This is a very beautiful sample problem from ISI MStat PSB 2008 Problem 3 based on Functional equation . Let’s give it a try !!

**Problem**– ISI MStat PSB 2008 Problem 3

Let \(g\) be a continuous function with \( g(1)=1 \) such that \( g(x+y)=5 g(x) g(y) \) for all \( x, y .\) Find \( g(x) \).

**Prerequisites**

Continuity & Differentiability

Differential equation

Cauchy’s functional equation

## Solution :

We are g is continuous function such that\( g(x+y)=5 g(x) g(y) \) for all \( x, y \) and g(1)=1.

Now putting x=y=0 , we get \( g(0)=5{g(0)}^2 \Rightarrow g(0)=0\) or , \(g(0)= \frac{1}{5} \) .

If g(0)=0 , then g(x)=0 for all x but we are given that g(1)=1 . Hence contradiction .

So, \(g(0)=\frac{1}{5} \) .

Now , we can write \( g'(x)= \lim_{h \to 0} \frac{g(x+h)-g(x)}{h} = \lim_{h \to 0} \frac{5g(x)g(h)-g(x)}{h} \)

\(= 5g(x) \lim_{h \to 0} \frac{g(h)- \frac{1}{5} }{ h} = 5g(x) \lim_{h \to 0} \frac{g(h)- g(0) }{ h} = 5g(x)g'(0) \) (by definition)

Therefore , \( g(x)=5g'(0)g(x)= Kg(x) \) , for some constant k ,say.

Now we will solve the differential equation , let y=g(x) then we have from above

\( \frac{dy}{dx} = ky \Rightarrow \frac{dy}{y}=k{dx} \) . Integrating both sides we get ,

\( ln(y)=kx+c \) c is integrating constant . So , we get \( y=e^{kx+c} \Rightarrow g(x)=e^{kx+c} \)

Solve the equation g(0)=1/5 and g(1)=1 to get the values of K and c . Finally we will get , \( g(x)=\frac{1}{5} e^{(ln(5)) x} =5^{x-1}\).

**But there is a little mistake in this solution** .

What’s the mistake ?

Ans- Here we assume that g is differentiable at x=0 , which may not be true .

**Correct Solution** comes here!

We are given that \( g(x+y)=5 g(x) g(y) \) for all \( x, y .\) Now taking log both sides we get ,

\( log(g(x+y))=log5+log(g(x))+log(g(y)) \Rightarrow log_5 (g(x+y))=1+log_5 (g(x))+log_5 (g(y)) \)

\( \Rightarrow log_5 (g(x+y)) +1= log_5 (g(x))+1+log_5 (g(y)) +1 \Rightarrow \phi(x+y)=\phi(x)+\phi(y) \) , where \( \phi(x)=1+log_5 (g(x)) \)

It’s a cauchy function as \(\phi(x)\) is also continuous . Hence , \( \phi(x)=cx \) , c is a constant \( \Rightarrow 1+log_5 (g(x))=cx \Rightarrow g(x)=5^{cx-1} \).

Now \(g(1)=1 \Rightarrow 5^{c-1}=1 \Rightarrow c=1 \).

Therefore , \(g(x)=5^{x-1} \)

## Food For Thought

Let \( f:R to R \) be a non-constant , 3 times differentiable function . If \( f(1+ \frac{1}{n})=1\) for all integer n then find \( f”(1) \) .

This solution is incomplete as it assumes that the function is differentiable at zero. Upon taking logarithm of both sides, the functional equation can be converted to a form of Cauchy’s functional equation (https://en.wikipedia.org/wiki/Cauchy%27s_functional_equation) and its solution can be inferred from it.