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Consider the function $f(x)=1-x-(1-\frac{a}{n}x)^n$ where $a$ and $n$ are both constants such that $\frac{a}{n}\in[0,1]$ ( it represents some probability indeed). One obvious solution is $x=0$. I want to prove that such a function has a real solution between $(0,1]$ and this solution has the form that dependening on $a$ and independent of $n$. The only naive technique that came to mind is to use Rolle's theorem. However, it seems that it is not a reasonable approach.

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I want to add some clarifications to my question. $n\ge 2$ is a positive integer. As a very special case, for $n=2$, we have $f(x)= 1-x-(1-\frac{a}{2}x)^2$. When $\frac{a}{2}>\frac{1}{2}$, this function has a solution $x=\frac{4(a-1)}{a^2}$. Since I set $n=2$, I don't know how to interpret the solution is independent of $n$.

This question is one part of my homework, I believe I have my $f(x)$ correct. Thanks for your time and sorry for any inconvenience.

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  • $\begingroup$ Is $n$ an integer? $\endgroup$
    – Arash
    Sep 22, 2013 at 21:29
  • $\begingroup$ How could the solution be independent of $n$? What do you mean by that? $\endgroup$ Sep 22, 2013 at 21:35
  • $\begingroup$ As a note, the cases $\frac{a}{n} = 0$ clearly leads to no real solution within $(0,1)$. The case $\frac{a}{n}= 1$ has solutions where $x-1$ is a root of unity, or 0, none of which are real and within $(0,1)$. Hence you might want to remove those. $\endgroup$
    – Calvin Lin
    Sep 22, 2013 at 22:22
  • $\begingroup$ @Langma Your answer is not "independent of n" as you have already "Set $n=2$". $\endgroup$
    – Calvin Lin
    Sep 22, 2013 at 22:50

4 Answers 4

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(As made in a comment to the question $a = 0,n$ leads to no real solutions in $(0,1)$, and will thus be ignored.)

The general answer is no, if $a$ is small enough. We can show that there exist values of $a$ close to 0 in which for $ x \in (0, 1]$,

$$ 1-x < (1 - \frac{a}{n}x ) ^n.$$

The reason is that when $x=0$, we have equality. By differentiation, we just need to get a value of $a$ where

$$-1 < (1 - \frac{a}{n} x)^{n-1} (-a)$$

for all $x \in (0,1)$. We are already given that $n \geq 2 > 1$, so observe that for $a <1 $ works since

$$ 1 > (1 - \frac{a}{n} x )^{n-1} > (1 - \frac{a}{n} x )^{n-1} a.$$


This same argument shows that a root exists when

$$\lim_{x \rightarrow 0} (1 - \frac{a}{n} x)^{n-1} (-a) < -1$$


some1.new's answer is a special case of this.

Robjohn gives a great answer without using calculus which shows $a<1$ would never work for $n \geq 1$.

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This is false. Set $a=1$ and $n=2$ which gives $\frac{a}{n}=\frac{1}{2}\in[0,1]$ and you'll have:

$$f(x) = 1-x - (1- \frac{1}{2}x)^2$$

But simple algebraic manipulation shows that

$$f(x) = 1-x - (1- \frac{1}{2}x)^2=-\frac{x^2}{4}$$

which has only one root at $x=0$ with multiplicity $2$.

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  • $\begingroup$ I guess the question is whether we can modify OP's desired claim and produce a solution. It does seem like for most $a$ and $n$, there will be a non-zero solution, even though it depends on $n$. $\endgroup$ Sep 22, 2013 at 21:53
  • $\begingroup$ @user2566092: I don't think so. The OP is asking us to prove his claim, while his prove is false and as you showed the solution can't be independent of $n$ as he wants it to be. It's very possible that his claim fails for higher values of $n$ as well. I don't know about you, but I personally have no interest in dedicating my time to find for what values of $n$ and under what conditions his claim could be true. Not to mention that this can be tremendously difficult to find necessary and sufficient conditions for his claim. $\endgroup$
    – user66733
    Sep 22, 2013 at 21:58
  • $\begingroup$ Fair enough, I agree, especially in light of the newest answer which shows that the existence of solutions depends on $a$ for arbitrary $n$. $\endgroup$ Sep 22, 2013 at 22:01
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The solution cannot be independent of $n$, even if $n$ is a positive integer. To see this, you can assume to the contrary and note that you get the same answer as when you take the limit of solutions as $n \to \infty$. But when $a$ is rational, the limit as $n \to \infty$ is the solution of $1 - x - e^{-ax} = 0$. Any non-zero solution to this is not algebraic, even though the solution is algebraic for any finite $n$.

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This cannot be true. Bernoulli's Inequality says that $\left(1-\frac anx\right)^n\ge1-ax$ when $\frac anx\le1$ and $n\ge1$.

Therefore, if $a\lt1$ and $0\lt x\le1$ $$ 1-x-\left(1-\frac anx\right)^n\le(a-1)x\lt0 $$

If $a=1$, Beroulli's Inequality also says that for $0\lt x\le1$ and $n\ge2$, $\left(1-\frac1nx\right)^n\gt1-x$.

Therefore, $$ 1-x-\left(1-\frac1nx\right)^n\lt0 $$

The only way with $a\le1$ that the equation can be true is if $x=0$ or $a=1$ and $n=1$.

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  • $\begingroup$ +1 Nice use of Bernoulli. Though, we can have $a>1$. OP just gives that $0 \leq a \leq n$. $\endgroup$
    – Calvin Lin
    Sep 22, 2013 at 22:05
  • $\begingroup$ @CalvinLin: Good point. I was considering $a\le1$. If $1\lt a\le n$, then all Bernoulli gives us is that $1-x-\left(1-\frac anx\right)^n\le(a-1)x$, which can allow a root. $\endgroup$
    – robjohn
    Sep 22, 2013 at 22:46
  • $\begingroup$ Yes, if you look at my answer, I believe that there are values slightly above 1 which would not work either. The relevance with yours is that the Bernoulli's inequality is extremely loose. $\endgroup$
    – Calvin Lin
    Sep 22, 2013 at 22:47

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