- #1

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(d/dx)(exp(2[pi]i(f(x))))=(x/c+c/x)(exp(2[pi]i(f(x)))),

where c is a constant? My thanks in advance.

- Thread starter Loren Booda
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- #1

- 3,077

- 4

(d/dx)(exp(2[pi]i(f(x))))=(x/c+c/x)(exp(2[pi]i(f(x)))),

where c is a constant? My thanks in advance.

- #2

suffian

(d/dx) [ e^{2π i f(x) } ] = (x/c + c/x) e^{2π i f(x) } (the given)

set y = e^{2π i f(x) } (subtitution)

(d/dx) [y] = (x/c + c/x) y

(1/y) (dy/dx) = x/c + c/x

∫(1/y)dy = ∫(x/c + c/x)dx

ln|y| = x^{2}/(2c) + c ln|x| + k

e^{ln|y|} = e^{x2/(2c) + c ln|x| + k}

|y| = K e^{x2/(2c) + c ln|x|} (extract k)

e^{2π i f(x) } = K e^{x2/(2c) + c ln|x|} (substitute original expression for y)

2^{π } i f(x) = x^{2}/(2c) + c ln|x|

ln( K e^{2π i f(x) } ) = ln(K e^{x2/(2c) + c ln|x|})

f(x) = -ix^{2}/(2^{π+1}c) - ic ln|x|/2^{π} + K

(same answer as hallsofivy except using 2^{π} instead of 2. note: i forgot to "un-e" the righthand side last time)

edit: made a serious typo plus changed to reflect that i stands for sqrt(-1)

set y = e

(d/dx) [y] = (x/c + c/x) y

(1/y) (dy/dx) = x/c + c/x

∫(1/y)dy = ∫(x/c + c/x)dx

ln|y| = x

e

|y| = K e

e

2

ln( K e

f(x) = -ix

(same answer as hallsofivy except using 2

edit: made a serious typo plus changed to reflect that i stands for sqrt(-1)

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- #3

- 3,077

- 4

suffian, yes, "i" is here the imaginary number. How does that change your result?

- #4

HallsofIvy

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(d/dx(exp(2if(x))))= 2i f'(x) exp(2if(x)) so this equation is the same as

2i f' (x)= x/c+ c/x which is a first order, non-linear differential equation for f. It is separable and can be written as

2i df= (x/c+ c/x)dx

Integrating both sides 2i f= (1/2c)x

( D is the constant of integration- an arbitrary complex number).

f(x)= (-i/4c)x

- #5

jeff

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Good, I'm glad that you haven't given up on your idea. Keep in mind though that it's usually faster to prove an idea wrong. A famous remark attributed to Feynman is that the job of a physicist is to prove themselves wrong as quickly as possible. No one wants to discover years later that their idea was not only doomed from the outset but that seeing this wasn't actually all that difficult, if only they'd been a little more objective. Also, don't underestimate how much can be learned by picking apart interesting, if ultimately wrong ideas, especially when they're your own.Originally posted by Loren Booda

(d/dx)(exp(2[pi]i(f(x))))=(x/c+c/x)(exp(2[pi]i(f(x)))),

where c is a constant? My thanks in advance.

There's always been a certain attraction in viewing things in dual terms. In physics, complementarity and more recently the dualities of string theory have given rise to all sorts of spectulation by people who like to play with abstract ideas (even if they don't understand them all that well at the time).

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