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

Related Program

ISI MStat PSB 2008 Problem 10
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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 \( x \in[0,1] \)

Similar Problems and Solutions

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ISI MStat PSB 2008 Problem 10
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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 \(f(x)=x\)

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ISI MStat PSB 2008 Problem 10
Outstanding Statistics Program with Applications

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