ (we don't even need $h$ to be continuous at other points) we get $$h(x)-h(0)=\sum_{k=0}^{\infty}r(\frac{x}{2^n})$$ (Note here that, if the series is divergent it implies that there is no solution which is defined at that $x$).

So, by calculating the series, you get the solution $$h(x)=c+\sum_{k=0}^{\infty}r(\frac{x}{2^n})$$ where $c=h(0)$.

Math Lover
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N. S.
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$y=mx^2+c$ is one solution where $c\in R$ $$mx^2-2\cdot m \frac{x^2}{4}+m \frac{x^2}{16}=x^2$$ $$m=\frac{16}{9}$$ $$y=\frac{16}{9}x^2+c$$

As mentioned by @lulu if $f(x)$ is a solution than $f(x)+c$ is also a solution.

Deepesh Meena
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In this case, since the equation is linear and given the "clue" that the coefficient of $x^2$ is non-zero, then consider $f(x) = a + b x + c x^2$. This yields the possible polynomial solutions, but not necessarily all possible solutions. \begin{align} x^2 &= (a + b x + c x^2) - 2 \, \left( a + \frac{b x}{2} + \frac{c x^2}{4} \right) + \left( a + \frac{b x}{4} + \frac{c x^2}{16} \right) \\ &= \frac{b x}{4} + \frac{ 9 \, c \, x^2}{16}. \end{align} This leads to $b=0$ and $c = 16/9$ and the general solution of the form $$f(x) = a + \frac{16 \, x^2}{9}.$$

Leucippus
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