Question

An odd function is symmetric about the vertical line $x=a\left(a > 0\right)$ and if $\displaystyle\sum _{r = 0}^{\infty} \left[f \left(1 + 4 a r\right)\right]^{r}=8$ , then find the numerical value of $8f\left(1\right)$ $\left(\right.$ where $0 < f\left(1\right) < 1\left.\right)$ .

Answer 1

Given that,
$f\left(- x\right)=-f\left(x\right)$
and $f\left(2 a - x\right)=f\left(x\right)$
By adding both equations,
$\Rightarrow f\left(- x\right)+f\left(2 a - x\right)=0$
Replace $x$ with $\left(- x\right)$
$f\left(x\right)+f\left(x + 2 a\right)=0....\left(1\right)$
Replace $x$ with $x+2a$
$\Rightarrow f\left(x + 2 a\right)+f\left(x + 4 a\right)=0....\left(2\right)$
Subtract equation $\left(1\right)$ and equation $\left(2\right)$
$\Rightarrow f\left(x\right)-f\left(x + 4 a\right)=0$
$\Rightarrow f\left(x\right)=f\left(x + 4 a\right)$
$\Rightarrow f\left(x\right)$ is a periodic function and its period is $4a$
$\Rightarrow f\left(1\right)=f\left(1 + 4 a\right)=f\left(1 + 8 a\right)$
Now,
$\displaystyle\sum _{r = 0}^{\infty }f\left(1 + 4 a r\right)^{r}=8$
$\Rightarrow \left[f \left(1\right)\right]^{0}+\left[f \left(1 + 4 a\right)\right]+\left[f \left(1 + 8 a\right)\right]^{2}+...........=8$
$\Rightarrow 1+f\left(1\right)+\left[f \left(1\right)\right]^{2}+\left[f \left(1\right)\right]^{3}+..........=8$
$\Rightarrow \frac{1}{1 - f \left(1\right)}=8$
$\Rightarrow f\left(1\right)=\frac{7}{8}$
$\Rightarrow 8f\left(1\right)=7$