Solve $ \sin \pi x = (x+1)^2$

Question

Is it really to solve it? I just need it because this solution is interval of integration. One solution is $-1$, but i need other

Answer 1

There is one obvious solution to the equation : $x=-1$. If you plot the function, you will notice that there is a second root (don't forget that you look for the intersections of $\sin (\pi x)$ and $(x+1)^2$ - so if one root, a second exists) which is close to $-2$. The problem is that the analytical solution cannot be obtained. Only numerical methods for solving $f(x)=0$ can be used. Newton method is a classical procedure : starting from a guess $x_0$, the iterates write $$x_{n+1}=x_n-\frac {f(x_n)}{f'(x_n)}$$ So, for you case, let us write $$f(x)=\sin \pi x - (x+1)^2$$ $$f'(x)=\pi \cos (\pi x)-2 (x+1)$$ and let us start iterating at $x_0=-2$; the successive iterates will then be $-1.80551$, $-1.78755$,$-1.78724$ which is the solution for six significant figures.

Using an higher-order Newton method (with quartical convergence), the following approximation of the root is obtained in one shot $$-2+\frac{6 (1+\pi ) (3+\pi )}{24+\pi (60+\pi (36+5 \pi ))} \simeq -1.78886$$

Answer 2

Okay, for $ x > 0$ and $x<-2$ there is no real solution because of bounds of sine function. The only solution are in $(-2, 0)$. We know $(x+1)^2$ is a parabola going upwards and the period of that sine function is 2. The sine function is is like curved upwards from below $x-$axis in $(-1, 0)$ and and above $x-$axis on $(-2,1)$. Since both are nice and smooth functions, if they intersect at most at two points.

Mathematica doesn't seem to find the other root. One root was found to be $-1$, I guess that means the other root can be found using Numerical methods.