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September 4[edit]

Proof that i =-∞[edit]

Can this proof be considered a mathematical fallacy?Almuhammedi (talk) 11:46, 4 September 2020 (UTC)[reply]

I'd more say that it's not even wrong. See also pseudomathematics, into which it squarely falls. –Deacon Vorbis (carbon • videos) 12:49, 4 September 2020 (UTC)[reply]

It's a basic misunderstanding of the mathematics of imaginary numbers. Pay it no mind. If I was to find its major fault, the issue is that the definition of i = sqrt -1 is axiomatic; it's just something we define. That is, we use the symbol i to represent the number that when squared will equal -1. It's not more complicated than that. Once we've defined that number, we don't need to invent anything new to deal with it, we treat it like we do any other numbers under normal rules of math we've always used. i is mostly useful in giving us a convenient basis to do rotational mathematics algebraically, but otherwise isn't THAT confusing. --Jayron 32 13:31, 4 September 2020 (UTC)[reply]

In a rigorous treatment,

i

is not just "something we define", because there is no guarantee that such a definition will give a useful or even consistent structure. We could "just define" names for 1/0 or 0/0, but these generally don't lead anywhere. See Complex_number#Formal_construction for some ways that the complex numbers can be constructed as a "natural" extension of the reals. AndrewWTaylor (talk) 14:58, 4 September 2020 (UTC)[reply]

The quadratic equation

x 2 + 1 = 0

has two distinct solutions, and

i

is one of the two. But which one?_;) --Lambiam 17:57, 4 September 2020 (UTC)[reply]

The one that isn't -i. ̣—Tamfang (talk) 02:42, 6 September 2020 (UTC)[reply]

Electrical engineers use

j

as the symbol for the the square root of

-1

. Perhaps this is the other solution. --Lambiam 05:16, 6 September 2020 (UTC)[reply]

This professor surprises me from time to time and I feel embarrassed to criticize some of his thesis because I hold no more than a bachelor degree. In the same website (and others seemingly peer reviewed) he discusses his own theory of hyperbolic universe that I am skeptic and would also like to ask about in the physics reference desk. Almuhammedi (talk) 14:47, 4 September 2020 (UTC)[reply]

Don't waste your time; this is all pure nonsense...bunk...crankery...cow excrement. None of this is "peer reviewed"; it's all posted to a site that let anyone post anything. –Deacon Vorbis (carbon • videos) 15:09, 4 September 2020 (UTC)[reply]

Prove or disprove: for any solution to analytical ode is analytical[edit]

Let $f:\mathbb {R^{\textrm {N}}\rightarrow R}$ be an analytical function. Prove or disprove, any solution to $f(x;y,y',...,y^{(n)})=0$ must be analytical.--Exx8 (talk) 22:40, 4 September 2020 (UTC)[reply]

Prove or disprove: the above is a homework question. Hint: consider

x^{3}y'-2y=0

. By the way, the term "analytic function" is more common. Also, it is sufficient to prove that something is true rather than that it must be true. The latter raises the question: OK, it must be true, but must it necessarily be true? --Lambiam 09:18, 5 September 2020 (UTC)[reply]

Actually, it is not a HW.--Exx8 (talk) 13:00, 5 September 2020 (UTC)[reply]

I'm guessing someone, somewhere has been given something like this as homework for the Picard–Lindelöf theorem. Basically find a counterexample for uniqueness when the conditions of the theorem are not met, if a given solution if an initial value problem can be extended in more than one way then at most one of these extensions is analytic. There are (at least) two ways this can happen. The first way is when all the curves in a family are tangent to the same line at a single point. One such family is y=cx². These are solutions to the ODE xy'=2y. But any solution can be grafted to another at the point where the two have the same tangent. So

y={\begin{cases}0,&{\text{if }}x\leq 0\ \\x^{2},&{\text{if }}x\geq 0\end{cases}}

is a solution which is not analytic. A second way it can happen is when a family of curves has an envelope, and the envelope itself is a solution to the underlying differential equation. For example the family y=(x-c)² has the envelope y=0. These are both solutions to the ODE (y')² = 4y. But again, two solutions can be grafted together where they have common tangent, so the function defined above is a non-analytic solutution to this ODE as well. Of course, if f≡0, which is analytic, then any function with sufficiently many derivatives is a solution, so I'm assuming that is excluded somehow.

It might be the case that, following Picard–Lindelöf, if the ODE is given as y' = f(x, y) where f is both analytic and Lipschitz continuous, then any solution is analytic. It certainly seems like it ought to be true, and if not I'd like to know why. @Lambiam: the NGram counter looks like a very interesting tool and I think I have a number of uses for it, so thanks. --RDBury (talk) 17:02, 5 September 2020 (UTC)[reply]

PS. The above function is also a solution to (y')² = 2xy', representing a third way for uniqueness to fail. --RDBury (talk) 18:17, 5 September 2020 (UTC)[reply]