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# ISI MStat PSB 2008 Problem 3 | Functional equation

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 $$f''(1)$$ .

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