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.

How to find the largest number in the sequence$$ \sqrt{50},2\sqrt{49},3\sqrt{48},\cdots 49\sqrt{2},50$$

I am interested in a "calculus-free" approach. Thanks,

share|improve this question
7  
If you square all of them, the largest of the squares will correspond to the largest of the square roots. –  J. M. Nov 8 '11 at 10:45
1  
@GerryMyerson Shame on me :) I did not read the question carefully enough. –  Sasha Nov 8 '11 at 13:42
add comment

3 Answers

up vote 14 down vote accepted

The $n$-th term in the sequence is $n\sqrt{51-n}=\sqrt{n^2(51-n)}$. So the question is: for which $n$ ($1\le n\le 50$), does $n^2(51-n)$ become the largest?

If you want to avoid calculus, you could use the AM-GM inequality: if $x,\,y,\,z\ge 0$, then $$\frac{x+y+z}{3}\ge\sqrt[3]{xyz},$$ with equality if and only if $x=y=z$.

If we set $x=y=n/2$ and $z=51-n$, we obtain: $$\frac{51}{3}\ge \sqrt[3]{\frac{n}{2}\cdot\frac{n}{2}\cdot (51-n)},$$ with equality if and only if $n/2=51-n$ or $n=34$.

It follows that $n^2(51-n)\le 4\cdot 17^3$, or $\sqrt{n^2(51-n)}\le 2\cdot 17^{3/2}$, where equality holds for $n=34$.

share|improve this answer
    
+1 Well spotted! –  Jyrki Lahtonen Nov 8 '11 at 13:19
    
Note that if the original question is for example $n\sqrt{50-n}$ instead of $n\sqrt{51-n}$, no integer $n$ satisfies the equality of the AM-GM inequality; in that case, a more careful evaluation, such as that used in Jyrki Lahtonen's answer, is needed. –  pharmine Nov 9 '11 at 4:39
add comment

I would like to apply calculus whenever it is possible. So, here is my trying (IMO this is not a better solution though):

Consider the function $f(x)=x^2(51-x)$ over $[1,50]$. Then as usual, $f'(x)=0\Rightarrow x=0,34$ and $f''(0)>0,f''(34)<0$ implies $f$ has a unique global maximum at $x=34$ and global minimum at $x=0$. So,...

share|improve this answer
    
Sorry the point $x=0$ is not within the domain, so need not be considered. –  Tapu Nov 8 '11 at 14:43
    
+1 for the effort. If you view this as a problem of a continuous variable $x$ as opposed to the described discrete variable $n$, then you might as well extend the range to $[0,51]$. I mean, then it is easier to evaluate your function at the end points, and with this type of "find the extremum" problem the answer is unlikely to be at the endpoint anyway :-) –  Jyrki Lahtonen Nov 8 '11 at 20:14
    
@JyrkiLahtonen: I did considered this problem as a continuous problem. Whats wrong with that? The given set of points is indeed a finite subset of the set of points I am considering. The point of maximum of this superset also belongs to the subset under consideration. So it should be the maximum. Is there anything I am missing? –  Tapu Nov 8 '11 at 20:40
    
No. Nothing wrong with your solution. But, if the zero of the derivative had not occurred at an integer point you would have had to work a bit more :-) –  Jyrki Lahtonen Nov 8 '11 at 20:43
    
Yes, thats why I wrote "IMO its not a better solution". However, I wish to mention that a particular problem can be tackled in a tricky way which may not be applicable to general problem. For this problem, may be I was lucky that it happens (indeed I was lucky today: if you don't mind, here is another example). I wish to thank you for having the patience of verifying and clarifying (yes some people just voted up/down probably without verifying). –  Tapu Nov 8 '11 at 21:14
add comment

I like the AM-GM argument a lot, but here's another more down-to-earth solution.

Write $a_n=n\sqrt{51-n}$, $1\le n\le 50$. Everything in sight is positive, so $a_n\le a_{n+1}$ if and only if $$ a_n^2\le a_{n+1}^2\Leftrightarrow n^2(51-n)\le (n+1)^2(50-n). $$ This latter inequality simplifies to the quadratic inequality $-3n^2+99n+50\ge0$. The plot of this function is a parabola opening downwards. Therefore the inequality holds between the zeros $n_1\approx-0.5$ and $n_2\approx 33.5$.

We have shown that $a_{n+1}$ is larger than $a_n$, when $1\le n\le 33$, and that $a_{n+1}$ is smaller than $a_n$, when $n\ge 34$. Therefore we can conclude that $a_{34}$ is the largest of this lot.

share|improve this answer
    
On second thought: I used $f(n+1)-f(n)$ as a crude substitute for $\lim_{h\to0}(f(n+h)-f(n))/h$, so it's a bit questionable, whether this really is a non-calculus approach? Well, Abhyankar once said in a plenary talk that he uses $[f(x+y)-f(x)]$ instead of a derivative. "Because it always exists!" –  Jyrki Lahtonen Nov 8 '11 at 14:21
    
Nice solution (+1). I have just use a "calculus approach". –  Tapu Nov 8 '11 at 14:48
    
Well, you used the difference calculus, but so what? This is neat! –  J. M. Nov 8 '11 at 14:59
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.