How Do You Estimate a Value for 1/tan(1/1000) Without a Calculator?

[TsAmE]

Homework Statement

Determine how large the number a has to be so that:

\int_{a}^{\infty} \frac{1}{1 + x^{2}} dx < 0.001

Homework Equations

None.

The Attempt at a Solution

I tried to evaluate the left hand side and got a final answer of:

a > \frac{\pi}{2} - \frac{1}{1000}

but the correct answer was 1000

I think these 2 steps in my working out might be the problem:

from tan(arctan(a)) > \frac{\pi}{2} - \frac{1}{1000} to:

arctan(a) > \frac{\pi}{2} - \frac{1}{1000}

a > \frac{\pi}{2} - \frac{1}{1000}

 

[phyzguy]

Why did you write tan(arctan(a))? Since the indefinite integral is just arctan(x), and arctan(x) ->pi/2 as x->infinity, you should just write pi/2-arctan(a)<.001, or arctan(a)>pi/2-.001. Do you know how to solve this?

 

[Mentallic]

The last 3 lines do not continue from each other at all. Firstly, how did you get tan(arctan(a))? How does this continue on to get just arctan(a) and then simply a?

 

[TsAmE]

Sorry I made a mistake in the post. I have corrected the question.

 

[TsAmE]

My last post was ambiguous. I meant I had corrected the mistake when I typed the question. Could someone please check my working?

 

[hunt_mat]

Perform the integral to obtain:
<br /> \int_{a}^{\infty}\frac{dx}{1+x^{2}}=\tan^{-1}(\infty )-\tan^{-1}a=\frac{\pi}{2}-\tan^{-1}a&lt;0.001<br />
Then re-arrange to obtain:
<br /> \tan^{-1}a&gt;\frac{\pi}{2}-0.001<br />
Take tan to obtain the answer...

 

[Mentallic]

My last post was ambiguous. I meant I had corrected the mistake when I typed the question. Could someone please check my working?

I don't see any change in your original post. Your last 3 lines of working are still wrong.
hunt_mat has given the answer.

 

[TsAmE]

I don't see any change in your original post. Your last 3 lines of working are still wrong.
hunt_mat has given the answer.

I reached the same answer as hunt as in:

arctan of &gt; \frac{\pi}{2} - \frac{1}{1000}

and tried solving for a only to get:

a &gt; \frac{\pi}{2} - \frac{1}{1000}

I don't see how I can get an answer of 1000 when there is a pi/2 - 0.001

My latex code is giving me problems when I try to edit it so that's probably why you didnt see my correction.

 

[hunt_mat]

So take tan of what I wrote down:
<br /> a&gt;\tan\Bigg(\frac{\pi}{2}-\frac{1}{1000}\Bigg) =\frac{1}{\tan (1/1000)}\approx 1000<br />
That is how you finish off the calculation.

 

[TsAmE]

So take tan of what I wrote down:
<br /> a&gt;\tan\Bigg(\frac{\pi}{2}-\frac{1}{1000}\Bigg) =\frac{1}{\tan (1/1000)}\approx 1000<br />
That is how you finish off the calculation.

With respect to the inequality why is a > and not < ? Do you swap the sign around when you use the inverse trig function?

Also a&gt;\tan\Bigg(\frac{\pi}{2}-\frac{1}{1000}\Bigg) = 0.0274... on my calculator as opposed to 1000?

 

[hunt_mat]

You use:
<br /> \tan x=\frac{\sin x}{\cos x}<br />
And then you note:
<br /> \sin\Bigg(\frac{\pi}{2}-\frac{1}{1000}\Bigg) =\cos\Bigg(\frac{1}{1000}\Bigg)<br />
and
<br /> \cos\Bigg(\frac{\pi}{2}-\frac{1}{1000}\Bigg) =\sin\Bigg(\frac{1}{1000}\Bigg)<br />
So you get the same as before, for small x, we have the following:
<br /> \tan x\approx x<br />
Regarding the inequality, tan x is an increasing function, so the inequality remains the same, regarding your calculator, I think you may have entered the sum incorrectly.

 

[Mentallic]

With respect to the inequality why is a > and not < ? Do you swap the sign around when you use the inverse trig function?

I don't understand why you think it would be < ?
You already had arctan(a)&gt;\pi/2-0.001 so why would it suddenly change sign?

Also a&gt;\tan\Bigg(\frac{\pi}{2}-\frac{1}{1000}\Bigg) = 0.0274... on my calculator as opposed to 1000?

Your calculator is set in degrees and you want radians.

 

[TsAmE]

Thanks I now see where I went wrong. One last thing I want to know is how would you know that \frac{1}{\tan (1/1000)}\approx 1000 without using a calculator, cause I am not allowed to use calculators in my tests.

 

[hunt_mat]

Use a Maclaurin series to show that:
<br /> \tan x =x+o(x),\quad \sin x=x+o(x),\quad\cos x=1-\frac{x^{2}}{2}+o(x^{2})<br />

 

[Mentallic]

Thanks I now see where I went wrong. One last thing I want to know is how would you know that \frac{1}{\tan (1/1000)}\approx 1000 without using a calculator, cause I am not allowed to use calculators in my tests.

For small x, x\approx tan(x) so \frac{1}{tan(1/1000)}\approx \frac{1}{1/1000}=1000