What is maximum value among $1, 2^{1/2}, 3^{1/3}, 4^{1/4},....$ ?
My approach: let $f(x)=x^{1/x}$ then I found out the derivative of $f$. Since $f(x)$ is maximum where $f'(x)=0$ and $f''(x)<0$
But it's not working. Is there any other way?
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Taking log should make the calculus easier. – Sandeep Silwal Apr 18 '14 at 03:18
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2http://math.stackexchange.com/questions/116112/find-the-maximum-of-fx-x1-x – lab bhattacharjee Apr 18 '14 at 03:19
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1The above link shows that the maximum of $f(x)$ occurs at $x=e$, so you need to figure out which of $2^{1/2}$ and $3^{1/3}$ is larger. – eeeeeeeeee Apr 18 '14 at 03:22
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I tried taking log. Answer is coming as x=10. where as the answer should ideally be 3^(1/3). – Rudstar Apr 18 '14 at 03:24
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2Perhaps you should show us your calculations, so we can see if you made a mistake. – Greg Martin Apr 18 '14 at 03:33
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As @GregMartin says, $x=10$ seems strange-how do you find it? $x=e$ seems natural because of logs. You could also just calculate a few terms, finding $2$ and $3$ exceed all others and trying to prove that. – Ross Millikan Apr 18 '14 at 03:54
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Finding $x = 10$ suggests that you're using a base-10 logarithm instead of a natural logarithm. – NovaDenizen Apr 18 '14 at 04:47
2 Answers
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Your approach is good but, may be, you have had problem with the derivatives. Even if it is easier using logarithms, let us use $$f(x)=x^{\frac{1}{x}}$$ Standard derivation leads, after factoring, to $$f'(x)=-x^{\frac{1}{x}-2} (\log (x)-1)$$ and $$f''(x)=-x^{\frac{1}{x}-4} (-3 x+\log (x) (2 x+\log (x)-2)+1)$$ The first derivative cancels if $\log (x)=1$, that is to say for $x=e$. For this value of $x$, the value of the second derivative is $-e^{\frac{1}{e}-3}$ which is negative. So $x=e$ is a maximum and $e$ is close to $3$.
This is the results for you discrete list of numbers since you can easily show that $2^{\frac{1}{2}}$ and $4^{\frac{1}{4}}$ are both equal to $\sqrt 2$
Claude Leibovici
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"$e$ is close to $3$" is not a rigorous justification that $3^{1/3}$ is the maximum. – Greg Martin Apr 18 '14 at 07:45
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Yes, it is since $2^{\frac{1}{2}}=4^{\frac{1}{4}}=\sqrt 2 \lt 3^{\frac{1}{3}}$ – Claude Leibovici Apr 18 '14 at 07:58
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What he's saying is that the REASON why you choose $n=3$ being bigger than $n=2$ is not simply because $e$ is close to $3$. This is not rigorous. – zhuli Apr 18 '14 at 08:12
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We have $f(x)=x^{1/x}$, so $\log(f(x))=\frac{\log(x)}{x}$. The derivative of this is, by the quotient rule:
$f'(x)=\frac{1-\log(x)}{x^2}=0 \iff \log(x)=1 \iff x=e.$
Therefore, the maximum of $f(x)$ is $e^{1/e}$. The closest number in your list to $e^{1/e}$ is $3^{1/3}$.
Fred
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$e^{1/e}$ is not on the list. Since this is the maximum, the number on the list with the greatest absolute value from $e^{1/e}$ is therefore the maximum of the sequence above. – Fred Apr 18 '14 at 04:01
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"The closest number in your list to $e^{1/e}$ is $3^{1/3}$" needs justification. – eeeeeeeeee Apr 18 '14 at 04:10
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Let $x$ be a number in the sequence $1,2^{1/2}, 3^{1/3},.. $ Since we know the maximum of $f(x)=x^{1/x}$ is $e^{1/e}$, then the maximum in the sequence is the number $x$ which minimises $|x-e^{1/e}|$. This is clearly $3^{1/3}$. – Fred Apr 18 '14 at 04:10
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Why is it "clear" that $|n^{1/n}-e^{1/e}|$ is minimized when $n=3$? Again, it needs justification. – eeeeeeeeee Apr 18 '14 at 04:13
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It might just be me, but this argument is not satisfactory. Yes, I can look at the graph and say 'it looks bigger at 3', but that is not at all rigorous! If I didn't care about rigor, I could get my calculator to give me decimal approximations to these numbers and call it a day. – eeeeeeeeee Apr 18 '14 at 04:21
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1Since the only maximum occurs at $e$, that means it's strictly decreasing from this point in either direction. That means the maximum can only be at either $n=2$ or $n=3$. Just evaluate both and find the bigger one. – zhuli Apr 18 '14 at 04:22