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Options: (a)$4$, (b)$4f(0)$, (c)$4-f(0)$, (d)$4+f(0)$, (e)$16+f(0)$.

CORRECT ANSWER USING REDUCTION

Deep thanks to @martini and @A.S. , soo much respect .

Since we don't exactly know the nature of $f(x)$ we will start of by considering our function as $f(x)=g(x)+cx^2$ where $g(x)$ represents all the other functions (including constant values) inside $f(x)$ and $cx^{2}$ includes the no. of $x^{2}$ terms.

Now we have $cx^{2}-2\frac{cx^{2}}{4}+\frac{cx^{2}}{16}=x^{2}$

Comparing coefficients of $x^{2}$ in first equation we have $c=\frac{16}{9}$.

And

$g(x)-2g(\frac{x}{2})+g(\frac{x}{4})=0$

Now to solve this recurring equation we substitute $x=\frac{x}{2},\frac{x}{4},\frac{x}{8}......$ and add it up .

Yielding $g(\frac{x}{2^i})=ig(\frac{x}{2})-(i-1)g(x)$

Now as $i\to\infty$. Since $g$ is continuous we will have

$g(x)=g(x/2)$.

Now solving the recurring equation $g(x)-g(x/2)=0$ by substituting $x=\frac{x}{2},\frac{x}{4},\frac{x}{8}......$ and adding it up .

Now as $i\to\infty$. Since $g$ is continuous we will have

$g(x)=g(0)$

Hence $f(3)=g(3)+(16/9)3^2$.

Now as $g(3)=g(0)= f(0)-(16/9)0^2$.

We finally have $f(3)=f(0)+16.$

Sujith Sizon
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2 Answers2

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Let $x = 3\cdot 2^{-n}$, giving $$ f(3\cdot 2^{-n}) = 9 \cdot 2^{-2n} + 2 f(3 \cdot 2^{-(n+1)}) - f(3 \cdot 2^{-(n+2)}) $$ Write brevitatis causa $a_n := f(3 \cdot 2^{-n})$, we have $$ a_{n+2} - 2a_{n+1} + a_n = 9 \cdot 2^{-2n} $$ The general solution of the homogenous recursion $$ a_{n+2} - 2a_{n+1} + a_n = 0 $$ is $$ a_n^{\text{hom}} = \alpha n + \beta $$ to find a solution of the inhomogenous equation, we make the ansatz $\gamma \cdot 2^{-2n}$, giving $$ \gamma 2^{-2n}(2^{-4} - 2\cdot 2^{-2} + 1) = 9 \cdot 2^{-2n} \iff \gamma = 16 $$ So we have $$ f(3 \cdot 2^{-n}) = a_n = 16\cdot 2^{-2n} + \alpha n + \beta $$ for some $\alpha, \beta \in \mathbf R$. As $f$ is continuous, we must have $a_n \to f(0)$. As $16\cdot 2^{-2n}$ converges, this implies $\alpha = 0$, or $$ f(3 \cdot 2^{-n}) = 16 \cdot 2^{-2n} + \beta $$ For $n \to \infty$ this gives $f(0) = \beta$, or $$ f(3 \cdot 2^{-n}) = 16 \cdot 2^{-2n} + f(0) $$ for $n=0$ we have $$ f(3) = 16 + f(0) $$

martini
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    If I let $x = 3$ (as we want to find $f(3)$), we get $$ f(3) = \text{some term with $f(3/2)$, $f(3/4)$} $$ so to find $f(3)$, we have to find $f(3/2)$. That is, we now let $x = 3/2$. Giving us some term with $f(3/4)$ and $f(3/8)$. So, in general, we have values of $f$ at $3\cdot 2^{-n}$ for different values of $n$. That means, using $x = 3\cdot 2^{-n}$, we get a recursion for $a_n = f(3\cdot 2^{-n})$. – martini Sep 30 '15 at 11:42
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    +1. This approach shows that, for every $x$, $$f(x)=f(0)+\tfrac{16}9x^2.$$ – Did Sep 30 '15 at 11:48
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    $a_n$ is just a shortcut for the value $f(3 \cdot 2^{-n})$, which is a real number. – martini Sep 30 '15 at 13:16
  • @martini I tried doing halfway with A.S.'s process by assuming another function $g(x)$ but when i find $g(3)$ and substitute it I'm getting another answer , i think I'm going wrong somewhere , could you please correct me ? Here is what I've done http://i.stack..com/j9ryq.jpg – Sujith Sizon Sep 30 '15 at 14:56
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    Your link does not work – martini Sep 30 '15 at 14:59
  • @martini http://i.stack.imgur.com/j9ryq.jpg – Sujith Sizon Sep 30 '15 at 15:05
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    You dropped a two in the $\vdots$-step. – martini Sep 30 '15 at 15:21
  • @martini where precisely ? I have actually added up all the equations and 4 terms get partially cancelled out and two terms remain as such – Sujith Sizon Sep 30 '15 at 15:36
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Look for a simple solution first to "homogenize" the equation. $f(x)=cx^2$ works for $c=16/9$. That means that $g(x)=f(x)-cx^2$ satisfies $g(x)-2g(x/2)+g(x/4)=0$. Rewriting $g(x/4)=2g(x/2)-g(x)$ and iterating yields $$g(x/2^i)=ig(x/2)-(i-1)g(x)$$ Take $i\to\infty$. Since $g$ is continuous at $0$, we must have $g(x)=g(x/2)=g(0)$. Hence $f(x)=f(0)+(16/9)x^2$. Hence $$f(3)=f(0)+16.$$

A.S.
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