How can I show that: $(2-x)^nx^{n-1}$ is decreasing with $n$ when $0
Prove that $(2-x)^nx^{n-1}$ decreases with $n$ for 0 ?
3 Answers
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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0@Michael: True, yours is simpler. – 2011-09-14
$(2-x)^nx^{n-1} = \frac1x \Big(x(2-x)\Big)^n.$ 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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0Oh, thanks! I do appreciate having a more simple way to do this! – 2011-09-14
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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0@Srivatsan Narayanan: Good point. – 2011-09-14