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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 be a continuous function with such that for all Find .

### Prerequisites

Continuity & Differentiability

Differential equation

Cauchy's functional equation

## Solution :

We are g is continuous function such that for all and g(1)=1.

Now putting x=y=0 , we get or , .

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

So, .

Now , we can write  (by definition)

Therefore , , for some constant k ,say.

Now we will solve the differential equation , let y=g(x) then we have from above . Integrating both sides we get , c is integrating constant . So , we get Solve the equation g(0)=1/5 and g(1)=1 to get the values of K and c . Finally we will get , .

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 for all Now taking log both sides we get ,  , where It's a cauchy function as is also continuous . Hence , , c is a constant .

Now .

Therefore , ## Food For Thought

Let be a non-constant , 3 times differentiable function . If for all integer n then find .

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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 be a continuous function with such that for all Find .

### Prerequisites

Continuity & Differentiability

Differential equation

Cauchy's functional equation

## Solution :

We are g is continuous function such that for all and g(1)=1.

Now putting x=y=0 , we get or , .

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

So, .

Now , we can write  (by definition)

Therefore , , for some constant k ,say.

Now we will solve the differential equation , let y=g(x) then we have from above . Integrating both sides we get , c is integrating constant . So , we get Solve the equation g(0)=1/5 and g(1)=1 to get the values of K and c . Finally we will get , .

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 for all Now taking log both sides we get ,  , where It's a cauchy function as is also continuous . Hence , , c is a constant .

Now .

Therefore , ## Food For Thought

Let be a non-constant , 3 times differentiable function . If for all integer n then find .

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

1. abhishek sinha says:

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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