Two Variables One Equation

Hi everyone!

Here is an interesting solution to a two-variable one-equation system.


We want to solve x+y=1x+y=1.

Subtracting xx from both sides, we see that y=1xy=1-x.

Subtracting yy from both sides of the original equation, we get that x=1yx=1-y.

Now we have a system of equations: {x=1yy=1x\left\{\begin{array}{l}x=1-y\\y=1-x\end{array}\right.

Plugging the second equation into the first, we get that x=1(1x)x=1-(1-x).

Now, plugging this equation into itself recursively, we see that x=1(1(1))x=1-(1-(1-\cdots))

To solve xx, we observe the following sequence: 1,1(1),1(1(1)),1(1(1(1))),1,1-(1),1-(1-(1)),1-(1-(1-(1))),\ldots

Agree that the limit of this sequence as it goes on to its infinite term is xx.

Also, note that this sequence is equivalent to the following sequence: 1,0,1,0,1,1,0,1,0,1,\ldots

As we all know from various proofs which I will not outline here (this sequence is pretty famous), the limit of this sequence is 12\dfrac{1}{2}.

However, this limit is equivalent to xx, as we know that x=1(1(1))x=1-(1-(1-\cdots)).

Therefore, our solution to our original system of equation is (x,y)=(12,12)\boxed{(x,y)=\left(\dfrac{1}{2},\dfrac{1}{2}\right)}, and we are done. \Box

Note by Daniel Liu
5 years, 7 months ago

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Here is a proof that 11+11+=121-1+1-1+\cdots=\dfrac{1}{2}.

Let S=11+11+S=1-1+1-1+\cdots.

Note that 2S=11+11++11+11+2S=1-1+1-1+\cdots +1-1+1-1+\cdots.

Pairing up the nnth term of the first sequence with the n+1n+1th term of the second sequence, we see that 2S=1+(1+1)+(1+1)+2S=1+(-1+1)+(-1+1)+\cdots.

Therefore, 2S=12S=1, and so S=12\boxed{S=\dfrac{1}{2}}.

Daniel Liu - 5 years, 7 months ago

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Daniel, what is the finite value of this series

n=010n=... \sum_{n=0}^\infty 10^n=\,...

Tunk-Fey Ariawan - 5 years, 7 months ago

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Let the summation be S S

We have S=100+101+102+103+104+ S = 10^0 + 10^1 + 10^2 + 10^3 + 10^4 + \dotsb

Multiplying both sides by 9, we get

9S=9(100+101+102+103+104+) 9S = 9(10^0 + 10^1 + 10^2 + 10^3 + 10^4 + \dotsb )

We can write 9 9 as 101 10 - 1

Hence, 9S=(101)(100+101+102+103+104+ 9S = (10 - 1)(10^0 + 10^1 + 10^2 + 10^3 + 10^4 + \dotsb

Expanding the RHS RHS

9S=[101+102+103+104+105+][100+101+102+103+104+] \Rightarrow 9S = [10^1 + 10^2 + 10^3 + 10^4 + 10^5 + \dotsb ] - [10^0 + 10^1 + 10^2 + 10^3 + 10^4 + \dotsb ]

9S=100 \Rightarrow 9S = - 10^0

subsequent terms get cancelled, only 1 -1 remains

We get S=19 S = - \frac {1}{9}

Deepansh Mathur - 5 years, 7 months ago

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@Deepansh Mathur S=100+101+102+103+S=100+10(100+101+102+103+)S=1+10S9S=1S=19. \begin{aligned} S&=10^0+10^1+10^2+10^3+\cdots\\ S&=10^0+10(10^0+10^1+10^2+10^3+\cdots)\\ S&=1+10S\\ -9S&=1\\ S&=-\frac{1}{9}. \end{aligned}

Tunk-Fey Ariawan - 5 years, 7 months ago

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@Deepansh Mathur Clearly,the expression is divergent as it is composed of only positive nos. So,there can't be a finite value.Also,there seems no flaw in the proof given by Deepanshu to me at least.But surely there's a fallacy somewhere.Can someone point it out?

Bhargav Das - 5 years, 7 months ago

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@Bhargav Das You might wanna check out this

link

Deepansh Mathur - 5 years, 7 months ago

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Yes... It is a famous Grandi series in which somehow I believe or I don't believe which made the sum of natural numbers be -1/12. But (1/2, 1/2) is not just one solution we are looking for. Say we have infinitely many solutions (integral, rational, etc.) that lies on this locus....

John Ashley Capellan - 5 years, 7 months ago

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