Skip to main content

Am I missing anything? Solving Equations


I want to know what $ a^4+b^4+c^4 $ would be when


$ a+b+c=1, a^2+b^2+c^2=2, a^3+b^3+c^3=3. $


I tried


Solve[{a + b + c == 1, a^2 + b^2 + c^2 == 2, 
a^3 + b^3 + c^3 == 3}, {a, b, c}]


a^4 + b^4 + c^4 /. {a -> ... }

I copied and pasted one instance from the result of Solve[] into {a -> ...}


The solution and the result of Simplify[] of it was not quite what I expected.


So as a test, I tried a+b+c /. {a -> ...}


Still it was strange, it didn't come up with 1 which is obvious correct answer.


So am I missing anything here?


Is there any better way?



Answer



Solve your equations with respect to another variable d == a^4 + b^4 + c^4 eliminating given ones



Solve[{ a + b + c == 1,       a^2 + b^2 + c^2 == 2, 
a^3 + b^3 + c^3 == 3, a^4 + b^4 + c^4 == d}, {d}, {a, b, c}]


{{d -> 25/6}}

This tutorial will be helpful Eliminating Variables.


Edit


For the sake of completeness we add another methods of dealing with the problem. Let's point out even simpler way than the first one, just add assumptions to Simplify:


Simplify[ a^4 + b^4 + c^4, {a + b + c == 1, a^2 + b^2 + c^2 == 2, a^3 + b^3 + c^3 == 3}]



25/6

Of course we can use FullSimplify as well. In this case, Simplify works nicely, but the first method is stronger and should work in more general cases.


SymmetricReduction is more appropriate when we would like to get symbolic results, i.e. here we would like to express a^4 + b^4 + c^4 in terms of these polynomials: {a + b + c, a^2 + b^2 + c^2, a^3 + b^3 + c^3}


First we need to get rid of unknown symmetric polynomials (we know only a + b + c):


 SymmetricPolynomial[#, {a, b, c}] & /@ Range[3]



{a + b + c, a b + a c + b c, a b c}

form the last reduction using the previous ones:


First @ SymmetricReduction[ a^# + b^# + c^#, {a, b, c}] & /@ Range[4] // Column


a + b + c

(a + b + c)^2 - 2 (a b + a c + b c)


3 a b c + (a + b + c)^3 - 3 (a + b + c) (a b + a c + b c)

4 a b c (a + b + c) + (a + b + c)^4 - 4 (a + b + c)^2 (a b + a c + b c)
+ 2 (a b + a c + b c)^2

Now we have a^4 + b^4 + c^4 expressed in terms of given polynomials:


str = 
First[ SymmetricReduction[ a^4 + b^4 + c^4, {a, b, c}]] //. {
(a b + a c + b c) -> 1/2 (a + b + c)^2 - 1/2 (a^2 + b^2 + c^2),
(a b c) -> 1/3 (a^3 + b^3 + c^3) - 1/3 (a + b + c)^3 + (a + b + c) (a b + a c + b c)}



 (a + b + c)^4 - 4 (a + b + c)^2 (1/2 (a + b + c)^2 + 1/2 (-a^2 - b^2 - c^2))
+ 2 (1/2(a + b + c)^2 + 1/2(-a^2 - b^2 - c^2))^2 + 4(a + b + c) (-(1/3)(a + b + c)^3
+ 1/3 (a^3 + b^3 + c^3) + (a + b + c) (1/2 (a + b + c)^2 + 1/2 (-a^2 - b^2 - c^2)))

of course


Simplify @ %



a^4 + b^4 + c^4

and finally:


str /. {a + b + c -> 1, (-a^2 - b^2 - c^2) -> -2, (a^3 + b^3 + c^3) -> 3}


25/6

Comments

Popular posts from this blog

functions - Get leading series expansion term?

Given a function f[x] , I would like to have a function leadingSeries that returns just the leading term in the series around x=0 . For example: leadingSeries[(1/x + 2)/(4 + 1/x^2 + x)] x and leadingSeries[(1/x + 2 + (1 - 1/x^3)/4)/(4 + x)] -(1/(16 x^3)) Is there such a function in Mathematica? Or maybe one can implement it efficiently? EDIT I finally went with the following implementation, based on Carl Woll 's answer: lds[ex_,x_]:=( (ex/.x->(x+O[x]^2))/.SeriesData[U_,Z_,L_List,Mi_,Ma_,De_]:>SeriesData[U,Z,{L[[1]]},Mi,Mi+1,De]//Quiet//Normal) The advantage is, that this one also properly works with functions whose leading term is a constant: lds[Exp[x],x] 1 Answer Update 1 Updated to eliminate SeriesData and to not return additional terms Perhaps you could use: leadingSeries[expr_, x_] := Normal[expr /. x->(x+O[x]^2) /. a_List :> Take[a, 1]] Then for your examples: leadingSeries[(1/x + 2)/(4 + 1/x^2 + x), x] leadingSeries[Exp[x], x] leadingSeries[(1/x + 2 + (1 - 1/x...

How to thread a list

I have data in format data = {{a1, a2}, {b1, b2}, {c1, c2}, {d1, d2}} Tableform: I want to thread it to : tdata = {{{a1, b1}, {a2, b2}}, {{a1, c1}, {a2, c2}}, {{a1, d1}, {a2, d2}}} Tableform: And I would like to do better then pseudofunction[n_] := Transpose[{data2[[1]], data2[[n]]}]; SetAttributes[pseudofunction, Listable]; Range[2, 4] // pseudofunction Here is my benchmark data, where data3 is normal sample of real data. data3 = Drop[ExcelWorkBook[[Column1 ;; Column4]], None, 1]; data2 = {a #, b #, c #, d #} & /@ Range[1, 10^5]; data = RandomReal[{0, 1}, {10^6, 4}]; Here is my benchmark code kptnw[list_] := Transpose[{Table[First@#, {Length@# - 1}], Rest@#}, {3, 1, 2}] &@list kptnw2[list_] := Transpose[{ConstantArray[First@#, Length@# - 1], Rest@#}, {3, 1, 2}] &@list OleksandrR[list_] := Flatten[Outer[List, List@First[list], Rest[list], 1], {{2}, {1, 4}}] paradox2[list_] := Partition[Riffle[list[[1]], #], 2] & /@ Drop[list, 1] RM[list_] := FoldList[Transpose[{First@li...

front end - keyboard shortcut to invoke Insert new matrix

I frequently need to type in some matrices, and the menu command Insert > Table/Matrix > New... allows matrices with lines drawn between columns and rows, which is very helpful. I would like to make a keyboard shortcut for it, but cannot find the relevant frontend token command (4209405) for it. Since the FullForm[] and InputForm[] of matrices with lines drawn between rows and columns is the same as those without lines, it's hard to do this via 3rd party system-wide text expanders (e.g. autohotkey or atext on mac). How does one assign a keyboard shortcut for the menu item Insert > Table/Matrix > New... , preferably using only mathematica? Thanks! Answer In the MenuSetup.tr (for linux located in the $InstallationDirectory/SystemFiles/FrontEnd/TextResources/X/ directory), I changed the line MenuItem["&New...", "CreateGridBoxDialog"] to read MenuItem["&New...", "CreateGridBoxDialog", MenuKey["m", Modifiers-...