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# Telescopic Continuity | ISI MStat 2015 PSB Problem 1

This problem is a simple application of the sequential definition of continuity from ISI MStat 2015 PSB Problem 1 based on Telescopic Continuity.

## Problem- Telescopic Continuity

Let $f: R \rightarrow R$ be a function which is continuous at 0 and $f(0)=1$
Also assume that $f$ satisfies the following relation for all $x$ :
$$f(x)-f(\frac{x}{2})=\frac{3 x^{2}}{4}+x$$ Find $f(3)$.

## Solution

$$f(3)-f(\frac{3}{2})=\frac{3 \times 3^{2}}{4}+3$$

$$f(\frac{3}{2})-f(\frac{\frac{3}{2}}{2})=\frac{3 \times \frac{3}{2}^{2}}{4}+\frac{3}{2}$$

$$f(\frac{3}{2^2})-f(\frac{\frac{3}{2^2}}{2})=\frac{3 \times \frac{3}{2^2}^{2}}{4}+\frac{3}{2^2}$$

$\cdots$

$$f(\frac{3}{2^n})-f(\frac{\frac{3}{2^n}}{2})=\frac{3 \times \frac{3}{2^n}^{2}}{4}+\frac{3}{2^n}$$

Add them all up. That's the telescopic elegance.

$$f(3)-f(\frac{3}{2^{n+1}})= \frac{3 \times 3^{2}}{4} \times \sum_{k = 0}^{n} \frac{1}{2^{2k}} + 3 \times \sum_{k = 0}^{n} \frac{1}{2^k} \rightarrow [*]$$

Observe that $a_n \to 0 \Rightarrow f(a_n) \to f(0) = 1$ since, $f(x)$ is continuous at $x=0$.

Hence take limit $n \to \infty$ on $[*]$, and we get $f(3) - f(0) = \frac{3 \times 3^{2}}{4} \times \frac{4}{3} + 3 \times 2 = 15$.

## Food for Thought

• Find the general function from the given condition.
• $$f(x)-f(\frac{x}{2})=g(x)$$ and g(x) is continuous, then prove that $g(0) =0$.
• What if $g(x)$ is not continuous?

This problem is a simple application of the sequential definition of continuity from ISI MStat 2015 PSB Problem 1 based on Telescopic Continuity.

## Problem- Telescopic Continuity

Let $f: R \rightarrow R$ be a function which is continuous at 0 and $f(0)=1$
Also assume that $f$ satisfies the following relation for all $x$ :
$$f(x)-f(\frac{x}{2})=\frac{3 x^{2}}{4}+x$$ Find $f(3)$.

## Solution

$$f(3)-f(\frac{3}{2})=\frac{3 \times 3^{2}}{4}+3$$

$$f(\frac{3}{2})-f(\frac{\frac{3}{2}}{2})=\frac{3 \times \frac{3}{2}^{2}}{4}+\frac{3}{2}$$

$$f(\frac{3}{2^2})-f(\frac{\frac{3}{2^2}}{2})=\frac{3 \times \frac{3}{2^2}^{2}}{4}+\frac{3}{2^2}$$

$\cdots$

$$f(\frac{3}{2^n})-f(\frac{\frac{3}{2^n}}{2})=\frac{3 \times \frac{3}{2^n}^{2}}{4}+\frac{3}{2^n}$$

Add them all up. That's the telescopic elegance.

$$f(3)-f(\frac{3}{2^{n+1}})= \frac{3 \times 3^{2}}{4} \times \sum_{k = 0}^{n} \frac{1}{2^{2k}} + 3 \times \sum_{k = 0}^{n} \frac{1}{2^k} \rightarrow [*]$$

Observe that $a_n \to 0 \Rightarrow f(a_n) \to f(0) = 1$ since, $f(x)$ is continuous at $x=0$.

Hence take limit $n \to \infty$ on $[*]$, and we get $f(3) - f(0) = \frac{3 \times 3^{2}}{4} \times \frac{4}{3} + 3 \times 2 = 15$.

## Food for Thought

• Find the general function from the given condition.
• $$f(x)-f(\frac{x}{2})=g(x)$$ and g(x) is continuous, then prove that $g(0) =0$.
• What if $g(x)$ is not continuous?

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