Shift theorem

<h2 id="statement">Statement</h2>
<p>The theorem states that, if <i>P</i>(<i>D</i>) is a polynomial of the <i>D</i>-operator, then, for any sufficiently <a href="/facts/Differentiable_function/jQqTgLk1">differentiable</a> function <i>y</i>,
</p>

P
        (
        D
        )
        (
        
          e
          
            a
            x
          
        
        y
        )
        ≡
        
          e
          
            a
            x
          
        
        P
        (
        D
        +
        a
        )
        y
        .
      
    
    {\displaystyle P(D)(e^{ax}y)\equiv e^{ax}P(D+a)y.}

<p>To prove the result, proceed by <a href="/facts/Mathematical_induction/KxAPqNrH">induction</a>. Note that only the special case
</p>

P
        (
        D
        )
        =
        
          D
          
            n
          
        
      
    
    {\displaystyle P(D)=D^{n}}

<p>needs to be proved, since the general result then follows by <a href="/facts/Linearity_of_differentiation/zWJ2Pqap">linearity</a> of <i>D</i>-operators.
</p><p>The result is clearly true for <i>n</i> = 1 since
</p>

D
        (
        
          e
          
            a
            x
          
        
        y
        )
        =
        
          e
          
            a
            x
          
        
        (
        D
        +
        a
        )
        y
        .
      
    
    {\displaystyle D(e^{ax}y)=e^{ax}(D+a)y.}

<p>Now suppose the result true for <i>n</i> = <i>k</i>, that is,
</p>

D
          
            k
          
        
        (
        
          e
          
            a
            x
          
        
        y
        )
        =
        
          e
          
            a
            x
          
        
        (
        D
        +
        a
        
          )
          
            k
          
        
        y
        .
      
    
    {\displaystyle D^{k}(e^{ax}y)=e^{ax}(D+a)^{k}y.}

D
                  
                    k
                    +
                    1
                  
                
                (
                
                  e
                  
                    a
                    x
                  
                
                y
                )
              
              
                
                ≡
                
                  
                    d
                    
                      d
                      x
                    
                  
                
                
                  {
                  
                    
                      e
                      
                        a
                        x
                      
                    
                    
                      
                        (
                        
                          D
                          +
                          a
                        
                        )
                      
                      
                        k
                      
                    
                    y
                  
                  }

=
                
                  e
                  
                    a
                    x
                  
                
                
                  
                    d
                    
                      d
                      x
                    
                  
                
                
                  {
                  
                    
                      
                        (
                        
                          D
                          +
                          a
                        
                        )
                      
                      
                        k
                      
                    
                    y
                  
                  }
                
                +
                a
                
                  e
                  
                    a
                    x
                  
                
                
                  {
                  
                    
                      
                        (
                        
                          D
                          +
                          a
                        
                        )
                      
                      
                        k
                      
                    
                    y
                  
                  }

=
                
                  e
                  
                    a
                    x
                  
                
                
                  {
                  
                    
                      (
                      
                        
                          
                            d
                            
                              d
                              x
                            
                          
                        
                        +
                        a
                      
                      )
                    
                    
                      
                        (
                        
                          D
                          +
                          a
                        
                        )
                      
                      
                        k
                      
                    
                    y
                  
                  }

=
                
                  e
                  
                    a
                    x
                  
                
                (
                D
                +
                a
                
                  )
                  
                    k
                    +
                    1
                  
                
                y
                .
              
            
          
        
      
    
    {\displaystyle {\begin{aligned}D^{k+1}(e^{ax}y)&\equiv {\frac {d}{dx}}\left\{e^{ax}\left(D+a\right)^{k}y\right\}\\&{}=e^{ax}{\frac {d}{dx}}\left\{\left(D+a\right)^{k}y\right\}+ae^{ax}\left\{\left(D+a\right)^{k}y\right\}\\&{}=e^{ax}\left\{\left({\frac {d}{dx}}+a\right)\left(D+a\right)^{k}y\right\}\\&{}=e^{ax}(D+a)^{k+1}y.\end{aligned}}}

<p>This completes the proof.
</p><p>The shift theorem can be applied equally well to inverse operators:
</p>

1
            
              P
              (
              D
              )
            
          
        
        (
        
          e
          
            a
            x
          
        
        y
        )
        =
        
          e
          
            a
            x
          
        
        
          
            1
            
              P
              (
              D
              +
              a
              )
            
          
        
        y
        .
      
    
    {\displaystyle {\frac {1}{P(D)}}(e^{ax}y)=e^{ax}{\frac {1}{P(D+a)}}y.}

<h2 id="related">Related</h2>
<p>There is a similar version of the shift theorem for <a href="/facts/Laplace_transform/0Ge4MIc9">Laplace transforms</a> (
  
    
      
        t
        <
        a
      
    
    {\displaystyle t<a}
  
):
</p>

e
          
            −
            a
            s
          
        
        
          
            L
          
        
        {
        f
        (
        t
        )
        }
        =
        
          
            L
          
        
        {
        f
        (
        t
        −
        a
        )
        }
        .
      
    
    {\displaystyle e^{-as}{\mathcal {L}}\{f(t)\}={\mathcal {L}}\{f(t-a)\}.}

<h2 id="examples">Examples</h2>
<p>The exponential shift theorem can be used to speed the calculation of higher derivatives of functions that is given by the product of an exponential and another function. For instance, if 
  
    
      
        f
        (
        x
        )
        =
        sin
        ⁡
        (
        x
        )
        
          e
          
            x
          
        
      
    
    {\displaystyle f(x)=\sin(x)e^{x}}
  
, one has that
</p><p>
  
    
      
        
          
            
              
                
                  D
                  
                    3
                  
                
                f
              
              
                
                =
                
                  D
                  
                    3
                  
                
                (
                
                  e
                  
                    x
                  
                
                sin
                ⁡
                (
                x
                )
                )
                =
                
                  e
                  
                    x
                  
                
                (
                D
                +
                1
                
                  )
                  
                    3
                  
                
                sin
                ⁡
                (
                x
                )
              
            
            
              
              
                
                =
                
                  e
                  
                    x
                  
                
                
                  (
                  
                    
                      D
                      
                        3
                      
                    
                    +
                    3
                    
                      D
                      
                        2
                      
                    
                    +
                    3
                    D
                    +
                    1
                  
                  )
                
                sin
                ⁡
                (
                x
                )
              
            
            
              
              
                
                =
                
                  e
                  
                    x
                  
                
                
                  (
                  
                    −
                    cos
                    ⁡
                    (
                    x
                    )
                    −
                    3
                    sin
                    ⁡
                    (
                    x
                    )
                    +
                    3
                    cos
                    ⁡
                    (
                    x
                    )
                    +
                    sin
                    ⁡
                    (
                    x
                    )
                  
                  )
                
              
            
          
        
      
    
    {\displaystyle {\begin{aligned}D^{3}f&=D^{3}(e^{x}\sin(x))=e^{x}(D+1)^{3}\sin(x)\\&=e^{x}\left(D^{3}+3D^{2}+3D+1\right)\sin(x)\\&=e^{x}\left(-\cos(x)-3\sin(x)+3\cos(x)+\sin(x)\right)\end{aligned}}}

</p><p>Another application of the exponential shift theorem is to solve <a href="/facts/Linear_differential_equation/nLEbIIVf">linear differential equations</a> whose <a href="/facts/Characteristic_equation_(calculus)/tBqj8HUJ">characteristic polynomial</a> has repeated roots.<a class="footnote-ref" id="fnref:1" href="#fn:1"><sup>1</sup></a>
</p>
<h2 id="notes">Notes</h2>

<ul><li>Morris, Tenenbaum; Pollard, Harry (1985). <a href="https://archive.org/details/ordinarydifferen00tene_0"><i>Ordinary differential equations : an elementary textbook for students of mathematics, engineering, and the sciences</i></a>. New York: Dover Publications. <a href="/facts/ISBN_(identifier)/15AdSPa9">ISBN</a> 0486649407. <a href="/facts/OCLC_(identifier)/Yqad8waq">OCLC</a> <a href="https://search.worldcat.org/oclc/12188701">12188701</a>.</li></ul>

<h2 id="references">References</h2>

<ol>
<li id="fn:1"><p>See the article homogeneous equation with constant coefficients for more details. <a href="/wiki/Linear_differential_equation#Homogeneous_equation_with_constant_coefficients" target="_blank">/wiki/Linear_differential_equation#Homogeneous_equation_with_constant_coefficients</a> <a href="#fnref:1" class="footnote-back-ref">↩</a></p></li>
</ol>

Shift theorem open-in-new

Shift theorem