Help with Solving an Inequation: |n| < (ε(n³+1))/2

[ShizukaSm]

Homework Statement

I was doing a certain problem, and in the end, I got this inequation which I can't solve. Probably due to lack of skill, can someone give me a hand on it?

|n|&lt;\frac{\epsilon(n^3 + 1)}{2}

The Attempt at a Solution

|n|(1-\frac{\epsilon*n^2}{2})&lt; \frac{\epsilon}{2}

I tried rearranging in this way, but still, no dice. I know that n must be positive, and I actually plan on plugging different values of epsilon, but still, I can't solve this inequation.

 

[STEMucator]

What is the problem exactly?

 

[ShizukaSm]

The number 27 on this list.

 

[Mark44]

Homework Statement

I was doing a certain problem, and in the end, I got this inequation which I can't solve. Probably due to lack of skill, can someone give me a hand on it?

|n|&lt;\frac{\epsilon(n^3 + 1)}{2}

The Attempt at a Solution

|n|(1-\frac{\epsilon*n^2}{2})&lt; \frac{\epsilon}{2}

I tried rearranging in this way, but still, no dice. I know that n must be positive, and I actually plan on plugging different values of epsilon, but still, I can't solve this inequation.

Since n > 0, you don't need the absolute values.

Are you trying to solve for n or for $\epsilon$? If you're trying to solve for $\epsilon$, that's pretty simple, but if you're trying to solve for n, you have a messy third-degree polynomial that makes factoring difficult, if not impossible. That would leave only numerical approximations or graphical solutions as potential approaches.

BTW, we call them inequalities in English.

 

[ShizukaSm]

Oh, I apologize, I'm not from an English speaking country, so Inequality, got it.

I'm trying to solve for ϵ.

 

[Mark44]

We typically use letters such as M or N to represent large numbers, and Greek letters $\epsilon$ and $\delta$ for small (close to zero) numbers.

You want to find a (large) number N such that 2n/(n^3 + 1) > N. For any given value of N you can use trial and error to find the smallest value of n that satisfies the inequality.

 

[ShizukaSm]

Wait, wait, I thought that :

<br /> \\a_n = \frac{2n}{n^3 + 1}\\<br /> |{a_n-L}| &lt; \epsilon \\<br /> L =\lim_{n-&gt;\infty} a_n = 0\\<br /> |{a_n}| &lt; \epsilon \\<br />

And thus, my problem was:

|\frac{2n}{n^3 + 1}|&lt;\epsilon\\

And N would be an particular "n" for a particular epsilon, isn't that the case?

 

[Mark44]

I'm sorry, I didn't look close enough at your sequence, and mistakenly thought that you needed to show that it diverged. Your use of $\epsilon$ was correct. I'm sorry for steering you the wrong direction.

An $\epsilon$ would normally be given, so your goal would be to find n so that the inequality was true. The usual way would be to use trial and error.

Note that, since n > 0, you don't need absolute values in the inequality.

 

[ShizukaSm]

Ah, that's fine, don't worry!

While I agree that I don't need absolute values, my problem still is with the inequality :/ How would I go about solving it?

 

[STEMucator]

So applying the definition :

\forall \epsilon &gt; 0, \exists N \space | \space n&gt;N \Rightarrow |a_n -L| &lt; \epsilon

So now :

|a_n - L| =|\frac{2n}{n^3 + 1} - 0| = |\frac{2n}{n^3 + 1}| = \frac{2n}{n^3 + 1} since n > 0.

Lets say we take the case where ε=1 as per your question. Okay, now that we have something plausible to work with consider :

\frac{2n}{n^3 + 1} &lt; 1
\frac{n}{n^3 + 1} &lt; \frac{1}{2}
n^2 + \frac{1}{n} &gt; 2

Can you continue from here?

 

[ShizukaSm]

Thanks for the input Zondrina!

I can't, that's exactly the point I made in OP, I don't know how to solve inequalities in this form. I usually try to factor them out, but I can't really find a factorization in this case. I can't apply the Rational Roots Theorem either, so, in short, I don't know how to proceed.

 

[Ray Vickson]

Homework Statement

I was doing a certain problem, and in the end, I got this inequation which I can't solve. Probably due to lack of skill, can someone give me a hand on it?

|n|&lt;\frac{\epsilon(n^3 + 1)}{2}

The Attempt at a Solution

|n|(1-\frac{\epsilon*n^2}{2})&lt; \frac{\epsilon}{2}

I tried rearranging in this way, but still, no dice. I know that n must be positive, and I actually plan on plugging different values of epsilon, but still, I can't solve this inequation.

In the context of the original problem you posted, you do NOT need to find the entire range of n values; you just need to find an N = N(ε) such that for all n ≥ N the inequality holds. Finding the *smallest* such N is hard; finding some N that works is a lot easier.

So, for n > 0 and ε > 0, we have $\epsilon \, n^3/2 < \epsilon \, (n^3+1)/2,$, so if we satisfy $n \leq \epsilon \, n^3/2,$ that value of n will also satisfy the original inequality. Solving $n \leq \epsilon \, n^3/2$ is a much easier problem.