# Show that the largest root of $f$ is greater than $5n$ where $n(\ge 3)\in \mathbb N$.

Given $$f=x^3+9x^2+24x-40n^3+40xn^2+94n^2-12x^2n-62nx-74n+20$$ has real roots, show that the largest root of $$f$$ is greater than $$5n$$ where $$n(\ge 3)\in \mathbb N$$.

I tried to do it by directly finding the roots in wolfram alpha for any $$n\ge 3$$, however I am getting the roots as complex which is against the hypothesis of the problem.

Also the roots obtained in Wolfram Alpha are very nasty or bad which is making my life my difficult.

Kindly help me out on how to find all the real roots of $$f$$ and show that they are $$>5n$$

• In your title and in your question text sentence, it states you're looking to prove the largest root is greater than $5n$, while in your last sentence you ask to "find all the real roots of $f$ and show that they are $>5n$". I assume the last sentence is in error, with my answer also assuming this. Nonetheless, please update your question text accordingly. Mar 27 '20 at 7:22
• @JohnOmielan; Thanks for the information, but it is not an error, I want to find(if possible) all the roots of $f$ too, can you kindly help me to find it out, Wolfram alpha is giving roots as complex Mar 27 '20 at 9:23
• Your main question was to show for $n \ge 3$ there's always a real root $\gt 5n$. You don't actually need to find the roots to determine this, as shown in my answer & the related one by Macavity. To determine the actual roots, you are then basically solving a cubic equation. This is a bit messy, with details on how to do this in the General cubic formula section of Wikipedia's "Cubic equation" article. You could use WolframAlpha, as you did, but, fortunately, the answer by Claude Leibovici gives you those roots & details. Mar 27 '20 at 9:29

Thanks to the containment, let me give a more complex one. For sure, the formulae given by Wolfram Alpha or any other CAS are really messy. However, if you use the trigonometric method for solving the cubic, the results are not so bad.

For $$\left(-40 n^3+94 n^2-74 n+20\right)+\left(40 n^2-62 n+24\right) x+(9-12 n) x^2+x^3=0$$ the roots are given by $$x_k=4n-3+\frac{2 \sqrt{8 n^2-10 n+3}}{\sqrt{3}}\times$$ $$\cos \left(\frac{2 \pi k}{3}-\frac{1}{3} \cos ^{-1}\left(3 \sqrt{3}\frac{ (n-1) \left(4 n^2-3 n+1\right)}{\left(8 n^2-10 n+3\right)^{3/2}}\right)\right)$$ with $$k=0,1,2$$. This is not so awful. The largest root (to prove) is $$x_0$$.

If you graph the roots as functions of $$n$$, you will notice that they are "almost" straight lines.

For the fun of it, compose Taylor series for large values of $$n$$; you will get $$x_0=\left(5+\sqrt{5}\right) n-\frac{75+11 \sqrt{5}}{20} -\frac{25-7 \sqrt{5}}{400\, n}-\frac{125+73 \sqrt{5}}{8000\, n^2}+O\left(\frac{1}{n^3}\right)$$

Consider the case where $$n=3$$; the exact solution is $$x_0=9+2 \sqrt{15} \cos \left(\frac{1}{3} \cos ^{-1}\left(\frac{56}{15 \sqrt{15}}\right)\right)\approx 16.7148$$ while the above truncated expansion gives $$\frac{808375+176747 \sqrt{5}}{72000}\approx 16.7166$$ This is not too bad.

Now, using the expansion $$\Delta=x_0-5n\sim n\sqrt{5} -\frac{75+11 \sqrt{5}}{20}$$ is positive as soon as $$n > \frac{11}{20}+\frac{3 \sqrt{5}}{4} \approx 2.22705$$

Thank you for the problem !

Edit

As I wrote in my answer, we need to prove that $$x_0$$ corresponds to the largest root of the cubic. The simplest way is to consider the series the series expansion for each of the roots. $$x_0=\left(5+\sqrt{5}\right) n-\frac{75+11 \sqrt{5}}{20} -\frac{25-7 \sqrt{5}}{400\, n}-\frac{125+73 \sqrt{5}}{8000\, n^2}+O\left(\frac{1}{n^3}\right)$$ $$x_1=\left(5-\sqrt{5}\right) n-\frac{75-11 \sqrt{5}}{20}-\frac{25+7 \sqrt{5}}{400 n}-\frac{125-73 \sqrt{5}}{8000\, n^2}+O\left(\frac{1}{n^3}\right)$$ $$x_2=2 n-\frac{3}{2}+\frac{1}{8 n}+\frac{1}{32 n^2}+O\left(\frac{1}{n^3}\right)$$ which clearly show the claim.

Using these truncated expressions, we properly find that $$x_0+x_1+x_2=12 n-9+O\left(\frac{1}{n^3}\right)$$ $$x_0\, x_1+x_0\, x_2+x_1\, x_2=40 n^2-62 n+24+O\left(\frac{1}{n^2}\right)$$ $$x_0\, x_1\, x_2=40 n^3-94 n^2+74 n-20+O\left(\frac{1}{n}\right)$$ which are exactly the coefficients of the cubic

• Thanks a lot sir but I dont understand two things in the solution provided Mar 28 '20 at 3:33
• Do you mind giving an explaination on how you got the roots and how you know that $x_0$ is the maximum value of the root Mar 28 '20 at 3:34
• @Math_Freak.Have a look at en.wikipedia.org/wiki/… . For the remaining, I shall try to show it in an edit. Just for your curiosity, I worked 50+ years with cubic equations (in thermodynnamics areas). Cheers :-) Mar 28 '20 at 4:42
• Oh okay, it just shows the amount of expertise you have, Thumbs Up!! Mar 28 '20 at 4:51
• @Math_Freak. You know, after so many years ... ! You will have an edit within a few minutes. Mar 28 '20 at 4:52

Hint: As the cubic is eventually positive, it is enough to show that $$f(5n)<0$$, i.e. $$9n^2+46n+20 < 15n^3$$. Can you show that, say using induction?

\begin{equation}\begin{aligned} f(x)&=x^3+9x^2+24x-40n^3+40xn^2\\ & \; \; \; +94n^2-12x^2n-62nx-74n+20 \end{aligned}\end{equation}\tag{1}\label{eq1A}

Note you have

\begin{equation}\begin{aligned} f(5n) & = (5n)^3 + 9(5n)^2 + 24(5n) - 40n^3 \\ & \; \; \; \; \; + 40(5n)n^2 + 94n^2 - 12(5n)^2n - 62n(5n) - 74n + 20 \\ & = 125n^3 + 225n^2 + 120n - 40n^3 + 200n^3 \\ & \; \; \; \; \; + 94n^2 - 300n^3 - 310n^2 - 74n + 20 \\ & = -15n^3 + 9n^2 + 46n + 20 \end{aligned}\end{equation}\tag{2}\label{eq2A}

At $$n = 3$$, you thus get

\begin{equation}\begin{aligned} f(15) & = -15(3)^3 + 9(3)^2 + 46(3) + 20 \\ & = -166 \end{aligned}\end{equation}\tag{3}\label{eq3A}

However, since the coefficient of the highest power of $$f(x)$$ is $$1$$ in $$x^3$$, this means that $$\lim_{x \to \infty}f(x) = \infty$$. Since $$f(x)$$ is continuous, there must be a root larger than $$5n$$ for when $$n = 3$$.

To confirm this is also true for all $$n \gt 3$$, one way is take the derivative of \eqref{eq2A}, as shown below

$$\frac{df(5n)}{dn} = -45n^2 + 18n + 46 \tag{4}\label{eq4A}$$

Using the quadratic formula to get the roots gives

\begin{equation}\begin{aligned} n & = \frac{-18 \pm \sqrt{18^2 - 4(-45)(46)}}{2(-45)} \\ & = \frac{3 \mp \sqrt{3^2 + 3(46)}}{15} \\ & = \frac{3 \mp \sqrt{3(3 + 46)}}{15} \\ & = \frac{3 \mp 7\sqrt{3}}{15} \\ & \approx -0.61, 1.01 \end{aligned}\end{equation}\tag{5}\label{eq5A}

The quadratic polynomial in \eqref{eq4A} being a concave-down parabola means its values are only positive with $$n$$ in the approximate range of $$(-0.61,1.01)$$, with it being negative everywhere else. Thus, for $$n \ge 3$$, the derivative is negative, so the value of \eqref{eq2A} would keep decreasing, confirming there's always a real root $$\gt 5n$$ for \eqref{eq1A}.

• I understand that I need to show that $9n^2+46n+20<15n^3$ but I dont understand how you go about proving it Mar 27 '20 at 9:37
• @Math_Freak I was suggesting using the derivative to determine how the function $f(5n)$ in ($2$) behaved for values of $n \gt 3$. This is a fairly standard way to investigate general function behavior. However, for example if you haven't studied derivatives, note they're really not needed in this case since, as Macavity suggests, you could use other means like induction to prove it instead. Mar 27 '20 at 9:40
• I have studied derivatives and I understand that if a function is increasing then $f^{'}(x)>0$ but I dont quite understand how to use it here, will u mind showing it elaborately Mar 27 '20 at 9:48
• Oh Okay ! Have a good sleep! sweet dreams Mar 27 '20 at 9:51
• @Math_Freak I've added more details which I trust should help explain how to prove it using a derivative. Mar 27 '20 at 20:01