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Algebra Level 4

$\large 1 = x^2+x^3+x^4+\ldots$

Find the sum of value(s) of $x$ satisfying the equation above. Give your answer to 3 decimal places.

The answer is 0.618.

8 solutions

Abhishek Sharma
Apr 27, 2015

$(x^2+x^3+x^4+\dots \infty) - 1 = 0$ For solution to exist, $|x|<1$ $\frac{{x}^{2}}{1-x} - 1 = 0$ $\frac{{x}^{2}}{1-x} =1$ ${x}^{2}=1-x$ ${x}^{2}+x-1 = 0$ $x=\frac{\sqrt{5}-1}{2}$ The other solution is rejected as it is less than -1 that is it doesn't satisfy $|x|<1$ .

Therefore, the answer should be $0.618$

Interesting thing to note is that there exists a solution for $x$ if $1 = x^n + x^{n+1} + x^{n+2} + \ldots$ for integer $n \geq 2$ .

Proof:

$x^n + x^{n+1} + x^{n+2} + \ldots$ converges to finite number only if $|x|<1$ .

Using formula for infinite GP we have,

$x^n + x^{n+1} + x^{n+2} + \ldots =\frac{{x}^{n}}{1-x}$

$\frac{{x}^{n}}{1-x}=1$

${x}^{n}=1-x$

Let $f(x)$ be ${x}^{n}$ and $g(x)$ be $1-x$ .

Plotting $g(x)$ between $-1$ and $1$ is easy.

To plot $f(x)$ we need to make cases.

Case 1

$n$ is odd .

Plotting $f(x)$ isn't that easy but we can surely plot $f(x)$ for $x=-1$ , $x=0$ and $x=1$ .

As $f(x)$ is continuous for all $x$ it will definitely intersect $g(x)$ between $0$ and $1$ .

Between $-1$ and $0$ $f(x)$ won't intersect $g(x)$ as it is continuous for all $x$ and possesses no maxima or minima between $-1$ and $0$ or you can simply say that $f(x)$ is increasing between $-1$ and $0$ .

Case 2

$n$ is even .

We plot $f(x)$ for $x=-1$ , $x=0$ and $x=1$ .

As $f(x)$ is continuous for all $x$ it will definitely intersect $g(x)$ between $0$ and $1$ .

Between $-1$ and $0$ $f(x)$ won't intersect $g(x)$ as it is continuous for all $x$ and possesses no maxima or minima between $-1$ and $0$ or you can simply say that $f(x)$ is decreasing between $-1$ and $0$ .

Therefore, there exists a solution for $x$ if $1 = x^n + x^{n+1} + x^{n+2} + \ldots$ for integer $n \geq 2$ and to be more precise the solution of this equation lies between $0$ and $1$ .

Moderator note:

Does there always exist a solution for $x$ if $1 = x^n + x^{n+1} + x^{n+2} + \ldots$ for integer $n \geq 2$ ?

In response to the Challenge Master's note, for $n \ge 2$ we end up with the equation $x^{n} = 1 - x.$ Considering the graphs of the functions on each side of this equation, we have $f(x) = x^{n}$ as a continuous, strictly increasing function for $x \ge 0$ with $f(0) = 0,$ and $g(x) = 1 - x$ as a line with slope $-1$ such that $g(0) = 1$ and $g(1) = 0.$ Thus $f(x)$ and $g(x)$ will have a unique point of intersection on the interval $(0,1),$ i.e., a unique solution exists for any $n \ge 2.$ This solution approaches $1$ as $n \rightarrow \infty.$

Brian Charlesworth - 6 years, 1 month ago

Hm, not quite. The domain that you should be looking at is $(-1, 1)$ and not just $(0,1)$ .

Calvin Lin Staff - 6 years, 1 month ago

For (-1,0), 1-x is always greater than or equal to zero, and x^n is never.

Aadil Bhore - 6 years, 1 month ago

@Aadil Bhore – That claim is not true if $n$ is an even integer. $x^2 \geq 0$ for all real $x$ .

Calvin Lin Staff - 6 years, 1 month ago

Could you please tell me why $|x|<1$ ? Thanks in advance! @Abhishek Sharma @Prasun Biswas @Pranjal Jain

User 123 - 6 years, 1 month ago

The infinite series is converging iff |x|<1.

Keshav Tiwari - 6 years, 1 month ago

OK, thanks a lot! But what about for an infinite geometric series (in general)? Are there any restrictions for the common ratio?

User 123 - 6 years, 1 month ago

@User 123 – Here's an explanation that might be simpler to understand. First, consider a finite geometric series with first term $T_1=a$ and common ratio $r=x\neq 1$ . Now, we know that, for sum upto $n$ terms,

$\sum_{k=1}^nT_k=a\cdot\frac{x^n-1}{x-1}$

Now, consider the limit of this sum as $n\to\infty$ for the infinite geometric series. Then, you have,

$\lim_{n\to\infty}\sum_{k=1}^nT_k=\frac{a}{x-1}\cdot\lim_{n\to\infty}(x^n-1)$

Now, the trick is to identify the domain of all possible values of $x$ $(x\neq 1)$ such that the limit on RHS converges. We know that,

$n\to\infty\implies \begin{cases}x^n\to 0\iff |x|\lt 1\\ x^n\textrm{ diverges }\iff |x|\gt 1~\lor~x=(-1)\end{cases}$

This directly implies that for the limit (and hence the sum) to be finite, we need $|x|\lt 1$ . Hence, we need $|x|\lt 1$ for the infinite geometric series to converge where $x\neq 1$ is the common ratio (in the case we showed).

For the specific case of $x=1$ where the GP sum formula can't be applied, you can manually show that the series will diverge using the ratio test.

Prasun Biswas - 6 years, 1 month ago

@Prasun Biswas – @Prasun Biswas Thanks so much!

User 123 - 6 years, 1 month ago

@Prasun Biswas – @Prasun Biswas Just a small correction (I think you must have made a typo): for $|x|=1$ , $\lim_{n \to \infty} x^n = 1$ . So only for $|x|>1$ , $x^n$ diverges.

User 123 - 6 years, 1 month ago

@User 123 – Note that the infinite GP series diverges for $x=1$ too. The limit case I showed verifies only for the case when the absolute value of the common ratio $\neq 1$ .

The case of $x=1$ is unique since the finite GP sum formula isn't valid for common ratio $r=x=1$ . We show that case separately. When you have $r=x=1$ , the sum becomes,

$\sum_{r=2}^\infty x^r=\sum_{r=2}^\infty 1\implies \textrm{ The sum diverges}$

Prasun Biswas - 6 years, 1 month ago

@User 123 – Ah yes, thanks for pointing it out! I have fixed it now. :)

Prasun Biswas - 6 years, 1 month ago

@Prasun Biswas – But what is $\lim_{n \to \infty} -1^n$ ? @Prasun Biswas

User 123 - 6 years, 1 month ago

@User 123 – That limit has no defined value because it oscillates among $1$ and $-1$ . Since $n\to\infty$ , it has no defined parity (odd/even). Hence, the limit does not exist.

Prasun Biswas - 6 years, 1 month ago

@Prasun Biswas – OK, thanks. I was really confused due to the oscillation between $1$ and $-1$ .

User 123 - 6 years, 1 month ago

You are right ! @Calvin Lin Sir please see this . :)

Keshav Tiwari - 6 years, 1 month ago

Thanks, I've corrected it.

Brilliant Mathematics Staff - 6 years, 1 month ago

Thanks. :)

Keshav Tiwari - 6 years, 1 month ago

@Keshav Tiwari – Going by theory of equations the sum of roots should be equal to -1. Why is it not so

Nisshith Sharma - 6 years, 1 month ago

@Nisshith Sharma – ^ indeed. The sum of the roots of the equation $x^{2}+x-1$ is -1. but the Infinite sum $x+x^{2}....=\frac{x}{1-x}$ iff $|x|<1$ . Hence only The above said root is valid .The other one one is rejected, giving us the required sum .:)

Keshav Tiwari - 6 years, 1 month ago

Typo in 6th and 8th lines.

In 6th line, it should be $x^2+x-1\color{#D61F06}{=0}$ .

In 8th line, it should be "The other solution is rejected as it is $\color{#D61F06}{\textrm{lesser}}$ than $\color{#D61F06}{(-1)}$ ".

Prasun Biswas - 6 years, 1 month ago

Fixed. Thank you.

Brilliant Mathematics Staff - 6 years, 1 month ago

Its such a easy problem and when u miss thia simple point of mod x less than 1....you just cant stop slapping yourself

Mridul Gupta - 5 years, 10 months ago

I wonder why this problem is level 5? Not that hard lol...

Micah Wood - 6 years, 1 month ago

It was previously at level 5 because
- Purushottam Abhisheikh set it there initially
- The given answer was wrong, and so that pushed it up further.

I see that it has now come down to level 3, which is reasonable.

Calvin Lin Staff - 6 years, 1 month ago

I did not set it at level 5. I actually set this question at level 2. And I apologize for my mistake and thanks to your Brilliant team for its correction.

Purushottam Abhisheikh - 6 years, 1 month ago

@Purushottam Abhisheikh – Yes, sorry about that. I see that it was set level 2.
After we fixed the answer, we reset it at level 3.
We did some promotion to make it popular, and it seems like a lot of people are now answering -1, so the problem is rising in it's rating.

Calvin Lin Staff - 6 years, 1 month ago

How did we get to step 3 where we replaced the value in braces to x2/1-x?

Sachin Vaya - 6 years, 1 month ago

This is an infinite Geometric Series with the common ratio as $x$ . The formula for the summation of an infinite Geometric Series is $\dfrac{a}{1-r}$ where $a$ is the first term and $r$ is the common ratio. Hope this helps... @Sachin Vaya

User 123 - 6 years, 1 month ago

Important to note that this formula is only applicable when it has already been shown that the series converges. Using analysis you can show that this is when $|r| \lt 1$ (even for complex $r$ )

Matt McNabb - 6 years, 1 month ago

@Matt McNabb – Yes, you're absolutely right.

User 123 - 6 years, 1 month ago

please resolve my confusion , i am late but..... for this expression we can write without taking conditions te sum of GP as... ${ x }^{ n }+{ x }^{ n-1 }+.......+{ x }^{ 2 }=\frac { { x }^{ 2 }(1-{ x }^{ n+1 }) }{ (1-x) }$ . So, we get... ${ x }^{ n+3 }-{ x }^{ 2 }-x+1=0$ . taking all solutions real or imaginary , from this expression we get sum of solutions as 0 , why didn't we conclude as such. Please reply..

Gaurav Jain - 5 years, 10 months ago

After manipulating an equation, you should check if you introduced any additional roots.

For example, if you square an equation, then you have to watch out for negative values. Similarly, when you apply the GP formuka, you have to verify that all of the $x$ values satisfy the condition $|x| < 1$ . Otherwise, they would not be valid solutions and hence have to be rejected.

Calvin Lin Staff - 5 years, 10 months ago

Thank you sir!.

Gaurav Jain - 5 years, 10 months ago

We have 1=x²+ x^3+.... So 1=x²(1+x+x²+....) so 1=x²(1+x+1) so x^3 +2x²-1=0 so x take three values {-1.618; -1; 0.618} so the sum of value(s) of x satisfying the equation above is -2 not 0.618

Abdelghani ßoris - 5 years, 9 months ago

i agree with you....this is the best solution...

মির্জা আহমেদ - 5 years, 7 months ago

did the same thing but then i used sum of roots = -1 . now i am thinking like i couldn't get any more stupid :p

Utkarsh Grover - 5 years, 8 months ago

1 = x^2 +x^3 +x^4 +... 1 = x^2( 1+ x+ x^2+x^3 +x^4 +...) 1 = x^2 (1+ x + 1) 1 = 2x^2 +x^3 (x + 1) (x^2 + x - 1) = 0 can somebody please explain why I am getting an additional root x = -1 , I am unable to find out where I went wrong

Yugesh Karnati - 4 years, 6 months ago

Because you manipulated the equation, you might have introduced an extraneous root. As such, you have to substitute them back in, and check if they satisfy the original equation.

As an explicit example, consider the equation $1 = x$ . In a similar manner, let's write $1 = x \times 1 = x \times x = x^2$ and get $(x-1)(x+1) = 0$ so $x= - 1$ is a root. This is a false conclusion due to our manipulation. A solution of $1 = x$ will be a solution of $1 = x^2$ but not vice versa.

Calvin Lin Staff - 4 years, 6 months ago

Abdullah Ahmed
Jul 31, 2016

$1$ $=$ $x^2$ $+$ $x^3$ $+$ $x^4$ $+....$

$or$ , $1$ $+$ $1$ $+$ $x$ $=$ $1$ $+$ $x$ $+$ $x^2$ $+$ $x^3$ $+$ $x^4$ $+....$

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

$Or,$ $(2+x)$ $*$ $(1-x)$ $=$ $1$

$or,$ $x^2$ $+$ $x$ $-$ $1$ $=$ $0$

$So,$ $x=\frac{\sqrt{5}-1}{2}$ or, $x=\frac{-\sqrt{5}-1}{2}$

You can check to both be extraneous, and that leaves us with only $x=\frac{\sqrt{5}-1}{2}$

$x$ = $0.618$

I also did like this. It is very elegant.

Mattia C - 4 years, 9 months ago

I dont get what happens in the third row at all.

Stas Glazkov - 3 years, 3 months ago

The sum of the elements from the element "ar" to the element "as" of a geometric progession whose constant is k is: S=ar(k^(r-s+1)-1)/(k-1) Then we have the sum from the zeroth element to the infinitesimal element of a geometric progression whose constant is x. Then: S=(x^-infinite-1)/(x-1)=1/(1-x). That is what happens in the third row.

Mattia C - 2 years, 10 months ago

Extraneous??

rusab baig - 2 years, 3 months ago

Suvan Shah
Jun 13, 2018

We know that: $x^2+x^3+x^4+\dots=1$

So,

$x^3+x^4+x^5+\dots=x$

Subtracting gives,

$1-x=x^2$

Rearranging and imputting into the quadratic formula gives x = 0.618 or x < 0. We know that the other number does not lead the series tending to 1, so this is the only solution.

Anonymous Anonymous
Jul 20, 2015

1 = x^2 +x^3 +x^4 +...

x^2 = x^4+x^5+x^6+...

1 = x^2 +x^3 +x^2

x^3 +2x^2 -1 = 0

Which has solutions of -1, (-1+sqrt(5))/2, (-1-sqrt(5))/2. -1 and (-1-sqrt(5))/2 you can check to both be extraneous, and that leaves us with only

(-1+sqrt(5))/2 = 0.618

That number is $1/\phi$ .

Brandon Stocks - 4 years, 12 months ago

Arthur Jaber Costato
May 10, 2015

I did it like that:

I know that: $\sum_0^{\infty} (\frac{x-1}{x})^{n}=x$ $x>1$

Let's call the sum of the problem $f(z)=z^2+z^3+z^4+...$ , and let's say $z=\frac{x-1}{x}$

By doing this, we have: $\sum_0^{\infty} {z}^{n}=x$ $1 + z + f(z) = x$

Since $z=\frac{x-1}{x}$ and $f(z)=1$ $1+\frac{x-1}{x}+1=x$ $x + (x-1)+x=x^2$

The solution for x is 2.61803398874989...

Then $z=\frac{x-1}{x}=0.6180339887...$

You said to post the answer as the sum of the values of x, which is -1. Not "post a single value of x"

elijah osborne - 4 years, 10 months ago

Lu Chee Ket
Oct 4, 2015

x^2 (1 + x^2 + x^3 + ...) = x^2/ (1 - x) = 1 {G.P. for x <> 1 or it is x^2 (1 - x)^(-1)}

=> x^2 + x - 1 = 0

=> x = (-1 + Sqrt(5))/ 2 = 0.618 {Modulus smaller than 1}

Answer: 0.618

Rahul Saxena
May 8, 2015

what was difficult about it???

7/10 for difficultly

The Flight Simulation - 3 weeks, 2 days ago

yeah, this question is really easy

Samarth Agarwal - 5 years, 10 months ago

The Flight Simulation
May 21, 2021

We've: S = x^2 + x^3 +x^4 +.... =1
We also have S* 1/x = x + x^2 +x^3+...

<=>S * 1/x = x+1 (1)

Solve (1), we'll have x=0.618

Just keep going

The answer is 0.618.

8 solutions

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