# How to find the solutions to $\frac{11}{2} x - \cos x = 0$?

Find the values of $x$ such that $\frac{11}{2} x - \cos x = 0$.

I really don't know how to find the solutions of this equation. I would appreciate if you could help me.

• The question basically asks to find the intersection point of two graphs.........y=(11/2)x and y=cosx – Sujan Dutta Aug 19 '16 at 5:13
• @SujanDutta ... and that's obviously solvable how? – fleablood Aug 19 '16 at 5:15

An equation which mix polynomial and trigonometric terms do not show explict solutions and numerical methods should be required.

If you consider the function $$f(x)=\frac{11}2x-\cos(x)$$ its derivative $$f'(x)=\frac{11}2+\sin(x)$$ is always positive so only one solution.

Apply Newton method which, starting from a guess $x_0$, will update it according to $$x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}$$ Let us be lazy and start using $x_0=0$. So, this simple method will generate the following iterates $$\left( \begin{array}{cc} n & x_n \\ 1 & 0.18181818181818181818 \\ 2 & 0.17891658537664712943 \\ 3 & 0.17891585607026614457 \\ 4 & 0.17891585607022005433 \end{array} \right)$$

Quite fast, isn't it ? You will learn that we could even do it faster.

Beside Newton, there are higher order iterative methods. Using them, still using $x_0=0$, the first iterate for a method of order $k$ is given in the following table $$\left( \begin{array}{ccc} k & x_1^{(k)} & \approx & \text{method} \\ 2 & \frac{2}{11} & 0.1818181818 &\text{Newton}\\ 3 & \frac{238}{1331} & 0.1788129226 &\text{Halley} \\ 4 & \frac{28814}{161051} & 0.1789122700 &\text{Householder}\\ 5 & \frac{10459726}{58461513} & 0.1789164437 &\text{no name}\\ 6 & \frac{1265622382}{7073843073} & 0.1789158127 &\text{no name}\\ 7 & \frac{2297105198362}{12839025177495} & 0.1789158574 &\text{no name}\\ 8 & \frac{5053631400166}{28245855390489} & 0.1789158562 &\text{no name}\\ 9 & \frac{21402128966834974}{119621197578720915} & 0.1789158561&\text{no name} \end{array} \right)$$

Moreover, "knowing" that $x$ has to be be quite small, we could have used the Taylor expansion of $\cos(x)$ around $x=0$ and limiting to $O\left(x^3\right)$ write $$f(x)\approx \frac{x^2}{2}+\frac{11 x}{2}-1$$ the solution of which being $x=\frac{\sqrt{129}-11}{2} \approx 0.178908$ which is already quite good.

Even better, using Padé approximants around $x=0$ and writing $$f(x)\approx \frac{P_{2}(x)}{Q_{n}(x)}$$ and solving $P_2(x)=0$ leads to quite accurate results as shown i the following table $$\left( \begin{array}{ccc} n & x & \approx \\ 2 & \frac{77-\sqrt{3265}}{111} & 0.1789173107 \\ 3 & \frac{3 \left(-1199+\sqrt{1645097}\right)}{1402} & 0.1789158655 \\ 4 & \frac{118635-\sqrt{13560037665}}{12226} & 0.1789158556 \\ 5 & \frac{-1624095+\sqrt{6130243687065}}{4761109} & 0.1789158560 \end{array} \right)$$

• Thanks alot, but i was actually helping to get an analitical solution, not a numerical one. But it's helpful, thanks. – Oscar Gutierrez Aug 20 '16 at 20:03
• @OscarGutierrez. Even $x=\cos(x)$ does not show analytical solution. – Claude Leibovici Aug 21 '16 at 4:35

We can approximate the value of $x$ without Newton's Method or any calculus at all. We can generalize the idea behind solving for $x$ in $x = \cos(x)$ by repeatedly hitting the cosine button on a calculator and noticing that it converges.

Rearranging the equation a bit, we are looking for the $x$ such that $$x = \frac{2}{11}\cos(x)\;.$$ Intuitively we know that such a value of $x$ exists since the graph of $y=\cos(x)$ and the line $y=\frac{11}{2}x$ intersect.

Now let's consider the function $f$ defined as $f(\_) = \frac{2}{11}\cos(\_)$. Suppose that we repeatedly apply $f$ to some input $a$, and let this infinite composition equal $x$. So we are going to say \begin{align} &f\left(f\left(\dotsb f(a)\right)\dotsb\right) \\=\;\;& \frac{2}{11}\cos\left(\frac{2}{11}\cos\left(\dotsb \frac{2}{11}\cos(a)\right)\dotsb\right) \\=\;\;& x \end{align} But if we suppose this converges as the number of times we apply $f$ approaches infinity (so $x$ actually exists and is a number), applying $f$ one more time shouldn't change the value of anything. So we now have \begin{align} &x \\=\;\;& f\left(f\left(\dotsb f(a)\right)\dotsb\right) \\=\;\;& f\left(f\left(f\left(\dotsb f(a)\right)\dotsb\right)\right) \\=\;\;& \frac{2}{11}\cos\left(\frac{2}{11}\cos\left(\frac{2}{11}\cos\left(\dotsb \frac{2}{11}\cos(a)\right)\dotsb\right)\right) \\=\;\;& \frac{2}{11}\cos(x) \end{align}

So the $x$ that this may converge to, the $x$ that we are looking for, is precisely the $x$ the solves the equation $x = \frac{2}{11}\cos(x)$. Furthermore we know such an $x$ exists by the reasoning of intersecting graphs above! And since there is no $a$ in the equation $x = \frac{2}{11}\cos(x)$, that must mean the value of $x$ doesn't depend on $a$. This means we can approximate the value of $x$ in $x = \frac{2}{11}\cos(x)$ by taking any number and repeatedly applying $f$ to it.

We can now easily calculate this value of $x$ by programming the function $f(\_) = \frac{2}{11}\cos(\_)$ into a calculator repeatedly apply $f$ to some (any) input. Calculating $f^n$ for different values of $n$ (mostly to compare to Newton's Method in Claude Leibovici's answer) I get:

\begin{array}{c|c} n & f^n(1) \\\hline 1 & 0.\color{#A00}{0982367828} \\ 2 & 0.1\color{#A00}{809415720} \\ 3 & 0.178\color{#A00}{8499431} \\ 4 & 0.17891\color{#A00}{79884} \\ 5 & 0.178915\color{#A00}{7870} \\ 6 & 0.17891585\color{#A00}{83} \\ \vdots & \\ 100 & 0.1789158560 \end{array}

So this doesn't converge as quickly as Newtons' Method, but I think that this can be punched into a calculator much quicker.

• Nice, reminds me when I discovered this when repeatedly pressing $\sqrt{Ans+1}$ on my calculator. math.stackexchange.com/questions/1599163/…. This is called the fixed point theorem, which you most likely are already are aware of. This link might be of interest to the OP. – Ahmed S. Attaalla Aug 21 '16 at 6:28