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2

Clear["Global`*"] g[x_] := x^3 f1[x_] := g[x^2] f2[x_] := g[x^3] Definition@f1 f1[x_] := g[x^2] FullDefinition@f1 f1[x_] := g[x^2] g[x_] := x^3 Head@f1 Symbol Information["f*"] a[1] = 1; a[2] = 2; ?a DownValues@a UpValues@a


6

Will a = -3; Print[Defer[\[FormalA] x + 5 + x^2] /. \[FormalA] -> a] -3 x + 5 + x^2 work for you?


4

What about Print[HoldForm[3 x + 5 + x^2]] ?


1

This particular problem can dealt with algebraically by completing squares, equating coefficients and subtracting constants. (I am making the assumptions this is a left hand side whose right hand side is 0). pol = -0.433284 - 0.758719 x + 0.00289158 x^2 - 0.443672 y + 0.00149027 y^2; cr = CoefficientRules[pol, {x, y}] {p, q, r, s, t} = cr[[All, 2]]; {h, ...


3

There are two problems. One is that you want SolveAlways to find values for the parameters for which the expressions are equivalent. The second is that, as stated, it's not possible because the system is overdetermined. You'll want to allow for a constant term in your second expression. expr1 = -0.433284 - 0.758719 x + 0.00289158 x^2 - 0.443672 y + ...


1

a=-1/(2x+1); b=-3/(x-1); c=2/(3x+2); d=2/(x-1); e=-3/(2x+1); f=-1/(3x+2); a+b+d -(1/(-1 + x)) - 1/(1 + 2 x)



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