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Find the number of real solutions to \(x = 1 - x + x^2 - x^3 + x^4 - x^5 + \dotsb.\)

 Jul 1, 2019

Best Answer 

 #5
avatar+23510 
+3

Find the number of real solutions to \(x = 1 - x + x^2 - x^3 + x^4 - x^5 + \dotsb\).

 

\(\begin{array}{|rcll|} \hline \mathbf{x} &=& \mathbf{1 - x + x^2 - x^3 + x^4 - x^5 + \dotsb} \\ \\ x &=& \underbrace{ 1 - x + x^2 - x^3 + x^4 - x^5 + \dotsb }_{\text{Infinite Geometric Series}} \\ && a=1,\ r= \dfrac{-x}{1}=\dfrac{x^2}{-x}=\dfrac{-x^3}{x^2}=\dotsb=-x \\ \\ && \boxed{sum = \dfrac{a}{1-r} \text{ as long as } -1 < r < 1 } \\ \\ && sum = \dfrac{1}{1-(-x)} \\ && sum = \dfrac{1}{1+x } \\\\ x &=& \dfrac{1}{1+x } \\ x(1+x) &=& 1 \\ x+x^2 &=& 1 \\ x^2+x-1 &=& 0 \\\\ x &=& \dfrac{-1\pm\sqrt{1^2-4\cdot (-1)}} {2} \\ &=& \dfrac{-1\pm\sqrt{5}} {2} \\\\ \mathbf{x_1} &=& \mathbf{ \dfrac{-1+\sqrt{5}} {2}} \\ r &=& -x_1 \\ &=& -\dfrac{-1+\sqrt{5}} {2} \\ &=& \dfrac{1-\sqrt{5}} {2} \\ r &=& -0.61803398875 \quad | \quad -1 < -0.61803398875 < 1 \checkmark \\\\ \mathbf{x_2} &=& \mathbf{ \dfrac{-1-\sqrt{5}} {2}} \\ r &=& -x_2 \\ &=& -\dfrac{-1-\sqrt{5}} {2} \\ &=& \dfrac{ 1+\sqrt{5}} {2} \\ r &=& 1.61803398875 \quad | \quad -1 < 1.61803398875 < 1 \text{ false, no solution!} \\ \hline \end{array}\)

 

Real solution is \(x=\mathbf{ \dfrac{-1+\sqrt{5}} {2}}\)

 

laugh

 Jul 2, 2019
edited by heureka  Jul 2, 2019
edited by heureka  Jul 3, 2019
 #1
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0

This is the expansion of: x = 1 / (x + 1) =x^2 + x - 1=0 . Use the quadratic formula to find x;

x = 1/2*(sqrt(5) - 1)

 

x = 1/2*(-1-sqrt(5))

 Jul 1, 2019
 #2
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0

I'm not sure if this is relevant but for x = 1/2*(-1-sqrt(5)) the RHS doesn't converge

Guest Jul 1, 2019
 #4
avatar+105195 
+2

The series     1  + x^2  + x^4 + .......  sums  to     1 / ( 1 - x^2)

The series     -[x  + x^3  + x^5  +  ] .........sums  to   -  [  x / (1 -x^2 )

 

So

 

x  =   1 /(1 - x^2)  - x / (1 -x^2)

 

x =  (1 - x) / (1 -x^2)

 

x ( 1 - x^2)  = (1 - x)

 

x - x^3  =  1 - x

 

x^3 - 2x + 1  =  0   (1)

 

x = 1  is a solution  to  (1)

 

Using synthetic division, we have

 

 

1  [  1  0   - 2    1  ]

           1    1    -1

      ____________

       1  1    -1     0

 

The remaining polynomial is

 

x^2  + x  - 1    =   0          complete the square on x

 

x^2 + x + 1/4   =  1 + 1/4

 

( x + 1/2)^2   =  5/4          take both roots

 

x + 1/2  = ±√5 /2

 

x  =   -1 ±√5

         _____         ( 2)

            2

 

But x = 1 makes the original equation undefined....so....the only solutions  are represented by (2)

 

 

cool cool cool

 Jul 1, 2019
 #5
avatar+23510 
+3
Best Answer

Find the number of real solutions to \(x = 1 - x + x^2 - x^3 + x^4 - x^5 + \dotsb\).

 

\(\begin{array}{|rcll|} \hline \mathbf{x} &=& \mathbf{1 - x + x^2 - x^3 + x^4 - x^5 + \dotsb} \\ \\ x &=& \underbrace{ 1 - x + x^2 - x^3 + x^4 - x^5 + \dotsb }_{\text{Infinite Geometric Series}} \\ && a=1,\ r= \dfrac{-x}{1}=\dfrac{x^2}{-x}=\dfrac{-x^3}{x^2}=\dotsb=-x \\ \\ && \boxed{sum = \dfrac{a}{1-r} \text{ as long as } -1 < r < 1 } \\ \\ && sum = \dfrac{1}{1-(-x)} \\ && sum = \dfrac{1}{1+x } \\\\ x &=& \dfrac{1}{1+x } \\ x(1+x) &=& 1 \\ x+x^2 &=& 1 \\ x^2+x-1 &=& 0 \\\\ x &=& \dfrac{-1\pm\sqrt{1^2-4\cdot (-1)}} {2} \\ &=& \dfrac{-1\pm\sqrt{5}} {2} \\\\ \mathbf{x_1} &=& \mathbf{ \dfrac{-1+\sqrt{5}} {2}} \\ r &=& -x_1 \\ &=& -\dfrac{-1+\sqrt{5}} {2} \\ &=& \dfrac{1-\sqrt{5}} {2} \\ r &=& -0.61803398875 \quad | \quad -1 < -0.61803398875 < 1 \checkmark \\\\ \mathbf{x_2} &=& \mathbf{ \dfrac{-1-\sqrt{5}} {2}} \\ r &=& -x_2 \\ &=& -\dfrac{-1-\sqrt{5}} {2} \\ &=& \dfrac{ 1+\sqrt{5}} {2} \\ r &=& 1.61803398875 \quad | \quad -1 < 1.61803398875 < 1 \text{ false, no solution!} \\ \hline \end{array}\)

 

Real solution is \(x=\mathbf{ \dfrac{-1+\sqrt{5}} {2}}\)

 

laugh

heureka Jul 2, 2019
edited by heureka  Jul 2, 2019
edited by heureka  Jul 3, 2019

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