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I'm working on these supremum and infimum problems and I just want to check to see if my approach is valid. The question asks to find the supremum and infimum of the set $\{\,(n+1)^2/2^n\mid n\in \mathbb{N}\,\}$.

My approach is to take the derivative of the equation and set it equal to $0$. Doing this I find that $n=-1$ and $n=\frac 2 {\log 2}-1$. I also found that $n=\frac 2 {\log 2}-1$ produces a local maximum and $n=-1$ produces a local minimum. Is it accurate to say that these are my infimum and supremum values or not?

I will take any help I can get. Thanks!

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    $\begingroup$ $-1$ and $\frac{2}{\log(2)}-1$ are not in $\mathbb{N}$. $\endgroup$
    – Michael L.
    Oct 7, 2017 at 16:03
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    $\begingroup$ When "take the derivative" you are assuming, at the very least, that the function exists for all real numbers- you are extending from N to R so no longer have the same function. $\endgroup$
    – user247327
    Oct 7, 2017 at 16:14
  • $\begingroup$ So how do I go about finding the infimum and supremum then? $\endgroup$
    – kitkat
    Oct 7, 2017 at 16:21
  • $\begingroup$ Did it cross your mind that you're taking a derivative with respect to a discrete variable? One can reasonably extend it to a function of a continuous variable, but why would one take this approach in the first place? Why continue thinking about negative values of $n\text{?}$ And about non-integer values of $n,$ after you've said $n\in\mathbb N \text{?}$ $\endgroup$ Oct 7, 2017 at 16:30
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    $\begingroup$ Yes, I understand that was a stupid mistake. But to be fair, I have been reviewing for an upcoming exam and my brain is a little fried at the moment which is why I posted it on here :). I think I understand it now though. $\endgroup$
    – kitkat
    Oct 7, 2017 at 16:34

4 Answers 4

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Yes, your approach is correct but, since $n \in N$ we need to modify it a little bit.

so $f(x) = \frac{(x+1)^2}{2^x}$ if first increasing and then decreasing(for $x $ non negative), this you check by taking derivative. So for $x \to \infty$, $f(x) $ goes to zero, so inifimum is $0$ .

For supremum your approach is correct, but $x$ shoud be integer, so just find the value of $x$ that is integral. precisely check $x_1 = \left[\frac{2}{\log2}-1\right]$ and $x2 = x1+1$. So $\frac{2}{\log2} = 2.88,\ x_1 = 1, \ x_2 = 2$ so for x2 we get the maximum value of $2.25$.

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$$A=\{\,(n+1)^2/2^n\mid n\in \mathbb{N}\,\}=\{1,2\}\cup\{\,0< (n+1)^2/2^n\mid ~~n>1~~n\in \mathbb{N}\,\}$$ Since $$\lim_{n\to \infty}(n+1)^2/2^n = 0$$ we have that for every $\varepsilon>0$ there exists $n_\varepsilon $ such that $n\ge n_\varepsilon $ implies

$$(n+1)^2/2^n\le \varepsilon $$

Hence taking $ x_\varepsilon=(n_\varepsilon +1)^2/2^{n_\varepsilon } $ We conclude that,

$$\color{blue}{x>0,~~\forall x\in A~~~and ~~~\forall~~\varepsilon>0,~~\exists x=x_\varepsilon\in A ~~\text{such that}~~ 0<x_\varepsilon\le \varepsilon}\\ \Longleftrightarrow \color{red}{\inf A =0.}$$

Now let check the Sup.splitting for $n=0,1,2,3,4,5$ we have $$A=\{1,2, \frac{9}{4},2,\frac{25}{16},\frac{36}{32}\}\cup\{\,(n+1)^2/2^n\mid ~~n>1~~n\in \mathbb{N}\,\}$$

Claim $$(n+1)^2<2^n~\forall n\ge 6$$

Therefore, $$\color{red}{\sup A =\max A =9/4 =2.25.}$$

Now let prove the claim by induction

  • for $n=6$ we have $(6+1)^2 =49<64=2^{6}$ the claim is true
  • Assume that, $(n+1)^2<2^n~\forall n\ge 6 $ and let show that $$(n+2)^2<2^{n+1}~\forall n\ge 6 $$

By assumption, $$(n+1)^2<2^n \implies (n+2)^2=(n+1+1)^2<\color{red}{(2^{n/2}+1)^2<2^{n+1}}$$ Let prove the last inequality

$$ (2^{n/2}+1)^2<2^{n+1} \Longleftrightarrow a^{n+1}-a^n-1>0 \forall n> 6 $$ Where $\color{blue}{a=\sqrt{2}}$.

Let $$f(x) = a^{x+1}-a^x-1~~~~x>6 $$

$$f'(x) = \ln a(a^{x+1}-a^x) = a^x\ln a(\sqrt{2}-1)>0 ~~~~x>6$$

Hence $f$ is strictly increasing therefore, and $$f(x)>f(6) = 8(\sqrt2-1)-1 >0\implies a^{n+1}-a^n-1>0~~\forall ~~n>6\\ \Longleftrightarrow (2^{n/2}+1)^2<2^{n+1}~~\forall ~~n>6 $$ Conclusion

$$(n+1)^2<2^n \implies (n+2)^2<\color{red}{(2^{n/2}+1)^2<2^{n+1}}$$ This prove the claim.

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Let $f(x)={(x+1)^2\over 2^x}$. Then $f'(x)=2^{-x}(x+1)[2-(x+1)\ln 2]$.

We see for $x>{2\over\ln 2}-1\approx 1.9$, $f'(x)<0$, i.e., $f(x)$ is decreasing for $x>1.9$.

So for $n\ge 2$, we have $f(n)>f(n+1)$.

Since $f(1)=2, f(2)=2.25$, we see $\sup_{n\in \mathbb N}{(n+1)^2\over 2^n}=2.25$.

Further, using L'Hospital's Rule we see $\lim_{x\rightarrow \infty}f(x)=0$. So $\lim_{n\rightarrow \infty}f(n)=0$. Thus $\inf_{n\in \mathbb N}{(n+1)^2\over 2^n}=0$.

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  • $\begingroup$ Don't see why the downvote? $\endgroup$
    – user32828
    Oct 7, 2017 at 18:07
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Sometimes when people post maximum likelihood problems posters seem to adhere to a dogmatic religion that says when you want to find a maximum or minimum value of something, you immediately start looking for places where the derivative is $0$ by knee-jerk reflex that precedes all attempts to understand the problem. In this case you're working with a discrete variable, and you rush into differentiation without remarking that the function can be extended to non-integer values of the argument, and then you continue thinking about critical values even after you find then at negative and non-integer values of $n$, after you've said $n\in\mathbb N.$ This is a more extreme version of that weakness.

Let us note that when $n$ is large, then incrementing $n$ by one unit from $n$ to $n+1$ has the effect of doubling the denominator in $(n+1)^2/2^n,$ whereas the increase in size of the numerator is small compared to the size of the numerator. Thus one can say that for large enough $n,$ every time you increment $n$ by $1$ unit, you multiply the term by a positive number less than, let us say $0.6.$ And of course we could make that smaller than $0.6$ (but always larger than $0.5$) by making $n$ large enough. Therefore the tail end of the sequence is dominated by a geometric sequence with a common ratio of $0.6,$ so it must go to $0$ as $n\to\infty.$ Therefore the infimum is $0.$

Since the sequence is decreasing for $n$ large enough, we need only look at finitely many terms to find the maximum, and the supremum is actually attained. $$ \begin{array}{c|l|l} n & (n+1)^2/2^n \\ \hline 1 & 2 \\ 2 & 9/4 & = 2.25 \\ 3 & 2 \\ 4 & 25/16 & = 1.5625 \\ 5 & 36/32 & = 9/8 = 1.125 \\ 6 & 49/64 & = 0.765625 \end{array} $$ Incrementing $n$ from $6$ to $7$ will multiply the numerator by $49/36 < 1.37$ but will double the denominator, and beyond $n=7$ it will multiply the numerator by something even less than that while doubling the denominator, so the sequence decreases beyond that point. Thus the supremum is the maximum, $2.25.$

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    $\begingroup$ Why did someone downvote this.. $\endgroup$
    – rae306
    Oct 7, 2017 at 17:11
  • $\begingroup$ I didn’t, but if I had, it would have been because of the tone of the first paragraph. As a teacher, there are mistakes that depress me, but I’m not being a good teacher when I vent about it. Understandable, but out-of-place. $\endgroup$ Oct 8, 2017 at 1:28
  • $\begingroup$ But be careful this answer may not be fully satisfactory to you but it clearly makes sense despite the tone paragraph even though might not been wish or hope of the Op to get such a long explanation $\endgroup$
    – Guy Fsone
    Oct 8, 2017 at 17:12

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