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 $f(3)$

Prerequisites

Telescopic Sum
Continuity of a Function

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 $f(3)$

Prerequisites

Telescopic Sum
Continuity of a Function

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?

Telescopic Continuity | ISI MStat 2015 PSB Problem 1

Problem- Telescopic Continuity

Prerequisites

Solution

Food for Thought

Related

Problem- Telescopic Continuity

Prerequisites

Solution

Food for Thought

Related

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