How do I solve this inequality? $\sin x < 2x^3$

Solution 1:

Introduction

To entirely clarify the situation for the OP, I will first explain a general method for solving inequalities, followed by a brief explanation of Newton's Method.

General method of solving inequalities

To find the region satisfied by an inequality, we can first find the points which will definitely not be satisfied by the inequality.

To explain this, I will demonstrate using an example which can be solved explicitly without resorting to numerical methods. Once we understand the basic process of solving inequalities using points, we can learn how to find the points with numerical methods.

Suppose we have the inequality

$$ x^2 < 4 $$

This means that whenever $x^2=4$, the inequality is not true. In fact, the solutions for $x^2 = 4$ form the boundaries for the region satisfied by the inequality. Solving the equation $x^2 = 4$ for $x$, we obtain $x=\pm 2$.

Plotting these points on a number line, we see the following:

Because these are the boundaries for $x^2 < 4$, we must use open circles, rather than closed circles to represent the points.

Now we can check which regions which satisfy the inequality. The three regions are

$$ x < -2 \\ -2 < x < 2\\ x > 2\\ $$

We check each region using any point from within the region. For example,

if $x < -2$ satisfies $x^2 < 4$, then $(-3)^2 < 4$. This is not true, so we know that the region $x < -2$ is not a solution to the inequality. We do similar checks for the remaining two regions, and find that the only region which satisfies the inequality is $-2 < x < 2$.

Thus we have solved the inequality $x^2 < 4$. We know that to solve an inequality, we simply find the roots of the equivalent equality, then test the regions around each root. This can be applied to any inequality, assuming we can find the roots of the equivalent equality.

Newton's Method

Sometimes we encounter equations for which we cannot easily find the roots. (Remember that we need to find the roots of the equivalent equality so we can find the regions for the inequality.) Newton's Method allows us to iteratively approximate the roots of a large number of functions.

Newton's Method is generally stated as follows, where $x_n$ is our first approximation for the root and $x_{n+1}$ is the slightly improved approximation from Newton's Method: (I'm stealing this from @nbubis so I don't have to type it.)

$$ x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)} $$

To use Newton's Method, we must start with a reasonably good guess for the roots. Note that Newton's Method can only find one root at a time, and it will generally home in on the root it is closest to.

Solution to the original question

We are trying to solve the inequality $\sin x < 2x^3$, so we must find the roots of the equivalent equality: $\sin x = 2x^3$. We can easily see that $x=0$ is a solution of the equality. Also, both $\sin x$ and $2x^3$ are symmetrical about the line $y=x$, so if x is a positive root, then the negative root -x also exists, as @nubis mentioned.

I don't quite follow @nubis' explanation that "$|x| < \sqrt[3]{1/2} \sim 0.79$". However, I think it is easy to see that when $x$ is positive, both $\sin x$ and $2x^3$ are positive in the region $0 < x < \pi$. Also, we know that $\sin x \leq 1$ at all times, while $2x^3 > 1$ when $x > 1$. Thus the positive root lies in the region $0 < x \leq 1$

Once again, we follow @nbubis' lead by approximating the root as $x = \frac{1}{2}$. (You don't have to use this approximation. The beauty of Newton's Method is that any reasonably close approximation will usually work. To get a feel for how the method works, try substituting 0.4, or 0.8 (or any number between 0 and 1) to see how it converges. You could also try something ridiculously far off, like 20, to see what happens then.)

Newton's Method states

$$ x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)} $$

We rearrange the equation $\sin x = 2x^3$ as $\sin x - 2x^3 = 0$.

Now

$$ f(x) = \sin x - 2x^3\\ f'(x) = \cos x - 6x^2\\ $$

and our first approximation of the root is $x_0 = 0.5$. So

$$ \begin{align} x_{n+1} &= x_n - \frac{f(x_n)}{f'(x_n)}\\ &= x_n - \frac{\sin x - 2x^3}{\cos x - 6x^2}\\ x_1&= 0.5 - \frac{f(0.5)}{f'(0.5)}\\ &= 0.5 - \frac{\sin (0.5) - 2(0.5)^3}{\cos (0.5) - 6(0.5)^2}\\ &= 0.8686039699 \end{align} $$

Substituting $x_1$ back into Newton's Method as our new approximation ($x_n$), we obtain a closer approximation for $x_2$ and so on. As @nbubis showed,

$$ \begin{align} x_2 &= 0.72759492\\ &...\\ x_5 &= 0.67996445\\ \end{align} $$

You can repeat this process until you either get bored, or you decide to search $\sin x = 2x^3$ in Wolfram Alpha. =P

We have now found the positive root of the equation. $x = 0.67996445$. We know from earlier arguments that the negative root is $-x$, so the negative root is $x = -0.67996445$.

We now have all the roots of the equation!

$$ \begin{align} x &= -0.67996445\\ x &= 0\\ x &= 0.67996445\\ \end{align} $$

These points define four regions, as shown below:

Now we simply follow the exact steps we used before to find the regions specified by an inequality once we know the roots of the equivalent equality.

We change the closed circles of the equality to the open circles of the inequality:

Then we check each region as before, and find the result:

$$ \begin{align} -0.67996445 < x &< 0\\ x &> 0.67996445\\ \end{align} $$

Conclusion

This all looks incredibly long and arduous. However, with practice (and when you are not including every step in an explanation =P) you can probably solve this in under five minutes.

Good luck with your future mathematical endeavours.

Solution 2:

First solve for equality. This equation can really only be solved numerically, but you can note a number of things without a computer:

• $x = 0$ is a solution.
• For any solution $x$, -$x$ is also one.
• Since $\sin x < 1$, any solution must have: $|x| < \sqrt[3]{1/2} \sim 0.79$

Numerically you will find that: $$x \sim -0.68, 0, 0.68$$ So that your inequality is satisfied when: $$x > 0.6799..$$ $$-0.6799.. < x < 0$$

Edit: To use Newton's method to solve $f(x) = \sin x - 2x^3 = 0$, we start with some guess value $x_0$, and the iterate: $$x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)} = x_n - \frac{\sin x - 2x^3}{\cos x - 6x^2}$$ Say we start with $x_0 =0.5$, then the next iterations leads to: $$x_1 = 0.86860396$$ $$x_2 = 0.72759492$$ $$x_3 = 0.68425682$$ $$x_4 = 0.68000441$$ $$x_5 = 0.67996445$$ As you can see, the more we iterate, the closer we get to the "true" result. This is what is meant by numerically - you can't find a closed expression for the solution, but you can find the numerical value.