At what value(s) of $x$ does $ \cos(x) = 7x$? For some reason, this question was posed to me on a calculus quiz, and I have completely drawn a blank on how to solve this problem.
If someone could explain how to solve this problem, it would be greatly appreciated.
Thanks.
At what value(s) of $x$ does $\cos(x) = 7x$?
 A: I will post an alternative iterative method.
Define a recurrence relation so that $a_0 = 1$ and $a_n = \dfrac{\cos(a_{n-1})}{7}$.
The solution to the problem is:
$$x = \lim_{n\to\infty} a_n.$$
The convergences of this series requires other math (Lipschitz constant satisfying certain conditions, etc.).
This may not work on a Ti-84 because of a lack of precision. 
A: We can approximate a solution $a$ of the equation $\cos x = 7x$ using Newton's Method. First observe that a solution to this is the same as a zero of the function $f(x) = 7x - \cos x$. So we'll apply Newton's method to $f(x)$, but we'll need to begin with a good starting value $x_0$.
Just by thinking about how the functions $y = 7x$ and $y = \cos x$ grow, you should be able to see immediately that any and all zeros of $f(x)$ must occur near $x = 0$. Indeed, since $-1 \leq -\cos x \leq 1$ for all $x$, you know that $7x - 1 \leq f(x) = 7x - \cos x \leq 7x + 1$. So if, for example, $x > 1/7$ you know that $0 = 7(1/7) - 1 < 7x - 1 < f(x)$, and similarly $f(x) < 0$ if $x < - 1/7$. So a safe bet for $x_0$ would be $x_0 = 0$ or $x_0 = 1$. Applying the formula $x_{n+1} = x_n - f(x_n)/f'(x_n)$ repeatedly will give you better and better approximations of one solution to this equation. (Now a question for you is: Is there only one such solution? Why?)
A: For variety...
Let $f(x) = \cos x - 7x$, and let $\alpha$ be the value $f(\alpha) = 0$. Let $g(x)$ be the inverse of $f(x)$ on an interval containing $\alpha$ and $0$.
Then, we have $g(0) = \alpha$ and $g(1) = 0$.
$$g'(x) = \frac{1}{f'(g(x))} = \frac{1}{-\sin g(x) - 7 } $$
so we have the differential approximation
$$ g(x) \approx g(1) + g'(1) (x-1) = \frac{1-x}{7}$$
$$ g(0) \approx \frac{1}{7} $$
and, in fact, we have
$$ f\left(\frac{1}{7} \right) \approx -0.01018673955$$
A: As the other answerers point out, you can approximate the solution using Newton's method. But note that there are an infinite number of solution in the complex plane and there are exact expressions involving the roots. E.g., if S is the set of all the (complex) solutions of $\cos(z) = 7 z$, then:
$$\sum_{z\in S}\frac{1}{z^2} = 50$$
$$\sum_{z\in S}\frac{1}{z^3} = \frac{707}{2}$$
$$\sum_{z\in S}\frac{1}{z^4} = \frac{7498}{3}$$
The solution closest to the origin is the real solution and this then makes the largest contribution the above summations. A good approximation to the real root x is obtained by putting $ 1/x^4 = \frac{7498}{3}$, this gives $x\approx 0.1414308$.
An advantage of this method is that you can easily obtain a very good approximation to the root expressed as an nth root of a rational number without having to use a calculator.
