- #1
fourier jr
- 765
- 13
everybody seems to know at least one of the "proofs" that 1=2 or 1=-1, etc but i had never seen this one before. check it out:
everybody knows that
[tex]log(1+x) = x-\frac{x^2}{2} + \frac{x^3}{3} - ...[/tex]
plug in x=1 & the series converges & we get
log2 = 1 - 1/2 + 1/3 - 1/4 + 1/5 - 1/6 + 1/7 - ...
2log2 = 2 - 1 + 2/3 - 1/2 + 2/5 - 1/3 + 2/7 - ...
take the terms together which have a common denominator (ie simplify) & we get
2log2 = 1 + 1/3 - 1/2 + 1/5 + 1/7 - 1/4 + 1/9 - ... = 1 - 1/2 + 1/3 -1/4 + ... = log2
hence 1 = 2
QED
here's a similar one
log2 = 1 - 1/2 + 1/3 - 1/4 + 1/5 - ...
= (1 + 1/3 + 1/5 + 1/7 + ...) - (1/2 + 1/4 + 1/6 + 1/8 + ...)
= {(1 + 1/3 + 1/5 + ...) + (1/2 + 1/4 + 1/6 + ...)} - 2(1/2 + 1/4 + 1/6 + ...)
= (1 + 1/2 + 1/3 + ...) - (1 + 1/2 + 1/3 + ...)
= 0
i guess the problem must have something to do with the 'simplification' & doing something to an infinite sum. off the top of my head those are my guesses
everybody knows that
[tex]log(1+x) = x-\frac{x^2}{2} + \frac{x^3}{3} - ...[/tex]
plug in x=1 & the series converges & we get
log2 = 1 - 1/2 + 1/3 - 1/4 + 1/5 - 1/6 + 1/7 - ...
2log2 = 2 - 1 + 2/3 - 1/2 + 2/5 - 1/3 + 2/7 - ...
take the terms together which have a common denominator (ie simplify) & we get
2log2 = 1 + 1/3 - 1/2 + 1/5 + 1/7 - 1/4 + 1/9 - ... = 1 - 1/2 + 1/3 -1/4 + ... = log2
hence 1 = 2
QED
here's a similar one
log2 = 1 - 1/2 + 1/3 - 1/4 + 1/5 - ...
= (1 + 1/3 + 1/5 + 1/7 + ...) - (1/2 + 1/4 + 1/6 + 1/8 + ...)
= {(1 + 1/3 + 1/5 + ...) + (1/2 + 1/4 + 1/6 + ...)} - 2(1/2 + 1/4 + 1/6 + ...)
= (1 + 1/2 + 1/3 + ...) - (1 + 1/2 + 1/3 + ...)
= 0
i guess the problem must have something to do with the 'simplification' & doing something to an infinite sum. off the top of my head those are my guesses