Change of variables

In <a href="/facts/Mathematics/pxTouaz4">mathematics</a>, a change of variables is a basic technique used to simplify problems in which the original <a href="/facts/Variable_(mathematics)/VbQlrvKz">variables</a> are replaced with <a href="/facts/Function_(mathematics)/CdReEKCh">functions</a> of other variables. The intent is that when expressed in new variables, the problem may become simpler, or equivalent to a better understood problem.
Change of variables is an operation that is related to <a href="/facts/Substitution_(algebra)/GwWdQf0m">substitution</a>. However these are different operations, as can be seen when considering <a href="/facts/Derivative/7Ito70Dh">differentiation</a> (<a href="/facts/Chain_rule/aMLYxP0x">chain rule</a>) or <a href="/facts/Integral/V7lyV997">integration</a> (<a href="/facts/Integration_by_substitution/NvXfTTaf">integration by substitution</a>).
A very simple example of a useful variable change can be seen in the problem of finding the roots of the sixth-degree polynomial:

x
          
            6
          
        
        −
        9
        
          x
          
            3
          
        
        +
        8
        =
        0.
      
    
    {\displaystyle x^{6}-9x^{3}+8=0.}

Sixth-degree polynomial equations are generally impossible to solve in terms of radicals (see <a href="/facts/Abel%E2%80%93Ruffini_theorem/tEFaRGvW">Abel–Ruffini theorem</a>). This particular equation, however, may be written

(
        
          x
          
            3
          
        
        
          )
          
            2
          
        
        −
        9
        (
        
          x
          
            3
          
        
        )
        +
        8
        =
        0
      
    
    {\displaystyle (x^{3})^{2}-9(x^{3})+8=0}

(this is a simple case of a <a href="/facts/Polynomial_decomposition/WkL9Yyxn">polynomial decomposition</a>). Thus the equation may be simplified by defining a new variable 
 
 
 
 u
 =
 
 x
 
 3
 
 
 
 
 {\displaystyle u=x^{3}}
 
. Substituting x by 
 
 
 
 
 
 u
 
 3
 
 
 
 
 
 {\displaystyle {\sqrt[{3}]{u}}}
 
 into the polynomial gives

u
          
            2
          
        
        −
        9
        u
        +
        8
        =
        0
        ,
      
    
    {\displaystyle u^{2}-9u+8=0,}

which is just a <a href="/facts/Quadratic_equation/9IMnkJAT">quadratic equation</a> with the two solutions:

u
        =
        1
        
        
          and
        
        
        u
        =
        8.
      
    
    {\displaystyle u=1\quad {\text{and}}\quad u=8.}

The solutions in terms of the original variable are obtained by substituting x3 back in for u, which gives

x
          
            3
          
        
        =
        1
        
        
          and
        
        
        
          x
          
            3
          
        
        =
        8.
      
    
    {\displaystyle x^{3}=1\quad {\text{and}}\quad x^{3}=8.}

Then, assuming that one is interested only in <a href="/facts/Real_number/R02gw5Pb">real</a> solutions, the solutions of the original equation are

x
        =
        (
        1
        
          )
          
            1
            
              /
            
            3
          
        
        =
        1
        
        
          and
        
        
        x
        =
        (
        8
        
          )
          
            1
            
              /
            
            3
          
        
        =
        2.
      
    
    {\displaystyle x=(1)^{1/3}=1\quad {\text{and}}\quad x=(8)^{1/3}=2.}

Change of variables open-in-new

Change of variables