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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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