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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 thatg(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

Subscribe to Cheenta at Youtube


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 thatg(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

Subscribe to Cheenta at Youtube


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