herewiss13 (![[personal profile]](https://www.dreamwidth.org/img/silk/identity/user.png)
herewiss13) wrote2005-11-06 09:49 am
herewiss13) wrote2005-11-06 09:49 am
Math!
Is there a pattern to the difference between inverse exponentials?
(i.e. (x^y)-(y^x)= ?)
Good, now that I'm alone in here...
The inspiration for this question comes from today's "number search" in Foxtrot (not that funny, really). I saw 2^3 and 3^2, did the math (for kicks) and discovered a difference of 1. This seemed non-random enough that I sought to derive the general case and got totally bogged down:
(x^y)-(y^x)= N
(x^y) = N + (y^x)
ylogx = logN + xlogy
(That was my first dodgy step, as it's been forever since I took math. My Dad (retired science teacher) remembers _just_ enough to assure this is ok.)
y/log y = (log x + log N)/x
And that's where I totally lost it. Actually, I first totally lost it when I forgot to divide log N by x. I nearly flunked 6th Grade Algebra because the teacher was an _incredibly_ bad 1st year...and it's haunted me ever since. I still have to fight to keep from dividing things by zero.
Actually, I tell a lie. I didn't lose it there, I just misplaced it. The next step is the fairly basic:
x(y/log y) = xlog x + xlog N
y/log y = (xlog x)/x + (xlog N)/x
y/log y = log x + log N
NOW we're getting somewhere (assuming this is a legitimate result). Also log, here, is the name as ln. So if we multiply all sides by 'e'...
(e^y)/y = x + N
(initially, I'd forgotten the 1/y), but fortunately I've discovered that in writing up this posting...and it only affects things a little.)
So if you want to know what x^y - y^x is:
N = [(e^y)/y] - x will tell you.
I'm about 90% confident of this proof, and 80% confident it can't be reduced further.
...just another lazy Sunday morning.
(i.e. (x^y)-(y^x)= ?)
Good, now that I'm alone in here...
The inspiration for this question comes from today's "number search" in Foxtrot (not that funny, really). I saw 2^3 and 3^2, did the math (for kicks) and discovered a difference of 1. This seemed non-random enough that I sought to derive the general case and got totally bogged down:
(x^y)-(y^x)= N
(x^y) = N + (y^x)
ylogx = logN + xlogy
(That was my first dodgy step, as it's been forever since I took math. My Dad (retired science teacher) remembers _just_ enough to assure this is ok.)
y/log y = (log x + log N)/x
And that's where I totally lost it. Actually, I first totally lost it when I forgot to divide log N by x. I nearly flunked 6th Grade Algebra because the teacher was an _incredibly_ bad 1st year...and it's haunted me ever since. I still have to fight to keep from dividing things by zero.
Actually, I tell a lie. I didn't lose it there, I just misplaced it. The next step is the fairly basic:
x(y/log y) = xlog x + xlog N
y/log y = (xlog x)/x + (xlog N)/x
y/log y = log x + log N
NOW we're getting somewhere (assuming this is a legitimate result). Also log, here, is the name as ln. So if we multiply all sides by 'e'...
(e^y)/y = x + N
(initially, I'd forgotten the 1/y), but fortunately I've discovered that in writing up this posting...and it only affects things a little.)
So if you want to know what x^y - y^x is:
N = [(e^y)/y] - x will tell you.
I'm about 90% confident of this proof, and 80% confident it can't be reduced further.
...just another lazy Sunday morning.