# Prove the function $f(x)=1-x-(1-\frac{a}{n}x)^n$ has a real solution between $(0,1]$

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.

• Is $n$ an integer? Sep 22, 2013 at 21:29
• How could the solution be independent of $n$? What do you mean by that? Sep 22, 2013 at 21:35
• 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. Sep 22, 2013 at 22:22
• @Langma Your answer is not "independent of n" as you have already "Set $n=2$". Sep 22, 2013 at 22:50

(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 $n \geq 1$.

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 $2$.

• 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 $n$. Sep 22, 2013 at 21:53
• @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. Sep 22, 2013 at 21:58
• Fair enough, I agree, especially in light of the newest answer which shows that the existence of solutions depends on $a$ for arbitrary $n$. Sep 22, 2013 at 22:01

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 $n$.

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 $n=1$.

• +1 Nice use of Bernoulli. Though, we can have $a>1$. OP just gives that $0 \leq a \leq n$. Sep 22, 2013 at 22:05
• @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.
– robjohn
Sep 22, 2013 at 22:46
• 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. Sep 22, 2013 at 22:47