Question

An interesting operator I discovered while playing around. I forget how I came up with it, but at any rate here it is. If you can think up a use for this I will give some things I think are related:

Answer 1

\[\Large e^{-ax}\frac{d}{dx}e^{ax}*\] So that's it, if you operate on a function y(x) say, 3 times you will get: \[\large e^{-ax}\frac{d^3}{dx^3}(e^{ax}y)=1a^3y^{(0)}+3a^2y^{(1)}+3a^1y^{(2)}+1a^0y^{(3)}\]

So we essentially have a hybrid sort of look at the binomial theorem with part of it as powers and the other part as derivatives. Hmm?

Answer 2

follow -.-
sadly ill go nw

Answer 3

Similarly the product rule for derivatives also obeys the binomial theorem\[\large (f*g)^{(3)}=1f^{(3)}g^{(0)}+3f^{(2)}g^{(1)}+3f^{(1)}g^{(2)}+1f^{(0)}g^{(3)}\]
so kind of interesting \[\Large (a+b)^3=1a^3+b^0+3a^2b^1+3a^1b^2+1a^0b^3\]

Not sure if there's some way we can relate all of these to get us something cool. or not. =/

Answer 4

wow! binomial theorem shows up in most unexpected places xD
from the pattern, this may hold i guess :
\[\large (fg)^n = \binom{n}{0}f^ng^0 +\binom{n}{1}f^{n-1}g^1 + \cdots +\binom{n}{n}f^0g^n \]

Answer 5

Yeah, it's fascinating because they cleanly stack together:

\[\Large (e^{-ax}\frac{d}{dx}e^{ax})...(e^{-ax}\frac{d}{dx}e^{ax})=e^{-ax}\frac{d^n}{dx^n}e^{ax}\]

Answer 6

the \(e^{-ax}\) factor just eats out the \(f\) when :
f = \(\large e^{ax}\)
g = \(\large y\)

Answer 7

giving us a beautiful looking binomial expansion in `y` xD

Answer 8

Oh good point I didn't see that!

Answer 9

Originally I had simply used a=1 by multiplying it by e^x. Then I noticed that I was looking at "half" of a binomial theorem haha.

Answer 10

Actually if y=e^(bx) then we get the binomial theorem from the product rule version X_X

Answer 11

somethign like this :
\[\large e^{-(a+b)x} (e^{ax}*e^{bx})^n = (a+b)^n\]

xD

Answer 12

i remember how we saw this :)

Answer 13

left side is nth derivative,
right side is a proper exponent

Answer 14

now go to sleep!

Answer 15

Hahaha that's actually not too surprising to see now. It's really just like in differential equations finding a characteristic equation... Hmm... Well... Not yet dan!

Answer 16

Can we somehow change it to calculate the binomial coefficients or something in a new way maybe?\[ e^{i(ix(a+b))} \frac{d^n}{dx^n}(e^{i(-ix(a+b))})=e^{i(ix(a+b))} \frac{d^n}{dx^n}(\cos(ix(a+b))-i \sin (ix(a+b)))\]

Answer 17

Errr... maybe less awkward:\[\Large e^{-x(a+b)}\frac{d^n}{dx^n}[\cosh(x(a+b))+\sinh(x(a+b))]\]

Answer 18

@ganeshie8 are you ready? please answer when your done with this problem

Answer 19

lolz i need to read this fully plz someone reply , so i could open this when im online again xD

Answer 20

*

Answer 21

I don't think the question is done yet actually. Maybe we can use this to help us find the nth derivative of tan(x) easier if we look at tanx=sinx*secx and then we can expand it as a binomial faster. Hmm.