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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.

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    $\begingroup$ Your question is not quite clear. Is, for instance, $y=3+\frac{1}{10}+\frac{4}{10^2}+\frac{1}{10^3}+\frac{5}{10^4}+\frac{9}{10^5}+\ldots$ a valid series solution to $\tan(y/4)=1$ in your terminology? $\endgroup$
    – yarchik
    Jan 31 '20 at 16:46
  • $\begingroup$ @yarchik: Hope the reference en.wikipedia.org/wiki/Closed-form_expression about the terminology will be useful. $\endgroup$
    – user64494
    Jan 31 '20 at 17:03
  • $\begingroup$ @user64494 It is not really clear, are you talking about this statement "Whether a number is a closed-form number is related to whether a number is transcendental." ? $\endgroup$
    – yarchik
    Jan 31 '20 at 17:24
  • $\begingroup$ @yarchik: The reference is about "the analytic solution". Hope I am (and was) clear. Don't hesitate to ask for further explanation in need. $\endgroup$
    – user64494
    Jan 31 '20 at 17:49
  • $\begingroup$ @yarchik What I'm talking about is what you mean.It's just that the equation I'm going to calculate is a little bit more complicated, but I just want to get the series solution in your form. $\endgroup$ Feb 1 '20 at 0:27
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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]
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  • $\begingroup$ Hope the reference en.wikipedia.org/wiki/Closed-form_expression about the terminology will be useful. $\endgroup$
    – user64494
    Jan 31 '20 at 16:28
  • $\begingroup$ The "set of operations and functions admitted in a closed-form expression may vary with author and context." For example, one can regard Root objects as essentially exact closed-form expressions for roots. They can be differentiated with respect to parameters, and implicitly understand the algebraic relations between roots. I think what the poster is after is a simple (e.g. hypergeometric) inverse series for $y=\sin{x}-e^x$ in which (at least) the coefficients have a simple "closed-form". In general the answer is no but as a non-trivial example, consider the inverse series for $y=x e^x$ $\endgroup$
    – TheDoctor
    Feb 3 '20 at 9:23
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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} *)
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  • $\begingroup$ If all real numbers can be represented by a series of terms, how to find the analytic solution of the series of terms of the transcendental equation. $\endgroup$ Jan 31 '20 at 6:48
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    $\begingroup$ @PleaseCorrectGrammarMistakes "If all real numbers can be represented by a series of terms" - Wait, do you just want to express a real number in terms of an infinite series of rational numbers? If so, you should edit your question to be more specific about this point. $\endgroup$ Jan 31 '20 at 22:58
  • $\begingroup$ @probably_someone Thank you very much.I've revised the statement of the problem. I hope I can make it clear this time. $\endgroup$ Feb 1 '20 at 1:22

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