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## ISI MStat PSB 2008 Problem 3 | Functional equation

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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 $g$ 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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## ISI MStat PSB 2007 Problem 3 | Application of L’hospital Rule

This is a very beautiful sample problem from ISI MStat PSB 2007 Problem 3 based on use of L’hospital Rule . Let’s give it a try !!

## Problem– ISI MStat PSB 2007 Problem 3

Let f be a function such that $f(0)=0$ and f has derivatives of all order. Show that $\lim _{h \to 0} \frac{f(h)+f(-h)}{h^{2}}=f”(0)$
where $f”(0)$ is the second derivative of f at 0.

### Prerequisites

Differentiability

Continuity

L’hospital rule

## Solution :

Let L= $\lim _{h \to 0} \frac{f(h)+f(-h)}{h^{2}}$ it’s a $\frac{0}{0}$ form as f(0)=0 .

So , here we can use L’hospital rule as f is differentiable .

We get L= $\lim _{h \to 0} \frac{f'(h)-f'(-h)}{2h} = \lim _{h \to 0} \frac{(f'(h)-f'(0)) -(f'(-h)-f'(0))}{2h}$

= $\lim _{h \to 0} \frac{f'(h)-f'(0)}{2h} + \lim _{k \to 0} \frac{f'(k)-f'(0)}{2k}$ , taking -h=k .

= $\frac{f”(0)}{2} + \frac{f”(0)}{2}$ = $f”(0)$ . Hence done!

## Food For Thought

Let $f:[0,1] \rightarrow[0,1]$ be a continuous function such $f^{(n)} := f ( f ( \cdots ( f(n \text{ times} ))$ and assume that there exists a positive integer m such that $f^{(m)}(x)=x$ for all $x \in[0,1] .$ Prove that $f(x)=x$ for all $x \in[0,1]$

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## ISI MStat PSB 2013 Problem 2 | Application of sandwich Theorem

This is a very beautiful sample problem from ISI MStat PSB 2013 Problem 2 based on use of Sandwich Theorem . Let’s give it a try !!

## Problem– ISI MStat PSB 2013 Problem 2

Let f be a real valued function satisfying $|f(x)-f(a)| \leq C|x-a|^{\gamma}$ for some $\gamma>0$ and $C>0$
(a) If $\gamma=1,$ show that f is continuous at a
(b) If $\gamma>1,$ show that f is differentiable at a

### Prerequisites

Differentiability

Continuity

Limit

Sandwich Theorem

## Solution :

(a) We are given that $|f(x)-f(a)| \leq C|x-a|$ for some $C>0$.

We have to show that f is continuous at x=a . For this it’s enough to show that $\lim_{x\to a} f(x)=f(a)$.

$|f(x)-f(a)| \leq C|x-a| \Rightarrow f(a)-C|x-a| \le f(x) \le f(a) + C|x-a|$

Now taking limit $x \to a$ we have , $\lim_{x\to a} f(a)-C|x-a| \le \lim_{x\to a} f(x) \le \lim_{x\to a} f(a) + C|x-a|$

Using Sandwich theorem we can say that $\lim_{x\to a} f(x) = f(a)$ . Since $\lim_{x\to a} -C|x-a| = \lim_{x\to a} C|x-a|=0$

Hence f is continuous at x=a proved .

(b) Here we have to show that f is differentiable at x=a for this it’s enough to show that the $\lim_{x\to a} \frac{f(x)-f(a)}{x-a}$ exists .

We are given that , $|f(x)-f(a)| \leq C|x-a|^{\gamma}$ for some $\gamma>1$ and $C>0$ ,

which implies $|\frac{f(x)-f(a)}{x-a} | \le C|x-a|^{\gamma -1}$

$\Rightarrow -C|x-a|^{\gamma -1} \le \frac{f(x)-f(a)}{x-a} \le C|x-a|^{\gamma -1}$

Now taking $\lim_{x\to a}$ we get by Sandwich theorem $\lim_{x\to a}\frac{f(x)-f(a)}{x-a} =0$ i.e f'(a)=0 .

Since , $\lim_{x\to a} C|x-a|^{\gamma -1} = \lim_{x\to a} -C|x-a|^{\gamma -1} = 0$ , for $\gamma >1$.

Hence f is differentiable at x=a proved .

## Food For Thought

$f : R \to R$ be such that $|f(x)-f(a)| \le k|x-y|$ for some $k \in (0,1)$ and all $x,y \in R$ . Show that f must have a unique fixed point .

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## ISI MStat PSB 2014 Problem 2 | Properties of a Function

This is a very beautiful sample problem from ISI MStat PSB 2014 Problem 2 based on the use and properties of a function . Let’s give it a try !!

## Problem– ISI MStat PSB 2014 Problem 2

Let $a_{1}<a_{2}<\cdots<a_{m}$ and $b_{1}<b_{2}<\cdots<b_{n}$ be real numbers such
that $\sum_{i=1}^{m}\left|a_{i}-x\right|=\sum_{j=1}^{n}\left|b_{j}-x\right| \text { for all } x \in \mathbb{R}$
Show that $m=n$ and $a_{j}=b_{j}$ for $1 \leq j \leq n$

### Prerequisites

Differentiability

Mod function

continuity

## Solution :

Let , $\sum_{i=1}^{m}\left|a_{i}-x\right|=\sum_{j=1}^{n}\left|b_{j}-x\right|=f(x) \text { for all } x \in \mathbb{R}$

Then , $f(x)=\sum_{i=1}^{m}\left|a_{i}-x\right|$ is not differentiable at $x=a_1,a_2, \cdots , a_m$ —(1)

As we know the function $|x-a_i|$ is not differentiable at $x=a_i$ .

Again we have , $f(x) = \sum_{j=1}^{n}\left|b_{j}-x\right|$ it also not differentiable at $x= b_1,b_2, \cdots , b_n$ —-(2)

Hence from (1) we get f has m non-differentiable points and from (2) we get f has n non-differentiable points , which is possible only when m and n are equal .

And also the points where f is not differentiable must be same in both (1) and (2) .

As we have the restriction that $a_{1}<a_{2}<\cdots<a_{m}$ and $b_{1}<b_{2}<\cdots<b_{n}$ .

So , we have $a_{j}=b_{j}$ for $1 \leq j \leq n$ .

## Food For Thought

$a<b \in \mathbb{R} .$ Let $f:[a, b] \rightarrow[a, b]$ be a continuous and differentiable on (a,b) . Suppose that $\left|f^{\prime}(x)\right| \leq \alpha<1$ for all $x \in(a, b)$ for some $\alpha .$ Then prove that there exists unique $x \in[a, b]$ such that $f(x)=x$