Calculus

Calculus Level 3

True or False?

\quad For 0 < x < π 2 0<x< \dfrac \pi 2 , the inequality tan ⁡ x < x + x 3 6 \tan x < x+ \dfrac {x^3}6 is satisfied.

True False

Aakhyat Singh by Aakhyat Singh , 20, India

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2 solutions

tan ⁡ x \tan{x} diverges at x = ( π 2 ) − x=(\frac{\pi}{2})^- to + ∞ +\infty . Therefore, as polynomials don't diverge for real values of x x , tan ⁡ x > > x + x 3 6 \tan{x}>>x+\frac{x^3}{6} for x x close to ( π 2 ) − (\frac{\pi}{2})^-

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A Former Brilliant Member - 3 years, 6 months ago
Sabhrant Sachan
Sep 12, 2017

Taylor Expansion of tan ⁡ x is given by : tan ⁡ x = x + x 3 3 + 2 x 5 15 + … x + x 3 3 > x + x 3 6 ∀ x ∈ R ⟹ tan ⁡ x > x + x 3 6 ∀ x ∈ ( 0 , π 2 ) \text{ Taylor Expansion of } \tan{x} \text{ is given by :} \\ \tan{x} = x+\dfrac{x^3}{3}+\dfrac{2x^5}{15}+\dots \\ x+\dfrac{x^3}{3} > x+\dfrac{x^3}{6} \hspace{4mm} \forall \hspace{4mm} x \in \mathbb{R} \\ \implies \tan{x} > x+\dfrac{x^3}{6} \hspace{4mm} \forall \hspace{4mm} x \in \left( 0,\dfrac{\pi}{2} \right)

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