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Let $a$ and $b$ satisfy $a \geq b>0, a+b=1$.

  1. Prove that if $m$ and $n$ are positive integers with $m<n$, then $a^{m}-$ $a^{n} \geq b^{m}-b^{n}>0$.
  2. For each positive integer $n$, consider a quadratic function $$g_{n}(x)=x^{2}-b^{n} x-a^{n} .$$ Show that $g(x)$ has two roots that are in between $-1$ and $1$ .

Hint: Use derivatives.

Question was given to me by my teacher.

I have successfully proved $(1)$ by considering a function $p(x) = a^x - b^x$ then working with its derivative involving the $\log$ function and provided data to reach the conclusion. (I admit that this idea is not original and is borrowed from the proof of $e^{\pi} > {\pi}^e$)

I couldn't figure out $(2)$. Writing $g(x) = x^2 - bx - a$ then its discriminant is $b^2 + 4a > 0$ by provided data so it does have distinct real roots. Checking the bounds $g(1) = 1 - b - a $ which is $0$ (by provided data) which doesn't lead anywhere and hence any attempt of IVT (or rather, Bolzano's Th.) is not possible.

So nothing conclusive is obtained but even if IVT held would it show that the roots are within $-1$ and $1$ ?

$\star$ Is there a theorem which shows that a polynomial must have two roots in some interval or is there some trivial deduction that I am missing on ?

Any hints/comments on how to proceed will suffice. Sorry if something is incorrect in the above reasoning.

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3 Answers 3

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Simple manipulation of the graph of the quadratic will suffice. Since the leading coefficient of $g_n$ is $1$, and the discriminant $b^{2n} + 4a^n$ is always positive, it suffices to prove that:

  • $g_n(-1) \ge 0$;

  • $g_n(1) \ge 0$;

  • $g_n$ attains its minimum somewhere in $(-1, 1)$.

We can finish our proof from these conditions. Since $g_n$ has a positive leading coefficient and discriminant, the minimum $g_n(b^n/2)$ is negative (or else $g_n$ will be either always positive or only attains $0$ at one point). Since $g_n(-1) > 0$ and $g_n(1) \ge 0$ but $g_n(b^n/2) < 0$ where $b^n/2 \in (-1, 1)$, $g_n$ has two roots $\alpha \in (-1, b^n/2)$ and $\beta \in (b^n/2, 1]$ by the intermediate value theorem.


The three conditions can be proved as follows. It can be shown that $\frac12 \le a < 1$ and $0 < b \le \frac12$. For any $n \in \mathbb{Z}^+$, \begin{gather*} g_n(-1) = 1 + b^n - a^n > 1 + 0 - 1 = 0, \\ g_n(1) = 1 - b^n - a^n \ge 1 - (a + b)^n = 0, \tag{*} \label{A} \end{gather*} where $\eqref{A}$ can be shown by expanding $(a + b)^n$ using the binomial theorem. The minimum of $g_n$ is attained at $b^n / 2 \in (0, 1/2] \subset (-1, 1)$.

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  • $\begingroup$ Can you elaborate a bit on "Since the leading coefficient of ... always positive, it suffices to prove that..." part ? It makes sense to me pictorially but I am not certain of the reasoning in words. I would especially like to see the reason behind why consider the two $\geq 0$ inequalities. $\endgroup$
    – noobman
    Mar 3 at 17:36
  • $\begingroup$ @noobman The first part sketches the general method we are following in the solution, and the third part of the answer is a more rigorous justification of the method sketched in the first part. I edited the third part - does it make more sense to you now? $\endgroup$
    – L. F.
    Mar 3 at 17:41
  • $\begingroup$ Aha. The third paragraph is the perfect answer to this question and my queries (which I think should be the at the top). Thanks for the clarification! $\endgroup$
    – noobman
    Mar 3 at 17:50
  • $\begingroup$ @noobman I changed the order. $\endgroup$
    – L. F.
    Mar 3 at 21:27
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$$ g(x) = x^{2} - b^{n}x - a^{n}. $$ There are always two real roots since $\Delta = b^{2n}+4a^{n}.$

Therefore

$$ x_{1,2} = \frac{b^{n} \pm\sqrt\Delta}{2} = \frac{b^{n}}{2}(1\pm\sqrt{1+4(1-b)^{n}b^{-2n}}). $$

Now, the conditions you gave force $b$ to be in the interval $[0;1]$.

Notice that plugging in the boundary values for $b$ will result in both of the roots being between $-1$ and $1$. So now you need to calculate the derivative of the functions representing roots (it suffices to consider only one function with a constant in front of the square root, then only consider what happens when the constant is $-1$ and what happens if it is $1$). Once that is found, you will be able to find a maximum value. I would assume judging by your post that you will now know how to proceed further. Is that right? If not, I will continue the explanation.

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    $\begingroup$ Yep I see the motivation behind the solution now. Thank you for the direction ! $\endgroup$
    – noobman
    Mar 3 at 17:55
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For each positive integer n, consider a quadratic function $g_n(x)=x^2−b^nx−a^n.$ Show that g(x) has two roots that are in between −1 and 1 .

$g_n(1)=1-(a^n+b^n) \implies g_n(1) \geq 0 ( a^n+b^n < (a+b)^n\implies a^n+b^n <1)$

Also, $g_n(-1)=1+b^n-a^n \implies g_n(2) > 0$ $ [(b^n-a^n)_{min}=-1$ which is possible only if $a=1$ and $b=0$ but since $b>0$ ,thus $(b^n-a^n) > -1$ ].

Lastly,speaking of $g_n(0)=-a^n \implies g_n(0)<0$.

Thus,the (sample)graph should look something like this: enter image description here

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  • $\begingroup$ "$g_n(1)=1-(a^n+b^n) \implies g_n(1) > 0$" but for $n=1$ I showed that $g(1)$ is $0$. \\ Sorry I lack the reasoning part here but why does showing the inequalities of $g_n(1)$ and $g_n(-1)$ when they are both greater than 0 show that two roots belong in $[-1;1]$ ? $\endgroup$
    – noobman
    Mar 3 at 17:40
  • $\begingroup$ @noobman 'why does showing the inequalities of gn(1) and gn(−1) when they are both greater than 0 show that two roots belong in [−1;1] '.. that can be explained by obseerving the graph.Basically,they are the conditions. $\endgroup$
    – user1012971
    Mar 3 at 17:43
  • $\begingroup$ "$g_n(1)=1−(a^n+b^n)⟹g_n(1)>0 $" I have edited my answer to correct the typo. $\endgroup$
    – user1012971
    Mar 3 at 17:44
  • $\begingroup$ I see it now, I have selected the answer (more or less the same as yours) by @L. F. due to it also explaining the connection via IVT. Thank you for the answer and the graphical visuals as well ! $\endgroup$
    – noobman
    Mar 3 at 17:57
  • $\begingroup$ You are welcome. $\endgroup$
    – user1012971
    Mar 3 at 18:28

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