Constraints Of A Geometric Progression Sum?

Algebra Level 4

Find the sum of all values of $x$ satisfying

$x=1-x+x^2-x^3+x^4-x^5+\cdots .$

-1

0

1

\frac{-1-\sqrt{5}}{2}

\frac{-1+\sqrt{5}}{2}

3 solutions

Chew-Seong Cheong
Nov 26, 2016

$\begin{aligned} x & = 1-x+x^2-x^3+x^4-x^5+... \\ & = (1 + x^2 + x^4 + ...) - (x + x^3 + x^5 + ...) \\ & = (1 + x^2 + x^4 + ...) - x(1 + x^2 + x^4 + ...) \\ & = (1-x)\color{#3D99F6}(1 + x^2 + x^4 + ...) \\ & = \frac {1-x}{\color{#3D99F6}1-x^2} & \small \color{#3D99F6} \text{for } |x| < 1 \\ & = \frac {1-x}{(1-x)(1+x)} \\ & = \frac 1{1+x} \end{aligned}$

$\begin{aligned} \implies x^2 + x - 1 & = 0 \\ \implies x & = \frac {-1+\sqrt 5}2 & \small \color{#D61F06} \text{Note that } \left| \frac {-1-\sqrt 5}2 \right| > 1 \text{ not acceptable} \end{aligned}$

There is only one solution of $x$ , therefore the sum of all solutions is $\boxed{\dfrac {-1+\sqrt 5}2}$ .

This is a test!

Thank you for the nice solution.

Hana Wehbi - 4 years, 6 months ago

I got wrong :(

This is my solution :

$\begin{aligned} x & = 1 -x+x^2 -x^3 + x^4 - x^5 + ... \\ x^3 & = x^2 -x^3 + x^4 - x^5 + ... \\ x - x^3 & = 1 - x \\ x^3 - 2x + 1 & = 0 \\ \end{aligned}$

By using Vieta Formula, the answer should be $0$

Note : I read some of the solution and i understood where does my solution goes wrong, but i just want to share it :)

Jason Chrysoprase - 4 years, 6 months ago

Why can't x=1 be a solution

Naveen Rock - 4 years, 6 months ago

Obviously not, if $x=1$ , then the answer will oscillates between $1$ and $0$ . Try to imagine that, does the expression will converge ?

Jason Chrysoprase - 4 years, 6 months ago

Ah, I had a feeling it wasn't 0 but I couldn't accurately show that $\dfrac{-1-\sqrt5}{2}$ was an unacceptable answer. Thanks for the solution!

Seth Christman - 4 years, 6 months ago

Hana Wehbi
Nov 25, 2016

Multiplying both sides by $(x+1$ ) yields into $x(x+1) = 1$ which is equal to

$x^2+x-1=0$ . Thus, the solution is $x= \frac{-1+\sqrt{5}}{2}$ . Also, there is an important comment mentioned below that must be considered here.

$x = \dfrac{-1 - \sqrt{5}}{2}$ is also a solution to $x^{2} + x - 1 = 0$ but since in this case $|x| \gt 1$ we can discard it, since the series on the RHS of the original equation only converges for $|x| \lt 1$ .

Brian Charlesworth - 4 years, 6 months ago

Yes, you are right. What you said must be added to my solution.

Hana Wehbi - 4 years, 6 months ago

Question, so for every converges, $|x|$ must bigger than $1$ ?

I still don't understand when or where should we use $|x| > 1$ but this explain where did i got wrong

This below is my solution

$\begin{aligned} x & = 1 -x+x^2 -x^3 + x^4 - x^5 + ... \\ x^3 & = x^2 -x^3 + x^4 - x^5 + ... \\ x - x^3 & = 1 - x \\ x^3 - 2x + 1 & = 0 \\ \end{aligned}$

By using Vieta Formula, the answer should be $0$

Jason Chrysoprase - 4 years, 6 months ago

With regards to the original equation, the series on the RHS is an infinite geometric progression with first term $1$ and ratio $-x$ , and hence for $|x| \lt 1$ is equal to $\dfrac{1}{1 + x}$ . For $|x| \gt 1$ it diverges and for $x = 1$ it oscillates between $1$ and $0$ and thus does not converge to a unique value. Thus in solving for $x$ we have to make sure that the individual roots satisfy the requirement that $|x| \lt 1$ . Vieta's tells us what the sum of the roots is, but it doesn't tell us what the individual roots are, which is what really matters in this problem.

I like your method of multiplying through by $x^{2}$ , but I think your final equation should be $x^{3} - 2x + 1 = 0$ . Note that this factors to

$x^{3} - 2x + 1 = (x - 1)(x^{2} + x - 1) = 0$ ,

so the roots of your equation are $1$ and the two roots to $x^{2} + x - 1 = 0$ as discussed above. These roots do add to $0$ , but as discussed above we must discard the roots $x = 1$ and $x = \frac{-1 - \sqrt{5}}{2}$ , leaving us with only $x = \frac{-1 + \sqrt{5}}{2}$ satisfying the condition $|x| \lt 1$ .

Brian Charlesworth - 4 years, 6 months ago

@Brian Charlesworth – That's explain, thx

Jason Chrysoprase - 4 years, 6 months ago

Tom Engelsman
Nov 25, 2016

The series equation computes to:

$x = \sum_{k=0}^{\infty} (-x)^k = \frac {1}{1 + x}$ (i).

Solving for $x$ in (i) yields:

$x^2 + x - 1 = 0$ ;

or $x = \frac{-1 \pm \sqrt(1 - 4(1)(-1))}{2}$ ;

or $x = \frac {-1 \pm \sqrt5}{2}$ .

It should be noted that only | $\frac{-1 + \sqrt5}{2}$ | < 1, which is required for convergence.

Nice solution. Thank you for posting it.

Hana Wehbi - 4 years, 6 months ago

My pleasure, Hana!

tom engelsman - 4 years, 6 months ago

Got my vote too :)

Hana Wehbi - 4 years, 6 months ago

I done exactly what u did but got wrong as i forgot -1-√5/2

genis dude - 4 years, 4 months ago

is not a solution

genis dude - 4 years, 4 months ago

Constraints Of A Geometric Progression Sum?

3 solutions

0 pending reports