I use this

sigma[t] = theta[t] - lambda[t];
sigmaF[t] = thetaF[t] - lambda[t];

rule1 = {Derivative[1][r][t] -> (-vm*Cos@(sigma[t])), 
   Derivative[1][lambda][t] -> -(vm*Sin@(sigma[t]))/(r[t]), 
   Derivative[1][thetaF][t] -> 0};
rule2 = {sigmat -> \[Sigma], sigmaft -> Subscript[\[Sigma], f], 
   vm -> Subscript[V, m], Derivative[1][theta][t] -> 
\!\(\*OverscriptBox[\(\[Theta]\), \(.\)]\)};

tgo = r[t]/
    vm*(1 + ((n + 2)*
         sigma[t]^2 + (n + 1)^2*(n + 2) sigmaF[t]^2)/(2*(2 n + 
          3)*(2 n + 5)) - (n + 1)*sigma[t]*
      sigmaF[t]/(2*(2*n + 3)*(2*n + 5)));

dtgo = ((D[tgo, t] /. rule1) // 
     FullSimplify) /. {(lambda[t] - theta[t]) -> 
     sigmat, (lambda[t] - thetaF[t]) -> sigmaft};
dtgo /. rule2 // 
 Collect[#, {Subscript[V, m], Cos[\[Sigma]], Sin[\[Sigma]], 
\!\(\*OverscriptBox[\(\[Theta]\), \(.\)]\), r[t]}] &

answer is

$$ \begin{aligned} &-\cos \sigma \cdot\left(1+\frac{(n+2) \sigma^{2}+(n+1)^{2}(n+2) \sigma_{f}^{2}}{2(2 n+3)(2 n+5)}-\frac{(n+1) \sigma \sigma_{f}}{2(2 n+3)(2 n+5)}\right) \\ &+\frac{R}{2(2 n+3)(2 n+5) V_{\mathrm{M}}}\left[(2 n+4) \sigma-(n+1) \sigma_{\mathrm{f}}\right] \dot{\theta} \\ &+\frac{1}{2(2 n+3)(2 n+5)}\left[(2 n+3) \sigma+\left(2 n^{2}+6 n+3\right)_{\sigma_{f}}\right] \sin \sigma \end{aligned} $$

use wolfram get this:

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By comparing the output with the original answer, you can see some differences

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the $\sigma_f$ not collect

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There are other problems

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I've listed some of them here, not all of them, but anyway, I want wolfram's output to be exactly the same as the answer

  • 4
    $\begingroup$ It sounds like your complaint isn't that Mathematica (MMA) is giving an answer that appears mathematically different from answer you are seeing in a textbook or paper—but rather than it's not presenting the output in the same way. If so, this is common. E.g., MMA orders its output in its own canonical order, which can lead to non-standard formatting. MMA will output y–x as –x+y, because it orders the variables alphabetically. You can fix this, and there are posts that address this, but it can be a lot of work if the formatting you want doesn't correspond to Mathematica's built-in functions $\endgroup$
    – theorist
    Jun 24 at 4:57
  • 1
    $\begingroup$ Here are some posts that should give you a flavor of how such manipulations are done: mathematica.stackexchange.com/questions/38426/… mathematica.stackexchange.com/questions/160476/… mathematica.stackexchange.com/questions/43581/… $\endgroup$
    – theorist
    Jun 24 at 4:58
  • $\begingroup$ Try: sol = dtgo /. rule2 // Collect[#, {Subscript[V, m], Cos[[Sigma]], Sin[[Sigma]], \!(*OverscriptBox[([Theta]), (.)]), r[t]}] &; Simplify /@ sol $\endgroup$ Jun 24 at 7:19
  • $\begingroup$ @theorist I feel difficult $\endgroup$ 2 days ago


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