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Algebra Level 3

$\large x^0 + x^1 + x^2 + x^3 + \ldots = 1, \ \ \ \ \ x = \ ?$

Try this one .

None of these choices 1 0 -1

10 solutions

Paul Ryan Longhas
Apr 10, 2015

$x^0 + x^1 + x^2 + x^3 + .................. = 1$

$\Rightarrow 1 + x + x^2 + ............ = 1$

$\Rightarrow x + x^2 + x^3 + x^4 .............. = 0$ . So, $x = 0$

But, if we check the solution

$0^0 + 0^1 + 0^2 + 0^3 +........... \neq 1$

We know that $0^0$ is not defined

Well... it is debatable whether $0^{0}=1$ or not. It is useful to define it as $1$ but not all mathematicians agree on that. See this Brilliant Discussion

Julian Poon - 6 years, 2 months ago

Yeah, I assumed $0^{0}$ = 1.

Nicholas Spinner - 6 years ago

Same here :(

Akhil Bansal - 5 years, 11 months ago

Yeah even my windows calculator showed 0^0=1 but still i chose yes

Apurv Rajput - 6 years ago

I know it's only in power towers that

$\large{\color{#3D99F6}{0^0=1}}$

$\huge{\color{#D61F06}{^na=a^{a^{a^{\cdot^{\cdot^{a}}}}}}}$

$\LARGE{^20=0^0=1}$

$\LARGE{^30 =0^{0^{0}}=0}$

$\LARGE{^40 =0^{0^{0^{0}}}=1}$

$\LARGE{^n0 =0^{0^{\cdot^{\cdot^{0}}}}=0}$

Therefor it will oscillate between 0 and 1 depending on the value of "n" odd or even

Ahmed Obaiedallah - 6 years ago

Sir I totally agree with you. In maths there are some things that is ridiculous but it's true. For example sum of Natural no. upto infinity is ( - 1/12) so who knows 0 raised to 0 is 1

Ankit Nigam - 6 years, 2 months ago

That isn't ridiculous actually, the $\sum _{ n=1 }^{ \infty }{ n } =-\frac { 1 }{ 12 }$ . The basic idea is to bring this entire function into the complex plane, so the "normal" rules don't apply.

Julian Poon - 6 years, 2 months ago

@Julian Poon – Yes sir totally agreed, the discussion is very interesting btw what is your opinion about 0 raised to 0

Ankit Nigam - 6 years, 2 months ago

@Ankit Nigam – By trying to assign a value to a point of "length" 0 (in the 0th dimension), I get the value 0, so I conclude that geometrically speaking, $0^{0}=0$

Julian Poon - 6 years ago

@Julian Poon – how is the summation of all natural numbers equal -1/12? I'm confused :P

Pratik Reddy - 6 years, 2 months ago

@Pratik Reddy – too much numberphile

Neil Yabut - 6 years, 1 month ago

@Pratik Reddy – 1-1+1-1+1...=x

1-1+1-1...=x

1=2x

x=1/2

1-2+3-4+5-6+...=y

1-2+3-4+5-...=y

1-1+1-1+1-1+...=2y

x=2y

1/2=2y

y=1/4

1+2+3+4+5+6+...=z

1-2+3-4+5-6+...=y

z-y=4+8+12+16+...

z-y=4(1+2+3+4+...)

z-y=4z

z-1/4=4z

-3z=1/4

z=-1/12

Thus 1+2+3+4+5+...=-1/12

Lee Isaac - 5 years, 9 months ago

Well 0^0 does not have a unique solution. For. Example x^x goes towards 1 as x goes to 0 (from the right ) but a function like (e^-1/x)^x will go towards 1/e as x goes towards 0 (from the right) Thus even in limits it has no unique solution for 0^0. Now as I said above in complex analysis specifically for power series 0^0 is defined as 1 for convention since x^0=1 for all x not zero. This will keep analytic continuity which over all is key in analysis.

Colin Gordon - 6 years ago

It is not correct to assume ${0}^{0}=1$ because $0/0$ is not defined. If we think in terms of calculus, For example, $\lim _{ a\rightarrow 0 }{ (a/a) } =1$ because the value of $a$ tends to $0$ ,not equal to zero. This means its infinitely small but not equal to zero.

$Conclusion$ : ${0}^{0}=1$ is wrong.

Saarthak Marathe - 5 years, 9 months ago

$0^{0}$ is undefined

There is nothing to debate on this

Check this proof;

We know from the basic laws of indices,

$\frac{a^{m}}{a^{n}}=a^{m-n}$

Then, $\frac{a^{b}}{a^{b}} = 1$ , where $b \neq 0$

Hence, we infer $a^{b-b}=1 \implies \boxed{a^{0}=1}$

Notice that this is possible only when $a \neq 0$ . If it had been $0$ , then the primitive expression would have been $\frac{0}{0}$ , which is indeterminate..

So its clear that $0^{0}$ is undefined.

Arkajyoti Banerjee - 5 years, 3 months ago

I've always heard that $0^{0}$ is undefined.

Brandon Stocks - 5 years, 1 month ago

$\large x^0=1,~~ if~~ \color{#D61F06}{x\ne0}.\\\large x^0 + x^1 + x^2 + x^3 + .................. = 1.$

$\large \Rightarrow 1 + x + x^2 + ............ = 1.$

$\large \Rightarrow x + x^2 + x^3 + x^4 .............. = 0.~~So~~, \color{#3D99F6}{x = 0}.$

$\large \text{So there is a contradiction. No solution.}$

Niranjan Khanderia - 6 years, 2 months ago

but we all know, that, any number raised to zero will always be equal to 1?

roland casuga - 6 years, 2 months ago

We know that any number raised to zero will be equal to 1, but we also know that 0 raised to any number will be equal to 0. By those rules $0^{0}$ is a contradiction. It's undefined, but the limit for $x^{x}$ as x approaches 0 is 1.

Daniel Ellesar - 6 years, 1 month ago

Can you explain why any number raised to $0$ is equal to $1$ ?

Rakshit Pandey - 5 years ago

We know that $a^0 = 1$ if and only if $a \neq 0$

Paul Ryan Longhas - 6 years, 2 months ago

The sum of all the natural numbers is not $-\frac {1}{12}$

Paulo Carlos - 6 years, 2 months ago

@Paulo Carlos – Let's assume that:

      1 - 1 + 1 - 1 + ... = x

      1 - 2 + 3 - 4 + ... = y

      1 + 2 + 3 + 4 + ... = z

     1 - 1 + 1 - 1 + ... = x
     1 - 1 + 1 - 1 + ... = 1 - x

x = 1 - x

x = 1/2

1 - 1 + 1 - 1 + ... = 1/2

   1 - 2 + 3 - 4 + ... = y

       1 - 1 + 1 - ... = x

1 -(y + 1/2) = y

y = 1/4

1 - 2 + 3 - 4 + ... = 1/4

1 + 2 + 3 + 4 + ... = z

1 - 2 + 3 - 4 + ... = y

    4   +   8 + ... = 4z

y + 4z = z

z = -1/12

Therefore sum of all natural numbers is -1/12

Jesse Nieminen - 6 years ago

@Jesse Nieminen – It's wrong man. You used the rules that should be used only with finite numbers...

Paulo Carlos - 6 years ago

@Jesse Nieminen – The only problem with that assumption is that the infinite sum 1 -1 + 1 - 1+.... does not equal 1/2! If we are to start with incorrect assumptions we can prove anything we want, if you take a partial sum analysis of the sum of all natural numbers it is quite obviously a divergent sum not a convergent one. The guys in the numberphile video are, unfortunately, spouting rubbish based on a partially understood proof.

Matthew Bracewell - 6 years ago

@Matthew Bracewell – Why is it incorrect?

Jesse Nieminen - 6 years ago

@Jesse Nieminen – It is incorrect as you have to make a slight fudge in order to make it work. In order for the sum to equal 1/2 you have to assume that you can take an average of the 1, 0, 1, oscillation of the answer. Now whilst this is an appropriate thing to do in order to use in a physical world as the numberphile guys do in order to make their string theory work it is inappropriate to do in a purely mathematical field. One of the main things they left out of the video was the special instance that this particular assumption was being used for i.e. a Riemann zeta function.

Matthew Bracewell - 6 years ago

@Matthew Bracewell – One of the other things I've just noticed having watched the video again is that he very clearly states that we cannot just add numbers up on a calculator in order to disprove his result of -1/12 as it is an infinite series, however he is quite happy to use the counter example in order to prove that 1-1+1-.. = 1/2 as the only way to make this assumption is to say that "at some point I'm going to have to stop this series" which means you have to take the average. If we were to continue using this same logic or rule then we would also state the same for the sum of all natural numbers which means that we wouldn't get -1/12 we would would get an incredibly large number.

Matthew Bracewell - 6 years ago

@Paulo Carlos – Is it now? :P -_-. It is ,brother. @Paulo Carlos

Mehul Arora - 6 years, 2 months ago

@Mehul Arora – The rules which are used to calculate the sum of all the natural numbers such that the result becomes $-\frac {1}{12}$ cannot be used with infinite numbers...

Paulo Carlos - 6 years, 2 months ago

Just because an indeterminate form has been encountered doesn't mean there has to be no answer. The limit of an expression that takes an indeterminate form can be any number depending on the expression .

$\lim _{ x\rightarrow 0 }{ \sum _{ n=0 }^{ \infty }{ { x }^{ n } } } =1$

so the $x$ in this particular expression must be $0$ .

André Cabatingan - 6 years, 2 months ago

As a matter of convention, it's not uncommon to define 0^0 = 1, so the answer is ambiguous. It would be better to make clear that 0^0 is undefined.

Kal Hourani - 6 years ago

isn't (-1)^0 is 1 ? then odd and even powers would cancel out each other right?

prashant keshvani - 6 years ago

1 - 1 + 1 - 1 + ... = x

1 - 1 + 1 - 1 + ... = 1 - x

x = 1 - x

x = 1/2

1/2 is not equal to 1

Jesse Nieminen - 6 years ago

The value isn't $\frac 12$ either. The sum $1-1+1-1+\ldots$ is a famous one, known as Grandi's series (google it). The sum has no defined value since it's a divergent sum. The value $\frac 12$ is only the Cesaro-summation value of that sum, not the actual value.

Prasun Biswas - 5 years, 11 months ago

Even though in general 0^0 does not exist. In complex analysis 0^0=1 in a power series as a convention. So it really should have both answers as correct.

Colin Gordon - 6 years ago

Sum to infinity:
\frac{1}{1 - x} = 1 Therefore 1 - x = 1
\Rightarrow x = 0

Jahangir Khan - 6 years, 2 months ago

0^0=1 confirmed by Google search http://lmgtfy.com/?q=0%5E0

Stern Huang - 6 years, 1 month ago

You can also use the Maclaurin series expansion where Sum x^n from 0 to infinity = 1-x, and notice this only converges from 0 to 1 exclusive.

Yassin Abouel Seoud - 6 years, 1 month ago

Why isn't 0^0 considered an indeterminate function? As far as I know, 0/0 is an indeterminate function and 0^0 can be converted to an indeterminate function. And there are multiple values for them

Nirmit Tripathii - 6 years ago

isn't -1^{0} is 1 ? then x^{2} would cancel out x^{3} and so on.

prashant keshvani - 6 years ago

Why can't it just be x=-1? Because (-1)^0 =1 and there are the same number of odd and even numbers so the -1s and 1s will cancel surely?

Ben Johnson - 6 years ago

I understand that that 0^{0} is technically non-existent. However, I am fairly sure that most algebra (at least at the level 3 section of this site) assumes that 0^{0} = 1. So, while I understand your premise and the "trick" of this problem, I disagree with the lack of clarification that 0^{0} = 1.

Sammy Berger - 6 years ago

Why can't it be -1?

Mehdia Nadeem - 5 years, 11 months ago

Write a comment or ask a question...

Mehdia Nadeem - 5 years, 11 months ago

Why can't it be -1?

Mehdia Nadeem - 5 years, 11 months ago

If it was $(-1)$ , the series we'd have in LHS is $1-1+1-1+1-1+\ldots$ which is a divergent series (its partial sum values alternate between $1$ and $0$ ).

Since the series is divergent, it cannot equal $1$ in RHS, hence we cannot have $(-1)$ . Read more about that series here .

Prasun Biswas - 5 years, 11 months ago

If it is ended with a even number then (-1) can be the answer

Sayantan Saha - 5 years, 8 months ago

I find this question suspect. Taking $x^{0}$ as a function, convention is to define the function piecewise with the mathematically derived case for $x\neq0$ and 1 for $x=0$ . This is no different from the definition of the factorial operation, i.e. $x!=x(x-1)!$ for $x=1,2,3,...$ , but $0!=1$ . This is to say, just because something is conventional doesn't mean it is not the truth on the matter. When something is convention, that amounts to working mathematicians defining the appropriate term, function, or operator in that way. After which, mathematics proceeds under that definitional constraint. Let's be quite honest, in mathematics definition is EVERYTHING!

Loren Wagner - 5 years, 7 months ago

0^0 is defined as 1 Search on net. Other wise assume x<1 As it cant be more than 1 obviously It's a gp till infinity therefore 1/(1-x)=1 Hence x=0

subh mandal - 5 years, 5 months ago

I don't agree... I've learned that the definition of $x^\alpha$ is threefold :

IF $x \in \mathbb{C}$ and $\alpha \in \mathbb{N}$ : THEN $x^\alpha \overset{\tiny \text{by def}}{=} \prod\limits_{k = 1}^n x$
IF $x \in \mathbb{C}$ and $\alpha \in \mathbb{Z}\backslash\mathbb{N}$ : THEN $x^\alpha \overset{\tiny \text{by def}}{=} \frac{1}{\prod\limits_{k = 1}^n x}$
IF $x \in \mathbb{R}^{*}_{+}$ and $\alpha \in \mathbb{R}$ : THEN $x^\alpha \overset{\tiny \text{by def}}{=} \exp(\alpha \ln(x))$

So, in this case : $\begin{aligned} &\sum\limits_{k \ge 0} x^k &= 1 \\ \Leftrightarrow \hspace{1em}&\sum\limits_{k \ge 1} x^k &= 0 \\ \Leftrightarrow \hspace{1em}&x \sum\limits_{k \ge 0} x^k &= 0 \\ \Leftrightarrow \hspace{1em} &x = 0 &\text{OR} \hspace{2em} \sum\limits_{k \ge 0} x^k = 0 && \text{(which is not the case)} \end{aligned}$

Hence : $x = 0$

You Kad - 4 years, 10 months ago

Some do define 0(exp 0)=1, removing the discontinuity in the complex plane. The question does not clarify which level to assume.

Will Heierman - 4 years, 7 months ago

0^0 can be de fined as 1 in many situations. For all intensive purposes, since 1×0 is 0, then 0/0 can give one then 0^0 can give 1

Isaac Thomas - 6 years, 2 months ago

what is 0^0

Yeldo Pailo - 6 years, 1 month ago

NOT DEFINED.

Niranjan Khanderia - 6 years, 1 month ago

Tootie Frootie
Apr 14, 2015

None. 0 can't be the answer because 0^0 is undefined. Plugging in -1 gets a very famous series that usually is considered to equal 1/2.

Why can't 0^0 be equal to 1, there are several pages which give 0^0 = 1, the google calculator being one.... 0^0 is even defined by the IEEE to be equal to 1...

Sutirtha Paul - 6 years, 1 month ago

but the truth is that 0^0 is undefined okk let me tell you why a"M / a^M = a^M-M here a = 0
0^M/0^M AND 0/0 IS NOT DEFINED

Tootie Frootie - 6 years, 1 month ago

In differential calculus, if f(x) = x^n then f'(x) = nx^(n-1) is valid at x = 0 only when $0^0 = 1$ . But really, it just depends on how you use it although there is a consensus around $0^0 = 1$

Sutirtha Paul - 6 years, 1 month ago

@Sutirtha Paul – but here we can simply consider the laws of exponents ....there is no need of calculus ....

Tootie Frootie - 6 years, 1 month ago

@Tootie Frootie – We can do that but $0^0 = 1$ is agreed upon and what I knew....

Sutirtha Paul - 6 years, 1 month ago

@Sutirtha Paul – at different levels we consider different formats and values we have to look at the structure of question ...and try to figure out what the question writer means

Tootie Frootie - 6 years, 1 month ago

Ayyappan Lakshminarayanan
Jun 3, 2015

It is an infinite geometric progression series. x ( 1+x+x^2+x^3+...)=1

x (1/(1-x)) = 1 ; x= 0.5

The term $x^0$ is missing.

Elizandro Max - 5 years, 11 months ago

you are dumb

Sparsh Setia - 4 years, 3 months ago

Ubaidullah Khan
Apr 13, 2015

When there is a summation series of exponents and the answer required is 1 then,
base should be 1 or -1.
Since, the answer 1 will come only if the base is 1 or -1.
Since these are not in the options, the answer is None Of these

Abhishek Kumar
Jun 6, 2015

Sum of infinite GP, s = a/(1-r) Here, s= 1 = x/(1-x) =>(1-x)=x =>2x=1 =>x=1/2

The first term is 1 not x

علاء رمضان - 4 years, 9 months ago

Arian Tashakkor
May 15, 2015

I like to think of the given polynomial as a geometric sequence.

Now the sum of all the terms in this polynomial can be evaluated using:

$S= \frac {(1)(1-x^n)}{1-x} = 1 \rightarrow 1-x^n = 1 - x \rightarrow x=x^n$

The answers to the last equality can be 0,1 and -1 but since the value $n$ is not determined in the question,the answer cannot be any of these.(Also if $x=0$ the first term is undefined)

I don't agree... I've learned that the definition of $x^\alpha$ is threefold :

IF $x \in \mathbb{C}$ and $\alpha \in \mathbb{N}$ : THEN $x^\alpha \overset{\tiny \text{by def}}{=} \prod\limits_{k = 1}^n x$
IF $x \in \mathbb{C}$ and $\alpha \in \mathbb{Z}\backslash\mathbb{N}$ : THEN $x^\alpha \overset{\tiny \text{by def}}{=} \frac{1}{\prod\limits_{k = 1}^n x}$
IF $x \in \mathbb{R}^{*}_{+}$ and $\alpha \in \mathbb{R}$ : THEN $x^\alpha \overset{\tiny \text{by def}}{=} \exp(\alpha \ln(x))$

Hence : $x = 0$

You Kad - 4 years, 10 months ago

Thomas Lee
Apr 16, 2015

$x^0 + x^1 + x^2 +x^3 + .... = 1$
$(1-x) [x^0 + x^1 + x^2 +x^3 + .... = 1]$
$1=1-x$
$x=0$
But since 0^0 is debatable, the answer is None of these choices.

the correct answer is -1 per arithmetic calculation and -1 raised to zero gives 1 :)

Harendra Kumar - 6 years ago

I did not understand step 2. Can you please explain ? Thanks.

Niranjan Khanderia - 6 years, 1 month ago

Multiply both sides of the equation by (1-x). Everything will cancel out except for x^0 which equals 1.

Thomas Lee - 6 years, 1 month ago

Thanks. I still did not still understand step 3 from step 2!!.

Niranjan Khanderia - 6 years, 1 month ago

@Niranjan Khanderia – In a small frame like x^0+x^1+x^2, if you multiply that by (1-x), you get x^0-x^3 because all the middle terms cancel out. If the middle terms are never ending, then you can assume that all the middle terms keep on canceling out into infinity, leaving only the first term.

Thomas Lee - 6 years, 1 month ago

@Thomas Lee – Thanks a lot. I should have understood that much before!!.

Niranjan Khanderia - 6 years, 1 month ago

Suhas Desai
Apr 13, 2015

Numbers r infinite so we canot guess the value of x

Agastya Chandrakant
Apr 13, 2015

I did not to the way others depicted: Used a/x-1=1....the sum of G.P. of Infinite terms. where a=1 and got x=2 which is not possible too. So clicked NOTA. I don't know why I am right or where I am wrong.

The sum of a G.P. of infinite terms is not $\frac{a}{x-1}$ , it is $\frac{a}{1-x}$

Tushar Marda - 6 years, 1 month ago

so unknowingly, i did wrong! and got write. First time NOTA saved me. though now i know i don't worth these points.

Agastya Chandrakant - 6 years, 1 month ago

Trevor B.
Apr 13, 2015

We know that $\displaystyle\sum_{n=0}^\infty x^n$ is the power series for $\dfrac{1}{1-x}$ if $|x|<1.$ Solving $\dfrac{1}{1-x}=1,$ we get that $x=0.$ However, when we back-substitute $x=0,$ we find that one of the terms in the sum is $0^0,$ which is an indeterminate form. Thus, there is no solution.

I feel cheated. Again. 0^0 may be indeterminate for some, but not all of us.

Fabio Bittar - 6 years, 2 months ago

Which hinges on whether we define 0^0 as an indeterminate form or equivalent to one.

If it were indeterminate, wouldn't the power series be undefined for x = 0? Not just for |x| > 1.

Mikal Olsen - 6 years, 2 months ago

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