# Difficult equation question

I have the equation $t\sin (t^2) = 0.22984$. I solved this with a graphing calculator, but is there any way to solve for $t$ without graphing?

Thanks!

• Not really, no. – Karolis Juodelė May 21 '14 at 16:15
• The tags need to be improved: this isn't about functional equations. – user111187 May 21 '14 at 16:48
• You need to know that we are lacking general methods to determine the exact number of solutions of such equations and isolate them grossly. – Yves Daoust May 22 '14 at 7:41

Using a Taylor series, $\sin(x)$ can be written as

$\sin(x)\approx x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \ldots$

Replacing $x$ for $t^2$ gives:

$\sin(t^2)\approx t^2 - \frac{t^6}{3!} + \frac{t^{10}}{5!} - \frac{t^{14}}{7!} + \ldots$

Plugging this into your original equation gives:

$t\sin(t^2)=0.22984$

$=t^3 - \frac{t^7}{3!} + \frac{t^{11}}{5!} - \frac{t^{15}}{7!} + \ldots=0.22984$

So you can see why solving this in a closed-form sense might be difficult.

That said, it's reasonable to think that there might be a value of $t$ less than one, in which case you can try neglecting the higher level terms (this is the small angle approximation).

This gives

$t^3=0.22984$

$t=0.61255046092664577035$

Graphically, you find a root at $\sim0.617544$. The difference is 0.8%.

The best you can hope for in this situation is to be able to calculate the solution out to as many digits as you are asked for. There are numerical methods to do just that, for instance Newton's Method.

• But there are infinite solutions... – evil999man May 21 '14 at 16:18
• I'm not sure what you mean by an "infinite solution". The subject of the question is solutions in the real numbers, which can certainly have infinite decimal expansions, if that is what you mean. The best you can hope for in that case is to compute approximations, as said in my answer. – Lee Mosher May 21 '14 at 16:37
• I think he/she means "infinitely many solutions". – Hans Lundmark May 21 '14 at 17:25

As is said, there is no closed form solution to this equation. No formula if you prefer.

In such cases, numerical methods are used, which means that different values for $t$ are tried, using specific strategies to get closer and closer to the solution.

It is useful to carry out the study of the function to get a rough idea where the solutions can be.

In this case, if you rewrite as $sin(p)=\frac{0.22984}{\sqrt p}$, to make more familiar functions appear, you see that you intersect a sinusoid with a kind of hyperbola, having an horizontal asymptote.

The first root can also be estimated from $sin(p)\approx p$ (for small $p$), so that $t\sin(t^2)\approx t^3$, and $t\approx\sqrt[3]{0.22984}=0.61255$. (Check: $0.61255\sin(0.61255^2)=0.22448$).

For large $p$, the equation becomes $sin(p)\approx 0$, meaning that you will find infinitely many solutions close to $p=k\pi$, i.e. $t=\sqrt{k\pi}$.

A yet better approximation is obtained by setting $p=q+k\pi$ (for small $q$), so that $sin(q+k\pi)\approx \pm q\approx\pm\frac{0.22984}{\sqrt{k\pi}}$ ($+$ for even $k$, $-$ for odd), i.e. $p\approx(-1)^k\frac{0.22984}{\sqrt{k\pi}}+k\pi$, and $t\approx\sqrt{(-1)^k\frac{0.22984}{\sqrt{k\pi}}+k\pi}$.

There could be additional solutions in between.