A Problem I Found In A Paper Bag Made Out Of A Mathematics Question Paper

Define the sequence $a_k = f(2^{2-k}) - f(2^{1-k})$ , which clearly is equal to $f(2(2^{1-k})) - f(2^{1-k}) = \boxed{2^{1-k}}$ .

Next, consider the sum: $\displaystyle \sum_{k=1}^{\infty} a_k = \sum_{k=1}^{\infty} 2^{1-k}$ $\displaystyle \implies \displaystyle \sum_{k=1}^{\infty} f(2^{2-k}) - f(2^{1-k}) = \sum_{k=1}^{\infty} 2^{1-k}$ $\displaystyle \implies f(2) - f(0) = \sum_{k=1}^{\infty} 2^{1-k}$ $\displaystyle \implies f(2) - 1 = 2$ $\displaystyle \implies f(2) = 3$

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For those, who might be wondering what $f(x)$ can be, it can be only $f(x) = x + 1$ :

We have: $f\left(x\cdot2^{2-k}\right)-f\left(x\cdot2^{\left(1-k\right)}\right)=x\cdot2^{\left(1-k\right)}$

$\displaystyle \implies \sum_{k=0}^{\infty}\left(f\left(x\cdot2^{2-k}\right)-f\left(x\cdot2^{\left(1-k\right)}\right)\right)=\sum_{k=0}^{\infty}x\cdot2^{\left(1-k\right)}$

$\displaystyle \implies f\left(x\right)-f\left(0\right)=x \implies \boxed{f(x) = x + 1}$

Let's differentiate expression: $\frac d{dx}\left(f(2x)-f(x)\right)=\frac{d}{dx}x$

Thus we have: $2f'(x)-f'(x)=1$ $\Rightarrow$ $f'(x)=1$

Next we integrate expression: $\int f'(x)=f(x)=x+C$

When $x=0$ we have $f(0)=0+C=1$

Consequently $C=1$

In this way: $f(x)=x+1.$

Let's consider $f(x)$ as a function with form $\frac{ax+b}{c}$ , where:

• $a$ is the coefficient of x,

• $b$ is a value that is going to be add to x,

• $c$ is the denominator of $ax+b$ ,

, so $f(0) = \frac{a(0)+b}{c}=\frac{b}{c}$ , and because $f(0)=1$ it means that $b$ and $c$ need to be equal to produce 1 in their division to satisfy the value of the function when $x=0$ . Eventually, $\frac{b}{c}$ is converted to 1 and simplifies the function to $ax+1$ . We can go further by solving the equality:

$(2ax+1)-(ax+1)=x$

$2ax+1-ax-1=x$

$ax=x$

This means that $f(x)=x+1$ and now we can solve $f(2)$ which equals $2+1=3$ . We also know that there cannot be another function that satisfies our necessities because this would imply at least a small change in results.

$Q.E.D$

PD: I'm just a 14 years old boy, and I would like to see your opinion on my first demonstration comment, thank you! :)

Let f(x)=x+1

I solved this with some algebraic manipulation;

if $f(2x)-f(x)=x ; f(2)=f(1)+1.$

If we attempt to write $f(1)$ in the same way,

$f(1)=0.5+f(0.5)$

Similarly,

$f(0.5)=0.25+f(0.25)$

and

$f(0.25)=0.125+f(0.125)$

and so on and so forth.

Imagine we keep writing each function as the sum of half its input and the function of half it's input, until we boil down to $f(0)$ .

Since $f(0)=1$ , we find that $f(1)= \displaystyle \sum_{n=1}^{\infty} 2^{-n}+f(0) = 1+1 = 2$

Therefore $f(2) = 2+f(0) = 2+1 = \boxed{3}$

Let linear function be ax + b f(2x)-f(x)=2ax+b-ax-b=x ax=x a=1 f(0)= 0+b =1 b=1 f(2)=ax+b =2+1=3