# Express $t$ through $a$ in $t-\sin(t)=a$.

Can I express $$t$$ throw $$a$$ in this expression?

$$t-\sin(t)=a$$

I don't know, whom should i tick. Thank you all.

I used half dividing method to fing $$t$$ in my task.

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• I think he means expressing $t$ solely through $a$, i.e. $t=f(a)$. – scoopfaze Jul 12 at 13:43
• You can do it, but this is impossible in terms of elementary functions. In oter words, there is no formula for it. – Crostul Jul 12 at 13:44
• @Crostul ok, so how can $t$ be expressed through $a$ using non-elementary functions? – Имя Фамилия Jul 12 at 13:47
• I think it is unlikely that there is a closed form method to solve $t - \sin t = a$ for $t$ in terms of $a$. – GEdgar Jul 12 at 13:58

The Wikipedia article Kepler's equation gives the equation as

$$M = E - e\sin E$$

where we want to find $$\,E\,$$ given $$\,M\,$$ and $$\,e.\,$$ In your equation $$\,e=1.\,$$ The Wikipedia article states:

Kepler's equation is a transcendental equation because sine is a transcendental function, meaning it cannot be solved for E algebraically. Numerical analysis and series expansions are generally required to evaluate E.

in attempting to define an inverse function we must first choose a suitable domain of definition.

let $$f(t) = t - \sin t$$. if you wish to view $$f$$ as a real-valued function of a real variable then we have $$f:\mathbb{R} \to \mathbb{R}$$. clearly $$f$$ has derivatives of all orders. the first derivative is $$f'(t) = 1 - \cos t$$ which is zero for $$t \in \{2\pi n|n \in \mathbb{N}\}$$.

this shows $$f$$ is $$1-1$$, and monotonic increasing. so $$f$$ has a continuous inverse $$g$$ defined on the whole of $$\mathbb{R}$$, but to define $$f$$ a diffeomorphism, the zero-set of $$f'$$ must be excluded from the domain of definition.

if you keep those considerations in mind (which rule out the simpler Maclaurin expansion in this case), you can apply the Lagrange inversion theorem, also known as the Lagrange–Bürmann formula.

There is no closed-form solution to the equation

$$a=t-\sin t$$

and a numerical solution is in order.

The function is invertible, it is odd, repeating modulo $$2\pi$$, and it suffices to solve for $$[0,\pi)$$.

A good approach will be with Newton's iterations,

$$t_{n+1}=t_n-\frac{t_n-\sin t_n-a}{1-\cos t_n}$$

with the initial value $$t_0=a.$$

Anyway, for small $$a$$ the denominator is tiny (near horizontal tangent), and it can be safer to invert the Taylor development

$$a\approx \frac{t^3}6$$ or $$t\approx\sqrt[3]{6a}.$$

In the general case, a better initialization is the empirical

$$t=\sqrt[3]{\pi^2t}.$$

As Yves Daoust answered, the simplest is Newton method.

If $$a$$ is small, you could use the usual series expansion of $$\sin(t)$$ and use series reversion to get $$t=x+\frac{x^3}{60}+\frac{x^5}{1400}+\frac{x^7}{25200}+\cdots \qquad \text{where} \qquad x=\sqrt[3]{6a}$$