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When solving an equation, the real root obtained by equating the numerator to zero is not equal to the real root obtained by equating the fraction to zero.

In the following, the fractional equation is defined as a function.

frac[\[Alpha]_, \[Eta]_, \[Lambda]_] = 
 1/(4 (-4 + \[Eta]) (-3 \[Alpha] + \[Eta])^2) (72 \[Alpha]^2 - 
    81 \[Alpha]^3 - 48 \[Alpha] \[Eta] + 54 \[Alpha]^3 \[Eta] + 
    8 \[Eta]^2 - 18 \[Alpha] \[Eta]^2 + 18 \[Alpha]^2 \[Eta]^2 - 
    9 \[Alpha]^3 \[Eta]^2 + 6 \[Eta]^3 - 6 \[Alpha]^2 \[Eta]^3 - 
    2 \[Eta]^4 + 
    2 \[Alpha] \[Eta]^4 + (-3 + \[Eta]) \[Sqrt](81 \[Alpha]^6 (-3 + \
\[Eta])^2 + 324 \[Alpha]^4 (-4 + \[Eta]) \[Eta] + 
        4 (-4 + \[Eta]) \[Eta]^5 - 
        8 \[Alpha] \[Eta]^4 (-20 + \[Eta] + \[Eta]^2) + 
        108 \[Alpha]^5 (12 - 6 \[Eta] - 3 \[Eta]^2 + \[Eta]^3) + 
        4 \[Alpha]^2 \[Eta]^3 (-144 - 27 \[Eta] + 
           12 \[Eta]^2 + \[Eta]^3) - 
        12 \[Alpha]^3 \[Eta]^2 (-84 - 21 \[Eta] + 3 \[Eta]^2 + 
           2 \[Eta]^3))) \[Lambda]

In the following, the numerator is defined as another function.

nume[\[Alpha]_, \[Eta]_] = (72 \[Alpha]^2 - 81 \[Alpha]^3 - 
   48 \[Alpha] \[Eta] + 54 \[Alpha]^3 \[Eta] + 8 \[Eta]^2 - 
   18 \[Alpha] \[Eta]^2 + 18 \[Alpha]^2 \[Eta]^2 - 
   9 \[Alpha]^3 \[Eta]^2 + 6 \[Eta]^3 - 6 \[Alpha]^2 \[Eta]^3 - 
   2 \[Eta]^4 + 
   2 \[Alpha] \[Eta]^4 + (-3 + \[Eta]) \[Sqrt](81 \[Alpha]^6 (-3 + \
\[Eta])^2 + 324 \[Alpha]^4 (-4 + \[Eta]) \[Eta] + 
       4 (-4 + \[Eta]) \[Eta]^5 - 
       8 \[Alpha] \[Eta]^4 (-20 + \[Eta] + \[Eta]^2) + 
       108 \[Alpha]^5 (12 - 6 \[Eta] - 3 \[Eta]^2 + \[Eta]^3) + 
       4 \[Alpha]^2 \[Eta]^3 (-144 - 27 \[Eta] + 
          12 \[Eta]^2 + \[Eta]^3) - 
       12 \[Alpha]^3 \[Eta]^2 (-84 - 21 \[Eta] + 3 \[Eta]^2 + 
          2 \[Eta]^3)))

Then, solving the equation

FullSimplify[Solve[frac[\[Alpha], \[Eta], \[Lambda]] == 0, \[Alpha]]]

will output three roots of alpha, and only one of the roots is real.

{{\[Alpha] -> 
   1/(9 (-3 + \[Eta])^2) (6 + 6 \[Eta] + 6 \[Eta]^2 - 
      2 \[Eta]^3 + (36 + 
         2 \[Eta] (-126 + \[Eta] (-243 + \[Eta] (276 + \[Eta] (-12 + 
                    5 (-6 + \[Eta]) \[Eta])))))/(9 (24 + 
            Sqrt[-(-3 + \[Eta])^6 \[Eta]^3 (-8 + \[Eta] (-23 + \[Eta] \
(2 + \[Eta]))) (-9 + \[Eta] (36 + \[Eta] (558 + \[Eta] (-744 + \[Eta] \
(327 + 4 (-15 + \[Eta]) \[Eta])))))]) + \[Eta] (-2268 + \[Eta] (-405 \
+ \[Eta] (15066 + \[Eta] (-36477 + \[Eta] (30942 + \[Eta] (-10431 + \
\[Eta] (990 + (153 - 26 \[Eta]) \[Eta]))))))))^(
       1/
        3) + (9 (24 + 
           Sqrt[-(-3 + \[Eta])^6 \[Eta]^3 (-8 + \[Eta] (-23 + \[Eta] \
(2 + \[Eta]))) (-9 + \[Eta] (36 + \[Eta] (558 + \[Eta] (-744 + \[Eta] \
(327 + 4 (-15 + \[Eta]) \[Eta])))))]) + \[Eta] (-2268 + \[Eta] (-405 \
+ \[Eta] (15066 + \[Eta] (-36477 + \[Eta] (30942 + \[Eta] (-10431 + \
\[Eta] (990 + (153 - 26 \[Eta]) \[Eta]))))))))^(1/3))}}

Name the above real root as the variable \[Alpha]1.

\[Alpha]1 = 
 1/(9 (-3 + \[Eta])^2) (6 + 6 \[Eta] + 6 \[Eta]^2 - 
    2 \[Eta]^3 + (36 + 
       2 \[Eta] (-126 + \[Eta] (-243 + \[Eta] (276 + \[Eta] (-12 + 
                   5 (-6 + \[Eta]) \[Eta])))))/(9 (24 + 
          Sqrt[-(-3 + \[Eta])^6 \[Eta]^3 (-8 + \[Eta] (-23 + \[Eta] \
(2 + \[Eta]))) (-9 + \[Eta] (36 + \[Eta] (558 + \[Eta] (-744 + \[Eta] \
(327 + 4 (-15 + \[Eta]) \[Eta])))))]) + \[Eta] (-2268 + \[Eta] (-405 \
+ \[Eta] (15066 + \[Eta] (-36477 + \[Eta] (30942 + \[Eta] (-10431 + \
\[Eta] (990 + (153 - 26 \[Eta]) \[Eta]))))))))^(
     1/3) + (9 (24 + 
         Sqrt[-(-3 + \[Eta])^6 \[Eta]^3 (-8 + \[Eta] (-23 + \[Eta] (2 \
+ \[Eta]))) (-9 + \[Eta] (36 + \[Eta] (558 + \[Eta] (-744 + \[Eta] \
(327 + 4 (-15 + \[Eta]) \[Eta])))))]) + \[Eta] (-2268 + \[Eta] (-405 \
+ \[Eta] (15066 + \[Eta] (-36477 + \[Eta] (30942 + \[Eta] (-10431 + \
\[Eta] (990 + (153 - 26 \[Eta]) \[Eta]))))))))^(1/3))

Then, solving the equation

FullSimplify[Solve[nume[\[Alpha], \[Eta]] == 0, \[Alpha]]]

will output four roots of alpha, and two of the roots are real.

{{\[Alpha] -> \[Eta]/
   3}, {\[Alpha] -> -(1/(
     9 (-3 + \[Eta])^2)) (-6 + 
      2 \[Eta] (-3 + (-3 + \[Eta]) \[Eta]) + (36 + 
         2 \[Eta] (-126 + \[Eta] (-243 + \[Eta] (276 + \[Eta] (-12 + 
                    5 (-6 + \[Eta]) \[Eta])))))/(9 (-24 + 
            Sqrt[-(-3 + \[Eta])^6 \[Eta]^3 (-8 + \[Eta] (-23 + \[Eta] \
(2 + \[Eta]))) (-9 + \[Eta] (36 + \[Eta] (558 + \[Eta] (-744 + \[Eta] \
(327 + 4 (-15 + \[Eta]) \[Eta])))))]) + \[Eta] (2268 + \[Eta] (405 + \
\[Eta] (-15066 + \[Eta] (36477 + \[Eta] (-30942 + \[Eta] (10431 + \
\[Eta] (-990 + \[Eta] (-153 + 26 \[Eta])))))))))^(
       1/
        3) + (9 (-24 + 
           Sqrt[-(-3 + \[Eta])^6 \[Eta]^3 (-8 + \[Eta] (-23 + \[Eta] \
(2 + \[Eta]))) (-9 + \[Eta] (36 + \[Eta] (558 + \[Eta] (-744 + \[Eta] \
(327 + 4 (-15 + \[Eta]) \[Eta])))))]) + \[Eta] (2268 + \[Eta] (405 + \
\[Eta] (-15066 + \[Eta] (36477 + \[Eta] (-30942 + \[Eta] (10431 + \
\[Eta] (-990 + \[Eta] (-153 + 26 \[Eta])))))))))^(1/3))}}

But the first one will make the denominator equal to zero. So there is only one eligible real root. Name the eligible real root as the variable \[Alpha]2.

\[Alpha]2 = -(1/(
   9 (-3 + \[Eta])^2)) (-6 + 
    2 \[Eta] (-3 + (-3 + \[Eta]) \[Eta]) + (36 + 
       2 \[Eta] (-126 + \[Eta] (-243 + \[Eta] (276 + \[Eta] (-12 + 
                   5 (-6 + \[Eta]) \[Eta])))))/(9 (-24 + 
          Sqrt[-(-3 + \[Eta])^6 \[Eta]^3 (-8 + \[Eta] (-23 + \[Eta] \
(2 + \[Eta]))) (-9 + \[Eta] (36 + \[Eta] (558 + \[Eta] (-744 + \[Eta] \
(327 + 4 (-15 + \[Eta]) \[Eta])))))]) + \[Eta] (2268 + \[Eta] (405 + \
\[Eta] (-15066 + \[Eta] (36477 + \[Eta] (-30942 + \[Eta] (10431 + \
\[Eta] (-990 + \[Eta] (-153 + 26 \[Eta])))))))))^(
     1/3) + (9 (-24 + 
         Sqrt[-(-3 + \[Eta])^6 \[Eta]^3 (-8 + \[Eta] (-23 + \[Eta] (2 \
+ \[Eta]))) (-9 + \[Eta] (36 + \[Eta] (558 + \[Eta] (-744 + \[Eta] \
(327 + 4 (-15 + \[Eta]) \[Eta])))))]) + \[Eta] (2268 + \[Eta] (405 + \
\[Eta] (-15066 + \[Eta] (36477 + \[Eta] (-30942 + \[Eta] (10431 + \
\[Eta] (-990 + \[Eta] (-153 + 26 \[Eta])))))))))^(1/3))

The two eligible real roots are found to be unequal.

FullSimplify[\[Alpha]1 - \[Alpha]2]

will output

(1/(9 (-3 + \[Eta])^2))(6 + 6 \[Eta] + 6 \[Eta]^2 - 
   2 \[Eta]^3 + (36 + 
      2 \[Eta] (-126 + \[Eta] (-243 + \[Eta] (276 + \[Eta] (-12 + 
                  5 (-6 + \[Eta]) \[Eta])))))/(9 (24 + 
         Sqrt[-(-3 + \[Eta])^6 \[Eta]^3 (-8 + \[Eta] (-23 + \[Eta] (2 \
+ \[Eta]))) (-9 + \[Eta] (36 + \[Eta] (558 + \[Eta] (-744 + \[Eta] \
(327 + 4 (-15 + \[Eta]) \[Eta])))))]) + \[Eta] (-2268 + \[Eta] (-405 \
+ \[Eta] (15066 + \[Eta] (-36477 + \[Eta] (30942 + \[Eta] (-10431 + \
\[Eta] (990 + (153 - 26 \[Eta]) \[Eta]))))))))^(
    1/3) + (9 (24 + 
        Sqrt[-(-3 + \[Eta])^6 \[Eta]^3 (-8 + \[Eta] (-23 + \[Eta] (2 \
+ \[Eta]))) (-9 + \[Eta] (36 + \[Eta] (558 + \[Eta] (-744 + \[Eta] \
(327 + 4 (-15 + \[Eta]) \[Eta])))))]) + \[Eta] (-2268 + \[Eta] (-405 \
+ \[Eta] (15066 + \[Eta] (-36477 + \[Eta] (30942 + \[Eta] (-10431 + \
\[Eta] (990 + (153 - 26 \[Eta]) \[Eta]))))))))^(1/3)) + (1/(
 9 (-3 + \[Eta])^2))(-6 + 
   2 \[Eta] (-3 + (-3 + \[Eta]) \[Eta]) + (36 + 
      2 \[Eta] (-126 + \[Eta] (-243 + \[Eta] (276 + \[Eta] (-12 + 
                  5 (-6 + \[Eta]) \[Eta])))))/(9 (-24 + 
         Sqrt[-(-3 + \[Eta])^6 \[Eta]^3 (-8 + \[Eta] (-23 + \[Eta] (2 \
+ \[Eta]))) (-9 + \[Eta] (36 + \[Eta] (558 + \[Eta] (-744 + \[Eta] \
(327 + 4 (-15 + \[Eta]) \[Eta])))))]) + \[Eta] (2268 + \[Eta] (405 + \
\[Eta] (-15066 + \[Eta] (36477 + \[Eta] (-30942 + \[Eta] (10431 + \
\[Eta] (-990 + \[Eta] (-153 + 26 \[Eta])))))))))^(
    1/3) + (9 (-24 + 
        Sqrt[-(-3 + \[Eta])^6 \[Eta]^3 (-8 + \[Eta] (-23 + \[Eta] (2 \
+ \[Eta]))) (-9 + \[Eta] (36 + \[Eta] (558 + \[Eta] (-744 + \[Eta] \
(327 + 4 (-15 + \[Eta]) \[Eta])))))]) + \[Eta] (2268 + \[Eta] (405 + \
\[Eta] (-15066 + \[Eta] (36477 + \[Eta] (-30942 + \[Eta] (10431 + \
\[Eta] (-990 + \[Eta] (-153 + 26 \[Eta])))))))))^(1/3))

Any idea of what is going on here?

Why do numerator equals zero and fraction equals zero give different results?

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1 Answer 1

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$Version

(* "13.1.0 for Mac OS X x86 (64-bit) (June 16, 2022)" *)

Clear["Global`*"]

frac[α_, η_, λ_] = 
  1/(4 (-4 + η) (-3 α + η)^2) (72 α^2 - 
     81 α^3 - 48 α η + 54 α^3 η + 8 η^2 - 
     18 α η^2 + 18 α^2 η^2 - 9 α^3 η^2 + 
     6 η^3 - 6 α^2 η^3 - 2 η^4 + 
     2 α η^4 + (-3 + η) √(81 α^6 (-3 + \
η)^2 + 324 α^4 (-4 + η) η + 4 (-4 + η) η^5 - 
         8 α η^4 (-20 + η + η^2) + 
         108 α^5 (12 - 6 η - 3 η^2 + η^3) + 
         4 α^2 η^3 (-144 - 27 η + 12 η^2 + η^3) - 
         12 α^3 η^2 (-84 - 21 η + 3 η^2 + 
            2 η^3))) λ;

nume[α_, η_] = (72 α^2 - 81 α^3 - 
    48 α η + 54 α^3 η + 8 η^2 - 
    18 α η^2 + 18 α^2 η^2 - 9 α^3 η^2 + 
    6 η^3 - 6 α^2 η^3 - 2 η^4 + 
    2 α η^4 + (-3 + η) √(81 α^6 (-3 + η)^2 \
+ 324 α^4 (-4 + η) η + 4 (-4 + η) η^5 - 
        8 α η^4 (-20 + η + η^2) + 
        108 α^5 (12 - 6 η - 3 η^2 + η^3) + 
        4 α^2 η^3 (-144 - 27 η + 12 η^2 + η^3) - 
        12 α^3 η^2 (-84 - 21 η + 3 η^2 + 2 η^3)));

Your nume is not the numerator of frac

Numerator[frac[α, η, λ]] - 
  nume[α, η] // FullSimplify

(* (-9 α^3 (-3 + η)^2 - 
   6 α^2 (-12 + (-3 + η) η^2) + 
   2 α η (-24 - 9 η + η^3) - 
   3 √(81 α^5 (16 + 9 α) - 
       162 α^4 (8 + α (4 + 3 α)) η + 
       9 α^3 (112 + 9 (-2 + α)^2 α) η^2 + 
       36 α^2 (-16 + 7 α + 3 α^3) η^3 - 
       4 α (-40 + 9 α (3 + α)) η^4 - 
       8 (2 + α + 3 (-2 + α) α^2) η^5 + 
       4 (-1 + α)^2 η^6) + η (-2 (-4 + η) η (1 + \
η) + √(81 α^5 (16 + 9 α) - 
         162 α^4 (8 + α (4 + 3 α)) η + 
         9 α^3 (112 + 9 (-2 + α)^2 α) η^2 + 
         36 α^2 (-16 + 7 α + 3 α^3) η^3 - 
         4 α (-40 + 9 α (3 + α)) η^4 - 
         8 (2 + α + 3 (-2 + α) α^2) η^5 + 
         4 (-1 + α)^2 η^6))) (-1 + λ) *)

nume is missing a factor of λ

Numerator[frac[α, η, λ]] == λ*nume[α, η] // FullSimplify

(* True *)

rootsF = Solve[frac[α, η, λ] == 0, α, Reals]

enter image description here

rootsN = Solve[Numerator[frac[α, η, λ]] == 0, α, Reals]

enter image description here

The last three roots of rootsN are the same as rootsF

Normal[α /. rootsF] === Normal[α /. Rest@rootsN]

(* True *)

The fourth (first) root of rootsN is cancelled by the corresponding root of the Denominator

Solve[Denominator[frac[α, η, λ]] == 0, α]

(* {{α -> η/3}, {α -> η/3}} *)
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  • $\begingroup$ Thank you for being so helpful. I did omit a factor of λ in the numerator function in the problem described above. However, this factor does not seem to affect the result of solving the equation, since the lambda is multiplied separately. You wrote in your reply "The last three roots of rootsN are the same as rootsF", but the result I got is still not equal. Can you please add more details? For example, the specific expressions of the roots. $\endgroup$
    – Willow.L
    Dec 7, 2022 at 1:41
  • $\begingroup$ The roots are shown above. Execute the code to get a form that you can work with. $\endgroup$
    – Bob Hanlon
    Dec 7, 2022 at 1:47

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