B Curiosity on this infinite product

AI Thread Summary
The discussion centers on the infinite product of twos, expressed as ##p_{n} = 2^n##, which diverges to infinity as ##n## approaches infinity. When considering the infinite case, the equation ##2 \cdot x = x## leads to the paradoxical conclusion that ##x = 0##, raising questions about the validity of this result. Participants highlight that assuming a limit exists for such divergent products is flawed, as it leads to contradictions, such as deriving that ##1 = 0## from other infinite sums. The conversation emphasizes the importance of recognizing the limitations of mathematical assumptions in infinite contexts. Ultimately, the discussion illustrates the complexities and potential pitfalls of manipulating infinite products and sums.
Ssnow
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Curiosity question on the infinite product ##2\cdot 2\cdot 2\cdots ##
Let us consider the infinite products ## p_{n}\,=\, 2\cdot 2\cdot 2 \cdot 2 \cdots 2 \,=\, 2^n## with ##n=1,\ldots ## . Clearly ##p_{n}\rightarrow +\infty## as ##n\rightarrow +\infty##. But if I put the infinity case ## 2\cdot 2\cdot 2 \cdot 2 \cdots \,=\, x## I have ##2\cdot x =x ## so ##x=0##. It is obvious I cannot put ##x=2\cdot 2\cdot 2 \cdot 2 \cdots ## and to try to seach the limit because the product diverges but has this "strange" algebraically formal result a conceptual reason to be (for example it is linked to the way to do the products ?) or it is only wrong and stop here ?
Thank you,
Ssnow
 
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Ssnow said:
Summary:: Curiosity question on the infinite product ##2\cdot 2\cdot 2\cdots ##

Let us consider the infinite products ## p_{n}\,=\, 2\cdot 2\cdot 2 \cdot 2 \cdots 2 \,=\, 2^n## with ##n=1,\ldots ## . Clearly ##p_{n}\rightarrow +\infty## as ##n\rightarrow +\infty##. But if I put the infinity case ## 2\cdot 2\cdot 2 \cdot 2 \cdots \,=\, x## I have ##2\cdot x =x ## so ##x=0##.
Or ##x=\infty##. If you rule out ##\infty## here, then you are biasing the result.
 
@FactChecker thanks, sure ##x=0 \vee x=\infty##. I ask for the absurd solution ##x=0## ...
Ssnow
 
This is a pretty classic thing where you can create fake math. The real issue is that you started off by assuming a limit exists. If there is a limit and it is L (L is a real number), then 2L=L so L=0. But this assumes the limit exists to begin with, which obviously it does not.

You can get more obvious contradictions. 1+1+1+..., If it has a limit of L then 1+L= L so 1=0.
 
@Office_Shredder I think ##1+L=L## imply that ##0L=-1## that is impossible!
In any case from something of false you can deduce everything ... :biggrin:
Thank you!
Ssnow
 
Seemingly by some mathematical coincidence, a hexagon of sides 2,2,7,7, 11, and 11 can be inscribed in a circle of radius 7. The other day I saw a math problem on line, which they said came from a Polish Olympiad, where you compute the length x of the 3rd side which is the same as the radius, so that the sides of length 2,x, and 11 are inscribed on the arc of a semi-circle. The law of cosines applied twice gives the answer for x of exactly 7, but the arithmetic is so complex that the...
Is it possible to arrange six pencils such that each one touches the other five? If so, how? This is an adaption of a Martin Gardner puzzle only I changed it from cigarettes to pencils and left out the clues because PF folks don’t need clues. From the book “My Best Mathematical and Logic Puzzles”. Dover, 1994.

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