Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

This is a very simple question. I want to prove that $\frac{4n+3}{7n-10} \to \frac{4}{7}$

I did the algebra, that is $\left | \frac{4n+3}{7n-10} - \frac{4}{7} \right | = \left |\frac{61}{49(n-\frac{10}{7})} \right | < \epsilon$

Now the answer key did something very interesting. They underestimated the fraction to get a neater upper bound. They did $\frac{61}{49(n-\frac{10}{7})} \leq \frac{61}{n} < \epsilon$.

Then they chose $N = \max\left \{ 10,61/\epsilon\right \}$.

I was lazy so I decided to not underestimated the fraction. Instead, I just took $n > N = \frac{61}{49\epsilon}+\frac{10}{7}$. Is my N big enough for the proof to work?

share|improve this question
    
I think you can answer this question yourself. If $n > \frac{61}{49\epsilon} + \frac{10}{7}$, then does it follow that $|\frac{61}{49(n - \frac{10}{7})}|<\epsilon$? –  Neal Oct 13 '12 at 1:26
1  
Wait, isn't estimating the fraction the lazy thing to do, since it involves doing less work with simpler expressions? :p P.S. I'm skeptical that your $N$ works when $\epsilon$ is very large, since you ignore the effect of the absolute value signs. –  Hurkyl Oct 13 '12 at 1:27
    
It does, it works "too perfectly". –  sidht Oct 13 '12 at 1:27
add comment

2 Answers 2

up vote 1 down vote accepted

$$n > \frac{61}{49\epsilon}+\frac{10}{7} \implies \frac{49\left(n-\frac{10}{7}\right)}{61} > \frac{1}{\epsilon}$$ Assuming $n > \frac{10}{7}$, you have the desired inequality, so it is correct. Although simplifying it via underestimation certainly is a lot cleaner.

share|improve this answer
add comment

It Should work. You have to consider the integer part of this choice of N($\epsilon$).

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.