How to obtain the exact solution of this equation in terms of an infinite series of rational numbers

Question

How to obtain the analytic solution of the number series form of this equation

Plot[E^x - Sin[x] - 3, {x, -2, 2}, AxesOrigin -> {0, 0}]
NSolve[E^x - Sin[x] - 3 == 0 && 0<x<2, x]
FindRoot[E^x - Sin[x] - 3, {x, 1}]
AsymptoticSolve[E^x - Sin[x] - 3 == 0, {x, 0, 7}]

At present, I can't get the analytic solution of this equation in the form of number series.

As mentioned in the comments below, the solution of equation $\tan\left(\frac{x}{4}\right)=1 $ can be expressed as $ x =4-\frac{4}{2}+\frac{4}{5}+\cdots +\left( -1 \right) ^{n+1}\frac{4}{2n-1}+\cdots =\sum\limits_{n=1}^\infty\left( -1 \right) ^{n+1}\frac{4}{2n-1}$ .This is what I want which is expressed exactly in terms of an infinite series of regular rational numbers(But this kind of expression with irregular coefficients like $x=3+\frac{1}{10}+\frac{4}{10^2}+\frac{1}{10^3}+\frac{5}{10^4}+\frac{9}{10^5}+\ldots $is not what I want).

I want to find a solution in terms of an infinite series of rational numbers of $e^x - \sin(x) - 3=0$ that can be expressed exactly like the equation $\tan(\frac{x}{4})=1$ with the help of MMA.

Answer 1

As Gerry Myerson suggested, for your transcendental equation both $\sin(x)$ and $e^x$ have very simple expansions into infinite series and a solution to $\sin(x)-e^x=3$ (which is slightly different to what you asked previously), and a series for your transcendental root can be obtained by the technique of reversion of series. But note that there are exist infinity of such reverted series as you have freedom to choose the series expansion point.

Consider the general problem $y=\sin(x)-e^x$. Series reversion of the series expanded about $\pi/2$ (to simplify the expansion and improve the convergence—and also to avoid the problematic expansion about $x=0$), is immediate:

transeries[y_] = Normal @ InverseSeries[Series[Exp[x] - Sin[x], {x, Pi/2, 4}], y]

Then you can see the terms in the series (which change as you increased the expansion order):

transeries[3] // Expand

or, numerically:

transeries[3.0]

Answer 2

I cannot address the issue of finding the series expansion for the value of the solution. However, the solution can be found with a slight modification of your use of NSolve, or FindRoot.

For Solve or NSolve restrict x to an interval. Solve provides an exact solution as a Root expression

(sol = Solve[E^x - Sin[x] - 3 == 0 && 0 < x < 2, x][[1]]) // InputForm

(* {x -> Root[{-3 + E^#1 - Sin[#1] & , 
    1.38183436742236889176584178212\
658926554`20.309465699250364}]} *)

Verifying the solution,

E^x - Sin[x] - 3 == 0 /. sol

(* True *)

As shown in the Root expression, the approximate numeric value is

sol // N

(* {x -> 1.38183} *)

Or with NSolve

NSolve[E^x - Sin[x] - 3 == 0 && 0 < x < 2, x][[1]]

(* {x -> 1.38183} *)

For FindRoot use a better initial estimate

FindRoot[E^x - Sin[x] - 3, {x, 1}]

(* {x -> 1.38183} *)