Yeah, so the analogy is that when <...> is inside another <...> or [], it reduces down like normal but with [a] instead
Hm, shoot, I missed something
When <...> is inside another <...> but not inside [], and has # directly to the right or somewhere on the left, then it reduces to [a]# instead of just a#
When <...> is outside, then it produces a#
When <...> is inside [], it behaves like it is outside
@SimplyBeautifulArt perhaps, and likely, but it's also (tagged as) recreational, so it might, like always, provide a space to vent, relax,let your minds play. Any way ... those with any sort of exams on the horizon, Good Luck! But you don't really need "luck" on your side to get passed exams, (or course a little luck never hurts!). Trust yourself to get through them successfully, and try to have a little fun, while you're at it! (yes, @Simply, that means history too!)
@SimplyBeautifulArt Hehehehehehehe! But seriously, I've no doubt that you know what you need to know for the history final. When you sit down to take it, think of it as a dialogue (like answering questions posted from anyone online (even though they're on paper)). At least that's always been a help to me. When I envision myself explaining something to another person, I'm better able to answer fluidly, reasonably, while at the same time a little less threatened.
aP... = (... ((aP,,,)P,,,)P,,, ... )P,,, with 'a' nestings and ,,, = reduced(...)
If k is the left-most thing that is a number and greater than zero, then: reduced(n#...#p#k+1#...) = n#...#p'#p'#...#p'#a#p#k#..., with 'a' iteration of #p', where p' = reduced(p)
reduced(0) = a
In a line of #'s, if the last entry is not a number greater than zero, drop it
<{?-}...> is the notation nester.
If the ... is empty, then we have reduced(<{?-}>) = ???...?a with 'a' nestings of ?
Context:
My friend (The same one who gave me this ODE) originally challenged me to obtain the general solution to this ODE:
$$\frac{d^3 y}{dx^3}+\alpha x\frac{dy}{dx}+\beta y=0 \tag{1}$$
Where $\alpha,\beta \in \mathbb{R}$.
I could not figure out any substitution without the use of power seri...
@projectilemotion 8 upvotes, and one downvote gives you a net 78 rep. I'd say that's pretty good on mse. Though I agree that the downvoter should have explained him/herself!
@SimplyBeautifulArt yes. Receiving a downvote is -2; giving a downvote on a question costs the downvoter nothing. "giving a downvote* on an answer costs 1 rep for the downvoter. But each and every downvote on a question or answer is -2 in rep for the questioner and/or answerer.
@projectilemotion Interesting. I've been very cordial to him; my last encounter with him was when he asked my to vote to close a list of some of his questions; I simply told him I'd have to wait to do so, because he changes his mind very rapidly sometimes, but I would do so if he felt the same a couple of days later. But it was all in all a very cordial interaction.
Exactly: he asks users to help to close his posts, and then does a 180 by asking for folks to reopen them. I'm pointed out what you observed to the mods (in the past, not recently)
@SimplyBeautifulArt and guys AND gals: Try to encourage meditation, to help calm people down, so there's no need for people to blow off hot steam! This world already has too many "hot-heads"!
Context:
My friend (The same one who gave me this ODE) originally challenged me to obtain the general solution to this ODE:
$$\frac{d^3 y}{dx^3}+\alpha x\frac{dy}{dx}+\beta y=0 \tag{1}$$
Where $\alpha,\beta \in \mathbb{R}$.
I could not figure out any substitution without the use of power seri...
This is the Realm of Simply Beautiful…
This is the Realm of Simply Beautiful…