Optimizing n for ε in a Square Root Expression

[peripatein]

Hello,

Is there a better way to find n such that for every n>n0 |sqrt(n^2+3)-sqrt(n^2-1)|<ε?
I have tried reformulating the expression on the left side so that only one of the square roots remains, to not much avail. The only way I could solve it was by squaring both sides. I did get the correct expression for ε for every ε<1, but it does not work for larger ε's, hence, alas, it is wrong.
Any word of advice, please?

 

[Mark44]

Hello,

Is there a better way to find n such that for every n>n0 |sqrt(n^2+3)-sqrt(n^2-1)|<ε?
I have tried reformulating the expression on the left side so that only one of the square roots remains, to not much avail. The only way I could solve it was by squaring both sides. I did get the correct expression for ε for every ε<1, but it does not work for larger ε's, hence, alas, it is wrong.
Any word of advice, please?

I don't see why you would need to worry about $\epsilon$ ≥ 1. In fact, you should probably be able to assume that $\epsilon$ < 1 - the whole goal of this exercise is to show that you can choose n large enough so that the difference of the two square roots is arbitrarily small.

 

[peripatein]

Thank you very much for replying.
n0 >= sqrt(epsilon^4-4epsilon^2+16)*1/(2epsilon)
Which not only seems very messy but also yields wrong results for epsilon greater than 1. The question states that epsilon is positive, so I presume my expression must hold for any epsilon, even if greater than 1.
I know that the right answer should probably be 1/(2epsilon), but how may I derive it?

 

[peripatein]

My bad, I meant that the right answer should probably be 2/epsilon!

 

[Mark44]

Thank you very much for replying.
n0 >= sqrt(epsilon^4-4epsilon^2+16)*1/(2epsilon)

If $\epsilon$ is suitably small, then $\epsilon^4$ and $\epsilon^2$ can be neglected. This makes the expression on the right side approximately equal to √16/$2\epsilon$, which gives you the (corrected) result below.

Which not only seems very messy but also yields wrong results for epsilon greater than 1. The question states that epsilon is positive, so I presume my expression must hold for any epsilon, even if greater than 1.

I know that the right answer should probably be (edit: corrected from later post) [STRIKE]1/(2epsilon)[/STRIKE] 2/$\epsilon$, but how may I derive it?

 

[peripatein]

But what if epsilon were equal to 40? In that case epsilon^4 and epsilon^2 may not be neglected and 2/epsilon still yields the right value for n0, whereas sqrt(epsilon^4-4epsilon^2+16)*1/(2epsilon) doesn't!

 

[Mark44]

But what if epsilon were equal to 40?

Why would anyone care? $\epsilon$ is almost always used to represent small (close to 0) numbers.

In that case epsilon^4 and epsilon^2 may not be neglected and 2/epsilon still yields the right value for n0, whereas sqrt(epsilon^4-4epsilon^2+16)*1/(2epsilon) doesn't!

 

[HallsofIvy]

But what if epsilon were equal to 40? In that case epsilon^4 and epsilon^2 may not be neglected and 2/epsilon still yields the right value for n0, whereas sqrt(epsilon^4-4epsilon^2+16)*1/(2epsilon) doesn't!

Since that is a decreasing function of n, and is 2 when n= 1, as soon as \epsilon is larger than 2, you can take n_0= 1.