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Two Variables One Equation

Hi everyone!

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


We want to solve \(x+y=1\).

Subtracting \(x\) from both sides, we see that \(y=1-x\).

Subtracting \(y\) from both sides of the original equation, we get that \(x=1-y\).

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

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

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

To solve \(x\), we observe the following sequence: \[1,1-(1),1-(1-(1)),1-(1-(1-(1))),\ldots\]

Agree that the limit of this sequence as it goes on to its infinite term is \(x\).

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

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

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

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

Note by Daniel Liu
3 years, 6 months ago

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

Let \(S=1-1+1-1+\cdots\).

Note that \(2S=1-1+1-1+\cdots +1-1+1-1+\cdots\).

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

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

Daniel Liu - 3 years, 6 months ago

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Daniel, what is the finite value of this series

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

Tunk-Fey Ariawan - 3 years, 6 months ago

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Let the summation be \( S \)

We have \( S = 10^0 + 10^1 + 10^2 + 10^3 + 10^4 + \dotsb \)

Multiplying both sides by 9, we get

\( 9S = 9(10^0 + 10^1 + 10^2 + 10^3 + 10^4 + \dotsb ) \)

We can write \( 9 \) as \( 10 - 1 \)

Hence, \( 9S = (10 - 1)(10^0 + 10^1 + 10^2 + 10^3 + 10^4 + \dotsb \)

Expanding the \( RHS \)

\( \Rightarrow 9S = [10^1 + 10^2 + 10^3 + 10^4 + 10^5 + \dotsb ] - [10^0 + 10^1 + 10^2 + 10^3 + 10^4 + \dotsb ] \)

\( \Rightarrow 9S = - 10^0 \)

subsequent terms get cancelled, only \( -1 \) remains

We get \( S = - \frac {1}{9} \)

Deepansh Mathur - 3 years, 6 months ago

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@Deepansh Mathur \[ \begin{align} S&=10^0+10^1+10^2+10^3+\cdots\\ S&=10^0+10(10^0+10^1+10^2+10^3+\cdots)\\ S&=1+10S\\ -9S&=1\\ S&=-\frac{1}{9}. \end{align} \]

Tunk-Fey Ariawan - 3 years, 6 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 - 3 years, 6 months ago

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@Bhargav Das You might wanna check out this

link

Deepansh Mathur - 3 years, 6 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 - 3 years, 6 months ago

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