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 $f''(1)$ .

Similar Problems and Solutions

ISI MStat PSB 2008 Problem 10 — Outstanding Statistics Program with Applications

Outstanding Statistics Program with Applications

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 $f''(1)$ .

Similar Problems and Solutions

ISI MStat PSB 2008 Problem 10 — Outstanding Statistics Program with Applications

Outstanding Statistics Program with Applications

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One comment on “ISI MStat PSB 2008 Problem 3 | Functional equation”

abhishek sinha says:

August 14, 2020 at 1:37 pm

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.

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