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I just don't get it, like at all.

$U_{n}$ is an iteration defined on $\mathbb{N}$, BTW.

The question was:

$$\begin{align} U_{n+1} &= \frac{8U_n - 8}{U_{n} + 2} = 8 - \frac{24}{U_{n} + 2}\\ U_0 &= 3 \end{align} $$ "Prove that $3 \leqslant U_n \leqslant 4$ by using mathematical induction"

Step 1: Test $n = 0$, yes, it's correct.

Step 2: Let's say the inequality is correct and test it for $n+1$

I used the first equation BTW. The result(for me) was:

$\tfrac{16}{6} \leqslant U_{n+1} \leqslant \tfrac{24}{5}$

Now, the exercise solved it by using the second equation, and its result was:

$\tfrac{16}{5} \leqslant U_{n+1} \leqslant 4 $

And since $16/5$ is bigger than 3 then it's correct and works.

How are there two different results and inequalities when you use each one of them?

Isn't $\tfrac{8U_{n} - 8}{U_{n} + 2}$ supposed to be equal to $8 - \tfrac{24}{U_{n} + 2}$ which means no matter which one you use, and there will be the same result?

I just don't get it. Which one is correct?

How are there two different results and inequalities when you use each one of them?

My solution:

An inequality for the numerator: \begin{align} 3 &\leqslant U_n \leqslant 4\\ 24 &\leqslant U_n \leqslant 32\\ 16 &\leqslant U_n \leqslant 24 \end{align} Now the equality for the dominator: \begin{align} 3 &\leqslant U_n \leqslant 4\\ 5 &\leqslant U_n + 2 \leqslant 6\\ \tfrac{1}{6} &\leqslant \tfrac{1}{U_{n} + 2} \leqslant \tfrac{1}{5} \end{align} By multiplying each side of the inequality then:

$\tfrac{16}{6} \leqslant U_{n+1} \leqslant \tfrac{24}{5}$

Peter Mortensen
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TechnoKnight
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    Well yes, since the equation is the same, you can prove the same stuff with it. So you must have just done something different with it than the book. Or you made a mistake somewhere. – Milten Apr 13 '20 at 22:12
  • @Milten I don't get it. I checked every step, they are all supposed to be correct. Can you please check them again to confirm? – TechnoKnight Apr 13 '20 at 22:16
  • @TedShifrin It's Un + 2, like Un then you add 2. I mean the second one. – TechnoKnight Apr 13 '20 at 22:28
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    @TechnoKnight Correct me if I'm wrong, but it seems like you're saying $\frac{ 8U_n - 8 } { U_n + 2 } \geq \frac{ 8 \times 3 - 8 } { 4 + 2 } $, where you're using $ 3 \leq U_n \leq 4 $ on the numerator / denominator respectively. That was my interpretation of what Riccardo said (before the edit), but you disagreed with them. As such, can you show your work? How did you get that inequality? – Calvin Lin Apr 13 '20 at 22:43
  • @CalvinLin I added the solution to the post. You can take a look at it. – TechnoKnight Apr 13 '20 at 23:05
  • I get for the maximal value $8 \cdot \frac{4 - 1}{4 + 2} = \frac{24}{6} = 4$ ($\frac{x - 1}{x + 2}$ is increasing in the relevant range) – vonbrand Apr 13 '20 at 23:56
  • This is different from how I was taught to do proof by induction. Step 1: show that the statement is true for the initial case, which here is $U_0$. Step 2: show that if the statement is true for some $U_{k}$ where $k>0$, that it is true for $U_{k+1}$. Using $n$ for both the indefinite subscript (any $U_{n}$) and the definite subscript (a specific $U_{n}$) is confusing, which I think is why I was taught to use $i, j, k$ for the definite subscript. – shoover Apr 14 '20 at 20:15
  • It should be pointed out that your upper bound is not tight enough to prove the desired result—24/5>4. – JOF14 Apr 14 '20 at 20:43

3 Answers3

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The book used the second inequality, and gets a tigher result (stronger bounds) than you. Why?

This is because, using the first expression, you bound separately numerator and denominator, and that's not the optimal thing to do. Namely, you known $m \leq u_n \leq M$, and then bound $$ \frac{8 m-8}{M+2} \leq \frac{8 u_{n}-8}{u_n+2} \leq \frac{8 M-8}{m+2} \tag{1} $$ This is correct, but not optimal. You can't see immediately it from the expression you use, but using the second expression is equivalent to bounding it as $$ \frac{8 m-8}{m+2} \leq \frac{8 u_{n}-8}{u_n+2} \leq \frac{8 M-8}{M+2} \tag{2} $$ which is better (but, again, it is not not obvious you can do this based on the first expression$^{\dagger}$).

${}^{(\dagger)}$ a way to realize you can indeed do this without using the second expression (as the book did) is to notice that the function $f\colon x\to \frac{8x-8}{x+2}$ is increasing, and so $f(m)\leq f(u_n) \leq f(M)$ whenever $m\leq u_n\leq M$.

Clement C.
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  • I think I get it now. So, just to know if I understand it right: I should choose the equality that doesn't need me to separately bound its numerator and denominator...right? So that I get the tighter(and more correct result/bound), right? – TechnoKnight Apr 13 '20 at 23:00
  • @TechnoKnight Yes, indeed! Glad that helped. – Clement C. Apr 13 '20 at 23:01
  • But what if it didn't give us the second equation and only the first one. Will I have to search for the second one or it will be okay to use the first one even though it gives a less tight result? – TechnoKnight Apr 13 '20 at 23:03
  • More or less, yes. Or try to see if the function $f$ such that $u_{n+1} = f(u_n)$ is monotone (non-increasing or non-decreasing). Here, as mentioned, $f(x) = \frac{8x-8}{x+2}$, so plotting it for $x\in[3,4]$ will give you a hint. @TechnoKnight – Clement C. Apr 13 '20 at 23:04
  • Got it, thank you so much! And thanks for everyone who helped me with this post! – TechnoKnight Apr 13 '20 at 23:06
  • @TechnoKnight No worries at all. – Clement C. Apr 13 '20 at 23:10
  • But to see that $f$ is monotone is easiest in the second form ... – Hagen von Eitzen Apr 14 '20 at 17:50
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I think I see what is going on. Let's focus on your left inequality. You basically do this: $$ U_{n+1} = \frac{8U_n-8}{U_n+2} \ge \frac{8\cdot3-8}{U_n+2}\ge \frac{8\cdot3-8}{4+2} = \frac{16}{6} $$ This is absolutely correct. However, we lose some tightness in the bound, because $\frac{8U_n-8}{U_n+2}$ is in fact increasing in $U_n$ (for $U_n>-2$). This is not obvious, but can be seen immediately in the rewritten form $8-\frac{24}{U_n+2}$. Therefore, you can replace $U_n$ with $3$ everywhere to get: $$ U_{n+1} = \frac{8U_n-8}{U_n+2} \ge \frac{8\cdot3-8}{3+2} = \frac{16}{5} $$

Milten
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  • I didn't do that. I actually first made an inequality for the numerator. Then I made an inequality for the denominator, then I "inversed" that inequality(I mean 1/x) and multiplied it each side of the inequality by each one of the second inequality. – TechnoKnight Apr 13 '20 at 22:41
  • @TechnoKnight It may be worth adding what you did to the question -- as it stands, you haven't, guessing (as Milten did) is our best move... – Clement C. Apr 13 '20 at 22:42
  • @ClementC. Ok. Just a minute. – TechnoKnight Apr 13 '20 at 22:43
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    @TechnoKnight Okay, so you did $8U_n-8 \ge 8\cdot3 - 8 = 16$ and $U_n+2\le 4+2=6$, which means the whole thing is $\ge \frac{16}{6}$, right? In essence, this is the same as I and Clement C. showed. – Milten Apr 13 '20 at 22:45
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    @Milten In your answer, the last inequality is wrong (and leads to a wrong equality), because of the typo: the numerator should be $8\cdot 3-8$, not $8\cdot 4-8$. I think that may have caused the confusion? – Clement C. Apr 13 '20 at 22:47
  • @ClementC. Thanks, edited! Sorry if it caused confusion, TechnoKnight. – Milten Apr 13 '20 at 22:51
  • @ClementC. I added my solution to the post. Please take a look and tell me where I went wrong. – TechnoKnight Apr 13 '20 at 22:51
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    @TechnoKnight As said in my answer (and Milten's), you did nothing "wrong". But since you bounded separately numerator and denominator, your bounds, albeit correct, are not as good as what you'd have obtained by bounding the fraction directly. – Clement C. Apr 13 '20 at 22:53
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You are not being careful with inequalities. Note that $$4=\frac{24}6 \le \frac{24}{U_n+2}\le \frac{24}5,$$ so $$8-\frac{24}5 \le 8-\frac{24}{U_n+2} \le 8-4 .$$ (Both taking the reciprocal and taking the additive inverse reverse inequalities.)

Ted Shifrin
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  • I didn't do that. I actually first made an inequality for the numerator. Then I made an inequality for the denominator, then I "inversed" that inequality(I mean 1/x) and multiplied it each side of the inequality by each one of the second inequality. – TechnoKnight Apr 13 '20 at 22:41
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    That's why you should first simplify the algebra and then use one value. What you did is not incorrect, but doesn't give the best inequalities. For example, if you look at $1\le x\le 2$ and consider $y=(x+1)/x$, you get $y\le 3/1$, which is correct, but saying $y=1+1/x\le 1+1 = 2$ is a stronger result. – Ted Shifrin Apr 13 '20 at 22:47
  • I have added my solution, please take a look at it. – TechnoKnight Apr 13 '20 at 22:51