This is a very beautiful sample problem from ISI MStat PSB 2008 Problem 3 based on Functional equation . Let's give it a try !!
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 \( g(x) \).
Continuity & Differentiability
Differential equation
Cauchy's functional equation
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 , \(g(x)=5^{x-1} \)
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 \( f''(1) \) .
In 2026, the following Cheenta students have been successful for Indian Statistical Institute's M.Stat Entrance. They ranked within the first 50 in the entire country in these entrances. I.S.I. M.Stat Entrance
In 2026, the following Cheenta students have been successful for Indian Statistical Institute's B.Stat Entrance and Chennai Mathematical Institute's B.Sc. Math Entrance. They ranked within the first 200 in the entire country in these entrances. Most of these students attended the problem solving workshops regularly, which happen 5 days every week. CMI B.Sc. Math Entrance […]
In 2025, 8 students from Cheenta Academy cracked the prestigious Regional Math Olympiad. In this post, we will share some of their success stories and learning strategies. The Regional Mathematics Olympiad (RMO) and the Indian National Mathematics Olympiad (INMO) are two most important mathematics contests in India.These two contests are for the students who are […]
Cheenta Academy proudly celebrates the success of 27 current and former students who qualified for the Indian Olympiad Qualifier in Mathematics (IOQM) 2025, advancing to the next stage — RMO. This accomplishment highlights their perseverance and Cheenta’s ongoing mission to nurture mathematical excellence and research-oriented learning.
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.