The Student Room Group
Uni students - Complete our survey >>
Uni students - Complete our survey >>
Uni students - Complete our survey >>

This discussion is now closed.

Check out other Related discussions

Something a bit trippy I saw in a book

Scroll to see replies

Zhen Lin
Ah. Very interesting. I didn't think of it that way. (In hindsight, I should have recognised it as composition of operators rather than some strange kind of product.) What about the commutativity of these operators? Are there any important rules?
Mainly "be careful, most operators don't commute!", I think. There are important senses in which operators behave very like matrices, so the same kind of stuff applies.

Again, I'm not an expert in this area - my impression has always been that you have to be very confident at solving these problems without playing tricks with treating operators as things you can divide by, do series expansions, etc. before you can reliably use the tricks without getting fingers burned. (and so I never really bothered much with the tricks).
Reply 21
Avatar for MC REN
MC REN
14
Adje
When you say you're 'used to seeing differential operators treated like that', what do you mean? Something like:

(x+x2+x3+x4) dx=1+2x+3x2+4x3 (x + x^2 + x^3 + x^4) \, dx = 1 + 2x + 3x^2 + 4x^3 ? Or something completely different? (Is dx even what you mean by a differential operator, or is it d/dx?)


No I meant d/dx, like what Zhen Lin said - Heaviside notation, I have no idea why you'd think I meant what you said ^
Reply 22
Avatar for Kolya
Kolya
17
DeanK2
However, I think Tom has shown why people need to be careful using this kind of shortcut / rule in which operators can be treated as real numbers (even though it can be made rigorous).
Indeed, perhaps best to avoid this argument when writing a C4 paper...
Reply 23
Avatar for Adjective
Adjective
OP
12
MC REN
No I meant d/dx, like what Zhen Lin said - Heaviside notation, I have no idea why you'd think I meant what you said ^


Because I often do idiotic things. :p:
Was messing around in computing, and found this:

sinh⁡x=∫∫sinh⁡x\sinh x = \int \int \sinh x
⇒sinh⁡x−∫∫sinh⁡x=0\Rightarrow \sinh x - \int \int \sinh x = 0
⇒(1−∫∫)sinh⁡x=0\Rightarrow (1 - \int \int) \sinh x = 0

⇒sinh⁡x=11−∫∫×0\Rightarrow \sinh x = \frac{1}{1 - \int \int} \times 0

Infinite geometric series: first term 1, common ratio ∫∫\int \int

sinh⁡x=(1+∫∫+∫∫∫∫+∫∫∫∫∫∫)×0\sinh x = (1 + \int \int + \int \int \int \int + \int \int \int \int \int \int) \times 0

sinh⁡x=(0+∫∫0+∫∫∫∫0+∫∫∫∫∫∫0)\sinh x = (0 + \int \int 0 + \int \int \int \int 0 + \int \int \int \int \int \int 0)

Let ∫0=1\int 0 = 1, then

sinh⁡x=x+x33!+x55!+x77!+...\sinh x = x + \frac{x^3}{3!} + \frac{x^5}{5!} + \frac{x^7}{7!} + ... .

If one were to replace sinh x with cosh x throughout, we'd need to define ∫0=0\int 0 = 0 and ∫∫0=1\int \int 0 = 1. Interesting.

Related discussions

Latest

Trending

Trending

Articles for you