This is more interesting than it may seem. But for now, no comment ;)
Previous iteration: #701531
118 sats \ 3 replies \ @senf 27 Sep
Same class of infinite number of solutions, I'd say.
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That's a win.
I'll post a video tomorrow elaborating on this concept for people unfamiliar with the works of George Cantor.
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Here is one of the many videos that illustrates this idea.
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That's my first thought, too. Unfortunately, I won't have time to dive into today's puzzle.
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Both have an infinity of solutions. And I'm guessing they have the same cardinality of solutions?
If cos(x)=0 then let y=sqrt(x) if x>0 or y=-sqrt(-x) if x<0. Then cos(y^2)=0.
Since there's a 1:1 mapping between these x's and these y's, the two sets have the same cardinality.
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I believe that both have infinite number of solutions. However, when looking into the chart of sin(x) and sin(x^2), it is quite obvious, that both the infinities have different cardinality.
I'd say that the number of solutions for cos(x)=0 has same cardinality as the set of all natural numbers. However, for the cos(x^2)=0, the cardinality would be as the set of all real numbers. At least, it is not greater.
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Cos (X) = Cos (X mod 360), therefore infinite solutions. The same applies to Cos (y), with y = x^2
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yes, it has one positive and one negative. while cos(x)=0 only has one.
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Are you sure there is only one solution for cos(x)=0?
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both have an infinite number of solutions. cos(x^2) = 0 has broader distribution of solutions compared to the linear and equally spaced solutions of cos(x) = 0. cos(x^2) = 0 is more dense in unique values.
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That's correct.
But can one argue that one infinity is larger than the other? They're both infinite...
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On what domain, sir?
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On what domain, sir?
Infinite domain. The real number line. I should have been clearer, indeed.
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