I have this math problem: $f(x)=\sin{x}+x^2$ and I have to find the function's zeros. Obviously, I could graph this function or use CAS to learn the zeros are $x\approx-0.8767\text{ or } x=0$. However, how can I use pencil and paper to work out the zeros? I was able to get $x=0$ as an answer by the following \begin{align} 0=&\sin x+x^2\\ -\sin x=&x^2\\ \sin^2x=&x^4 \end{align}

\begin{align} 0&=(\sin x+x^2)^2=\sin^2x+2x^2\sin x+x^4\\ 0&=x^4+2x^2\sin x+x^4=2x^4+2x^2\sin x\\ 0&=x^2(\sin x+x^2)\\ x^2&=0\quad\text{or}\quad \sin x + x^2=0\\ x&=0 \end{align}

I do not know if my method resulting in 0 is a happy coincidence or not, since the second quantity is the same as the original function. I'm sorry if my high school-level knowledge of math offends you. Maybe it's because I'm in high school.

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    $\begingroup$ You have cancelled a factor of 0 in the above work, so this isn't really a valid computation. You can see that f(0) = 0 immediately so I'm not sure what you are trying to necessarily achieve by insisting that it is derived by explicit manipulations. $\endgroup$ – Thompson Aug 16 '16 at 21:32
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    $\begingroup$ You divided by 0 to get x^2 = 0. Not allowed. $\endgroup$ – fleablood Aug 16 '16 at 21:35
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    $\begingroup$ @fleablood No need to be quite so rude. Maybe this person didn't spot that he was dividing by zero. $\endgroup$ – ÍgjøgnumMeg Aug 16 '16 at 21:41
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    $\begingroup$ @fleablood Probably a case of loading times for certain comments. Being rude is "inexcusable". Wind your neck in; this site is for learning, not berating people for not understanding straight away. $\endgroup$ – ÍgjøgnumMeg Aug 16 '16 at 21:49
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    $\begingroup$ Consider this anology: Solve for $(x - 57) + x^2 = 0$. $(x-57) + x^2 = 0 \implies -x^2 = (x-57) \implies x^4 = (x - 57)^2$. Note $0 = (x-57) + x^2 \implies 0 = [(x-57) + x^2]^2 = (x-57)^2 + 2(x-57)x^2 + x^4$. But $(x-57)^2 = x^4$ so $0 = 2(x-57)^2x^2 + 2(x-57)^2 = (x-57)[x^2 + (x -57)] $. So either $x-57 = 0$ or $x^2+ (x -57) = 0$. So $x-57= 0$. So $x = 57$. That is the same argument that you gave. The error is that you can not assume $x-57$ must equal 0. $\endgroup$ – fleablood Aug 16 '16 at 22:08

Let's see if I can firstly explain why this is a wrong approach and secondly explain how you should approach such a problem.

Start with a different problem,


We want to find the zeroes of this equation.

Spectacularly, we find that $x^2g(x)=0$ if $g(x)=0$. We go on to factor, and we get that $x^2=0$ or $g(x)=0$. Since it is so much harder to solve for $g(x)=0$, we simply deal with the first case, that is, $x^2=0$. Solving this will give us $x=0$.

Thus, we conclude that $x=0$ is a root of some function $g(x)$ (which could be anything in this case), but this surely cannot be the case, since it is very easy to make some function $g(x)$ that does not have a root at $x=0$, or maybe a function that has no roots at all!

Basically compare the above argument to yours, looking only at your last few steps and having $g(x)\implies f(x)$.

And... so you wish to use paper and pencil to solve this problem? That is a mighty task, as it is generally impossible to find the exact closed form solution of trig/non-trig combination type equations, but with some astute observation, we can get as close as possible.

$$\sin(x)=\frac x1-\frac{x^3}{1\times2\times3}+\frac{x^5}{1\times2\times3\times4\times5}-\dots$$

So in essence, we are solving for $x$ in

$$0=\left(\frac x1-\frac{x^3}{1\times2\times3}+\frac{x^5}{1\times2\times3\times4\times5}-\dots\right)+x^2$$


So $x=0$, or the other mess equals $0$.

$$x=0\text{ or }0=x+\left(\frac11-\frac{x^2}{1\times2\times3}+\frac{x^4}{1\times2\times3\times4\times5}-\dots\right)$$

Now, at this point, I note that it is actually not possible to solve for $x$ throughout that mess. The best you could do now is to find some $x$ as best you can that satisfies part of the mess:

$$0\approx x+\left(\frac11-\frac{x^2}{1\times2\times3}\right)$$


By seeing that $\sin(x)+x^2>0$ for all $x>0$, it should become more clear that $x\approx3-3\sqrt{5/3}$ is more of what we are looking for. And while its not the best approximation, it is improvable by solving the following:

$$0=x+\left(\frac x1-\frac{x^3}{1\times2\times3}+\frac{x^5}{1\times2\times3\times4\times5}\right)$$

which is simply closer to the full expansion of sine. For higher accuracy, look towards more terms of the sine function. (they aren't as neat as the quadratic above though)

Not that $\sin(3-3\sqrt{5/3})+(3-3\sqrt{5/3})^2=0.00415\dots$ isn't very bad though.

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    $\begingroup$ One way to improve the approximation is to note that the root is near $x=−\pi/4$. Hence one can instead consider the Taylor series approximation around this point; taking the linear approximation (easily done in WolframAlpha) and solving for $x$ yields $x\approx -\pi/4-\frac{8\sqrt{2}-\pi^2}{8\pi-8\sqrt{2}} \approx -0.889899.$ (By comparison, $3-3\sqrt{5/3}$ is roughly $-0.872983$. So the quadratic approximation near 0 does a bit better than the linear approximation near $-\pi/4$). $\endgroup$ – Semiclassical Aug 17 '16 at 13:49

The equation being not algebraic but trascendental the usual way to solve it is by numerical approximation which can be made of different forms.

The function $g(x)=x^2$ has a double zero at $x=0$,which is the point of its minimum and the function $f(x)=x^2+\sin x$ has a zero simple at the same point $x=0$.

At the negative neighborhood of $0$ the sinus is negative and $f(x)=x^2+\sin x\lt 0$ for $x_0\lt x\lt 0$ and $f(x)\gt 0$ for $x\lt x_0$ where $x_0$ is the other root of $f(x)=0$.

The minimum of $f(x)$ is taken approximately at the midpoint of the segment $\overline {x_00}$ because of the small variation of quantities (so $x_0$ is almost symmetric of $0$ respect to the point of minimum of $f(x)$) hence we can doubling the abscissa satisfying $f'(x)=2x+\cos x=0$. This point is $x\approx -0.439$. Thus $$x_0\approx -0.878$$

  • $\begingroup$ The method you suggest effectively trades one transcendental equation ($f(x)=0$) for another ($f'(x)=0$) and then pulling the root of the latter out of a hat. It's not clear to me that this is an improvement... $\endgroup$ – Semiclassical Aug 17 '16 at 14:06
  • $\begingroup$ @Semiclassical But you can apply this and other numerical methods to any degree of accuracy, something not as easily doable by methods like Taylor's Theorem. $\endgroup$ – Simply Beautiful Art Aug 17 '16 at 21:07
  • $\begingroup$ That presumes you've made clear what your method is, and that's sort've my point: How exactly do you obtain the root of $2x+\cos x=0?$ You need some method for that, just as one does for the original $x^2+\sin x=0$ equation. As it stands, all you do is assert that $f'(-0.439)=0$. $\endgroup$ – Semiclassical Aug 17 '16 at 21:47
  • $\begingroup$ Read please (so $x_0$ is ALMOST symmetric........... respect to the point of the minimum of $f(x)$). $\endgroup$ – Piquito Aug 19 '16 at 1:43

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