Question

If $f\left( x \right).f\left( y \right) = f\left( {x + y} \right)$ , $\forall $ real x, y and $f\left( 1 \right) = \dfrac{1}{2}$ , then $\sum\limits_{r = 0}^\infty {f\left( r \right) = } $ ?

Answer 1

Hint: We can find the $f\left( 0 \right)$ using the given equation by substituting $x = 0$ and $y = 1$ . Similarly, we can find the values of $f\left( 2 \right)$ , $f\left( 3 \right)$ , etc. Then we can write the summation of these values and it forms a geometric progression. Then we can find the value of the required summation by finding the sum of the infinite GP.

Complete step-by-step answer:
We have the relation $f\left( x \right).f\left( y \right) = f\left( {x + y} \right)$ … (1)
Let us take $x = 0$ and $y = 1$ ,
 $ \Rightarrow f\left( 0 \right).f\left( 1 \right) = f\left( {0 + 1} \right)$
On dividing throughout with $f\left( 1 \right)$ , we get,
 $ \Rightarrow f\left( 0 \right) = \dfrac{{f\left( 1 \right)}}{{f\left( 1 \right)}}$
 $ \Rightarrow f\left( 0 \right) = 1$
Now we can substitute $x = 1$ and $y = 1$ ,in equation (1)
 $ \Rightarrow f\left( 1 \right).f\left( 1 \right) = f\left( {1 + 1} \right)$
It is given that $f\left( 1 \right) = \dfrac{1}{2}$ , so we get,
 $ \Rightarrow f\left( 2 \right) = \dfrac{1}{2} \times \dfrac{1}{2}$
 $ \Rightarrow f\left( 2 \right) = {\left( {\dfrac{1}{2}} \right)^2}$
Now we can substitute $x = 2$ and $y = 1$ ,in equation (1)
 $ \Rightarrow f\left( 2 \right).f\left( 1 \right) = f\left( {2 + 1} \right)$
It is given that $f\left( 1 \right) = \dfrac{1}{2}$ , and we have $f\left( 2 \right) = {\left( {\dfrac{1}{2}} \right)^2}$ so we get,
 $ \Rightarrow f\left( 3 \right) = {\left( {\dfrac{1}{2}} \right)^2} \times \dfrac{1}{2}$
 $ \Rightarrow f\left( 3 \right) = {\left( {\dfrac{1}{2}} \right)^3}$
So generally, we can write,
 $ \Rightarrow f\left( n \right) = {\left( {\dfrac{1}{2}} \right)^n}$ for every natural number n.
Now we can write the summation,
Let $S = \sum\limits_{r = 0}^\infty {f\left( r \right)} $
On expanding the summation, we get,
 $ \Rightarrow S = f\left( 0 \right) + f\left( 1 \right) + f\left( 2 \right) + .... + f\left( n \right)$
From the above results, we can substitute the value,
 $ \Rightarrow S = 1 + \dfrac{1}{2} + {\left( {\dfrac{1}{2}} \right)^2} + {\left( {\dfrac{1}{2}} \right)^3} + .... + {\left( {\dfrac{1}{2}} \right)^n}$
Now this forms an infinite geometric series with a=1 and common ratio $d = \dfrac{1}{2}$ . We know that for an infinite geometric series, the sum is given by the formula ${S_\infty } = \dfrac{a}{{1 - r}}$ .
 $ \Rightarrow S = \dfrac{a}{{1 - r}}$
On substituting the values of a and d, we get,
  $ \Rightarrow S = \dfrac{1}{{1 - \dfrac{1}{2}}}$
On simplification, we get,
 $ \Rightarrow S = \dfrac{2}{{2 - 1}}$
$ \Rightarrow S = 2$
Therefore, the required summation $\sum\limits_{r = 0}^\infty {f\left( r \right)} = 2$.

Note: We used the concept of functions to solve the problem. As the summation has r=0, we must find the value of the function at 0. We must find the values of the function for the 1st few natural numbers to understand the trend and formulate a general equation for $f\left( n \right)$ . For expanding the summation, we give the values 0 to n in the place of r. We know that geometric progression is a series where any term is given by multiplying a common ratio with the previous term. As the GP is an infinite GP and the common ratio is less than zero we used the equation ${S_\infty } = \dfrac{a}{{1 - r}}$ to find its sum.