Close approximation

Geometry Level 3

( tan ⁡ − 1 ( 2000 ) − tan ⁡ − 1 ( 1000 ) ) × 1000 = ? \large \left(\tan^{-1}(2000)-\tan^{-1}(1000)\right)\times1000 = \, ?

Round your answer to 3 decimal places.


The answer is 0.5.

Subh Mandal by subh mandal , 22, India

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2 solutions

Anirudh Sreekumar
Mar 23, 2017

tan ⁡ − 1 ( 2000 ) − tan ⁡ − 1 ( 1000 ) = tan ⁡ − 1 ( 2000 − 1000 1 + 2000 × 1000 ) = tan ⁡ − 1 ( 1000 1 + 2000 × 1000 ) \begin{aligned} \tan^{-1}(2000)-\tan^{-1}(1000)&=\tan^{-1}\left(\dfrac{2000-1000}{1+2000\times1000}\right)\\&=\tan^{-1}\left(\dfrac{1000}{1+2000\times1000}\right)\end{aligned}

we can see that the angle, θ = tan ⁡ − 1 ( 1000 1 + 2000 × 1000 ) \theta=\tan^{-1}\left(\dfrac{1000}{1+2000\times1000}\right) is really small

so,we can use the small angle approximation θ ≈ tan ⁡ θ \theta\approx\tan\theta

1000 θ ≈ 1000 tan ⁡ θ ≈ 1000 × ( 1000 1 + 2000 × 1000 ) ≈ 0.5 \begin{aligned}1000\hspace{1mm} \theta&\approx1000\tan\theta\\&\approx1000\times\left(\dfrac{1000}{1+2000\times1000}\right)\\&\approx0.5\end{aligned}

2 Helpful 0 Interesting 0 Brilliant 0 Confused
Subh Mandal
Mar 22, 2017

2 Helpful 0 Interesting 0 Brilliant 0 Confused

While the limit gives a roadmap for the problem, we would need to do some error analysis to support setting h = 1 1000 h = \frac{1}{1000} to approximate the integral.

In this case, we can use simple estimates to show ∣ ∫ 1 2 d x x 2 − ∫ 1 2 d x x 2 + h 2 ∣ ≤ h 2 \left|\int_1^2 \frac{dx}{x^2} - \int_1^2 \frac{dx}{x^2+h^2}\right| \leq h^2 so the error is at most h 2 = 1 0 − 6 h^2 = 10^{-6} , which is well under the required input accuracy.

Brian Moehring - 4 years, 2 months ago

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