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How can I show that: $$(2-x)^nx^{n-1}$$ is decreasing with $n$ when $0<x<1$? I think this is generally true, but specifically I am concerned with $n$ as an integer $\geq 2$ and showing that the maximum of the function is when $n=2$ (its minimum) for all $x$. When I take the derivative with respect to $n$, I just get $$(2 - x)^n x^n \log(2 - x) + (2 - x)^n x^n \log x ,$$ but I don't know how to show that that is negative either. I guess it comes down to showing that the absolute value of $\log(2-x)$ is less than the absolute value of $\log(x)$... but I don't know how to do that with logs, or if that's necessarily the right approach.

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3 Answers

up vote 3 down vote accepted

Your idea works, you just have to push it a little farther. Take the derivative with respect to $n$, then consolidate like terms and put the two logarithms together for

$$(2-x)^nx^n \log ((2-x)x).$$ The factor $(2-x)^n x^n$ is positive so it remains to be seen that $0<(2-x)x<1$ for $x\in(0,1)$. We can subtract one and factor to get $-1<-(x-1)^2<0$, which is obviously true in our case.

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Thanks so much! –  Angada Sep 14 '11 at 22:16
    
This answer is certainly more complicated than necessary. –  Michael Hardy Sep 14 '11 at 22:21
    
@Michael: True, yours is simpler. –  anon Sep 14 '11 at 22:26
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$$(2-x)^nx^{n-1} = \frac1x \Big(x(2-x)\Big)^n.$$ Everything in parentheses here is positive. If you can show that the expression raised to the power $n$ is between $0$ and $1$, you're done. $y=x(2-x)$ is a parabola opening downward with $x$-intercepts at $0$ and $2$, and parabolas are symmetric, so the vertex is half-way between $0$ and $2$. That's the highest point. When $x=1$, then $y=1$. So $y<1$ if $x=\text{anything else}$.

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Oh, thanks! I do appreciate having a more simple way to do this! –  Angada Sep 14 '11 at 22:35
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Is $n$ a whole number or a real? Usually it would be a whole number. If so, to show it is decreasing with $n$, you need to show that the multiplicative factor, $(2-x)x$ is less than $1$. This, with the fact that the basic term is greater than $0$, is enough.

If $n$ is real, you just need to show that $\log x +\log(2-x) \lt 0$ as the other terms are positive and distribute out. You can check this with a derivative test.

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Ross, you do not have to worry about real $n$ explicitly. Notice that the function is an exponential function in $n$ with base $x(2-x) < 1$. So it is clearly strictly decreasing with $n$. (Of course, here I am assuming elementary properties of exponential functions. Now I am not sure that you are :-)) –  Srivatsan Sep 14 '11 at 22:13
    
Thank you very much! It is an integer in my case, but that is good to know. –  Angada Sep 14 '11 at 22:16
    
@Srivatsan Narayanan: Good point. –  Ross Millikan Sep 14 '11 at 22:38
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