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I want to solve this without using calculator.

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  • $\begingroup$ This can be done with a geometric series. With two terms the answer is $a =999.9996666$ which is a slight improvement over the bound $a = 1000$. $\endgroup$
    – Brad
    Jul 11, 2014 at 12:14

4 Answers 4

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Use that $$ \frac{1}{1+x^2}<\frac{1}{x^2}. $$

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  • $\begingroup$ we need to the smallest value of $a$, so that the equation is satisfied, but that does not answer this particular question $\endgroup$
    – Holy cow
    Jul 11, 2014 at 12:26
  • $\begingroup$ In that case Riccardo's answer is exact. $\endgroup$ Jul 11, 2014 at 14:29
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$\displaystyle\int_a^{+\infty}\dfrac{1}{1+x^2}=\dfrac{\pi}{2}-\tan^{-1}(a)$, so the inequality becomes: $\dfrac{\pi}{2}-\tan^{-1}(a)\lt 10^{-3}$, so: $a\gt\cot\left(\dfrac{1}{1000}\right)$

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  • $\begingroup$ Nice answer, if it wasn't constrained by the requirement to compute this without a calculator? $\endgroup$
    – Chinny84
    Jul 11, 2014 at 12:14
  • $\begingroup$ @Chinny84: the integral is a standard integral and the inequality isn't so difficult to solve. $\endgroup$ Jul 11, 2014 at 12:17
  • $\begingroup$ I know. I was talking about the cot function but I guess since x<<1 then you can use the first few terms :)! +1. Thanks $\endgroup$
    – Chinny84
    Jul 11, 2014 at 12:21
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Hint: The integral is a standard trigonometric integral.

Extra hint: The arctan of infinity is $\pi/2$.

Last hint: at small values of $x$, $\cos(x)$ becomes close to 1, and $\sin(x)$ is close to $x$.

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Ok... I not sure if this is right, if anyone points out the flaws that would help me as well! Took so long to type up using this new math formatting!

The curve $$ 1/{1+x^2} $$

gets closer to $$ \frac {1}{x^2} $$ as x gets larger, so if you want an approximation this might work $$ \int^{\infty} _a \frac {1}{x^2}dx = \frac{-1}{\infty} - \frac{-1}a $$ $$ \frac{-1}{\infty} - \frac{-1}a < \frac1{1000} $$ $$ \frac1a<\frac1{1000} $$ $$ a>1000 $$

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    $\begingroup$ If you shift the curve to the right, you end up with $1/(1+(x-1)^2)$, which is not quite what you wrote. $\endgroup$
    – user105475
    Jul 11, 2014 at 12:25
  • $\begingroup$ But $1/(1+x^2)<1/x^2$, so $a>1000$ works. This is the approach Julián Aguirre suggested. $\endgroup$ Jul 11, 2014 at 12:26
  • $\begingroup$ Bhoot, you're right, silly mistake there! $\endgroup$
    – Kevin
    Jul 11, 2014 at 12:30

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