Two is equal to One
Voting Style:  Open  Point System:  7 Point  
Started:  9/2/2012  Category:  Science  
Updated:  4 years ago  Status:  Post Voting Period  
Viewed:  1,471 times  Debate No:  25411 
In R2 I will present a proof that 2 = 1. Please don't worry. The proof is flawed. In R2 it will be task of Con to spot the flaw. If she is able to do this, she wins the debate. If she is unable to spot the flaw in R2, I win this debate. If Con is unable to spot the flaw in R2 (or provides an errenous explanation), I will provide the correct explanation in R3. Con need not do anything in R1 and R3. I will be using Taylor series expansion and logarithms. Nothing in much detail. But you need to be comfortable with basics of these two concepts and series calcuations to accept this debate. Hoping to get a sincere opponent. I accept this challenge  it'll be nice to actually be able to argue mathematics on here for once, since I've tried starting quite a few math debates before. I'll post my solution as soon as I find an error in whatever proof you offer, and I commit to working this problem alone. Good luck! 

Thanking bencbartlett for accepting this debate. I agree that it feels nice to debate mathematics. I am numbering the steps in order to assist discussion. Taylor expansion of ln (1+x) is: 1> ln (1+x) = x  x^2/2 + x^3/3  x^4/4 + x^5/5 + Putting x = 1. 2> ln (1 + 1) = ln (2) = 1  1/2 + 1/3  1/4 + 1/5 + Rearranging the terms 3> ln (2) = (1 + 1/3 + 1/5 +++)  (1/2 + 1/4 + 1/6 +++) 4> Let P = 1 + 1/3 + 1/5 +++ 5> Let Q = 1/2 + 1/4 + 1/6 +++ From steps 3, 4 and 5 6> ln (2) = P  Q Consider Q 7> Q = 1/2 + 1/4 + 1/6 + 1/8 + 1/10 +++ Taking 1/2 as common from LHS. 8> Q = 1/2 [1 + 1/2 + 1/3 + 1/4 + 1/5 +++] Rearranging terms 9> Q = 1/2 [ (1 + 1/3 + 1/5 +++) + (1/2 + 1/4 + 1/6 +++) ] Subsituting series in step 9 with help of step 4 and 5, 10> Q = 1/2 [ P + Q] 11> 2Q = P + Q 12> Q = P Substituting Q in step 6. 13> ln (2) = P  P 14> ln (2) = 0 Taking antilogs on both sides 15> 2 = 1 Best of Luck to my esteemed opponent. My opponent’s argument breaks down at step 2. The Taylor expansion for ln(x) has a radius of convergence of 1, around the point x = 1, such that 0<x<2 (as opposed to 0≤x≤2) as I will show below, meaning that the Taylor expansion of ln(x1) likewise has a radius of convergence of 1 around the point x = 0, such that 1<x<1. By inserting x=1 into the Taylor expansion for ln(x1), he is causing the series to become divergent, thereby inducing the error in the proof that appears to make 2=1. Proof that the Taylor expansion for ln(x+1) fails at x=1: 1) Differentiating ln(1+x), we have: d/dx (ln(1+x)) = 1/(1+x).
2) Trivially, 1/(1+x) = 1/(1(x)) = 1/(1y), where y≡x.
3) Taking the Maclaurin expansion of 1/(1y), we have: 1/(1y) = 1+y+y^2+y^3+y^4+…
4) Note that the result in step 3 is the expression of a geometric series, which has limits of evaluation at y<1. Since y≡x, y<1 is equivalent to x<1. Substituting 1 for x, we have 1=1<1, which is false, since 1=1 (again, note the difference between 1<1 and 1≤1). For example, substituting x=1 into the previous expression yields: 1/(1+x) = 1/(1y) = 1+(1)+1+(1)+1+(1)+… This expression is divergent, resulting in the error that causes 2 to appear to equal 1. This concludes the proof.


My opponent is correct that the series is divergent. As a result many algebric operations I have performed are not valid. Since my opponent has spotted the mistake, I will concede. Thanks to him for debating this topic with me. 
baggins  bencbartlett  Tied  

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baggins  bencbartlett  Tied  

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baggins  bencbartlett  Tied  

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baggins  bencbartlett  Tied  

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baggins  bencbartlett  Tied  

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baggins  bencbartlett  Tied  

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Any flaw which invalidates the proof is sufficient.