Let f: R R be a continuous function satisfying f(0)=1 and f(2 x)-f(x)=x for all x R.

Question

Let $f: \mathbb{R} \rightarrow \mathbb{R}$ be a continuous function satisfying $f(0)=1$ and $f(2 x)-f(x)=x$ for all $x \in \mathbb{R}$. If $\lim _\limits{n \rightarrow \infty}\left\{f(x)-f\left(\frac{x}{2^n}\right)\right\}=G(x)$, then $\sum_\limits{r=1}^{10} G\left(r^2\right)$ is equal to

Accepted Answer

$$\begin{aligned}
& f(0)=1, f(2 x)-f(x)=x \
& \text { Replace } x \rightarrow \frac{x}{2} \
& f(x)-f\left(\frac{x}{2}\right)=\frac{x}{2}\quad\text{..... (1)}
\end{aligned}$$

Again, Replace $x \rightarrow \frac{x}{2}$

$f\left(\frac{x}{2}\right)-f\left(\frac{x}{2^2}\right)=\frac{x}{2^2}\quad\text{...... (2)}$

: $\qquad$ : $\qquad$ :

$$\begin{aligned}
& f\left(\frac{x}{2^{n-1}}\right)-f\left(\frac{x}{2^n}\right)=\frac{x}{2^n} \
& \text { Adding }(1)+(2)+(3)+\ldots+(\mathrm{n}) \
& \text { We get } f(x)-f\left(\frac{x}{2^n}\right)=\frac{x}{2}+\frac{x}{2^2}+\ldots .+\frac{x}{2^n} \
& \lim _{n \rightarrow \infty}\left(f(x)-f\left(\frac{x}{2^n}\right)\right)=\lim _{n \rightarrow \infty}\left(\frac{x}{2}+\frac{x}{2^n}+\ldots+\frac{x}{2^n}\right)
\end{aligned}$$

$$\begin{aligned}
& f(x)-f(0)=\frac{\frac{x}{2}}{\frac{1}{2}} \
& \Rightarrow G(x)=x \
& \Rightarrow G\left(r^2\right)=r^2 \
& \Rightarrow \sum_{r=1}^{10} G\left(r^2\right)=\sum_{r=1}^{10} G\left(r^2\right) \
& =\frac{(10)(11)(21)}{6}=(55) 7 \
& \Rightarrow 385
\end{aligned}$$

Solutions · 1