Feynman's Calculus

quetzalcoatl9

not to nitpick, but shouldn't it be [tex]\frac{d}{dt}[/tex] ?

lurflurf

Yes it was a typo.
[tex]\frac{d}{dt}F(t)=F'(t)=-2F(t)[/tex]
while
[tex]\frac{d}{dx}F(t)=0[/tex]

saltydog

You know what, I just don't see that. Can someone help me? When I take the derivative I get:

[tex]\frac{d}{dt}\int_0^{\infty}\exp(-x^2-t^2/x^2) dx=-2t\int_0^{\infty}\frac{1}{x^2}e^{-(x^2+\frac{t^2}{x^2})}dx[/tex]

and thus I don't see how the derivative is -2F(t).

lurflurf

That step (F'(t)=-2F(t)) was a little sudden so upon request I added a few intermidiate steps a few posts up. You can see the equality more easily if a substitution like u=t/x is made.

saltydog

Very good Lurflurf. I see that now. Also, I took the minus sign out of parenthesis so I think I was Ok with that.

Castilla

For the change of variable I am using these functions:
u = u(x) = t/x (t fixed)
<´¨¨´¨¨´¨k¨´¨´´´´j´´¨¨¨¨z¨´s¨s¨s´s´s´´´<<sp(u) = exp(-t¨Z¨Z´´ä¨zñS´¨pa´zo¨¨¨Qq´qqá´´AááÁÁAÍ´SD´S
and I obtained the result, but with 2 instead of -2.

Castilla

Sorry for that post.

Castilla.

nanoWatt

So if one integrates from -infinity to infinity, can you always change the range (or whatever it's called) of integration to 0 to infinity, and multiply the remaining integral by 2? Or is that a property of the sin that makes it symmetrical to where the negative side doesn't affect it? In other words, why doesn't a -2 come out?

How does this exp(i z) work? Is it like i to the z power?

I may have seen some exp functions with three variables like exp (x y z) or so. Does that mean anything?

Mute

The property you described depends on whether a function is even or odd. An even function is a function such that f(-x) = f(x), while an odd function is one such that g(-x) = -g(x).

So, for example, cosine is an even function, since cos(-x) = cos(x), while sine is an odd function, since sin(-x) = -sin(x). The exponential function is neither odd nor even, as e^(-x) does not equal either e^(x) (for any arbitrary value of x) or -e^(x).

Because an even function looks the same in the x > 0 half plane as it does in the x < 0 half plane, if you have symmetric limits about x = 0, then integrating from -L to L of an even function is just like integrating from 0 to L twice. This is only true for even functions. For odd functions, the contribution from the negative half plane will cancel out that from the positive half plane, so the result will be zero.

To summarize:

[tex]\int_{-L}^{L} dx f(x) = 2 \int_{0}^{L} dx f(x)~\mbox{if f(x) is even}[/tex]
[tex]\int_{-L}^{L} dx f(x) = 0~\mbox{if f(x) is odd}[/tex]




Have you ever heard of the imaginary unit i? It is the number defined such that [itex]i^2 = -1[/itex]. With this number one can define the complex exponential function, which has the property that

[tex]e^{ix} = \cos x + i \sin x[/tex]

Since exponentials typically have nicer properties than sines or cosines, by considering the integral of exp{ix}/x you might be able to get the integral for sin(x)/x in an easier fashion (and, you'll probably also get the integral of cos(x)/x out of it too, if that happens to be finite, which I don't think it is).

nanoWatt

Thanks for the breakdown.

So how would you expand exp (i z)?
Is that the same as [tex] e^{iz} [/tex] ?

Mute

exp{z} is just another notation for e^z.

If z is a complex number z = x + iy (where x and y are real), then

e^{z} = e^{x}e^{iy} = e^{x}(cos(y) + i sin(y))

And e^{i z} = e^{-y + ix} = e^{-y}(cos(x) + i sin(x))

swensonj

It seems to me that the signs on the last two terms of the right-hand side are reversed; should the formula be
[tex]\frac{d}{dx}\int_{a(x)}^{b(x)} f(x,t)dt= \left(\int_{a(x)}^{b(x)}\frac{\partial f(x,t)}{\partial x}dt\right)+ \frac{db(x)}{dx}f(x,b(x))- \frac{da(x)}{dx}f(x,a(x))[/tex]
?

I guess so, but I haven't figured out how to prove this formula yet....

Hurkyl

It's actually pretty straightforward; e.g. you might simply apply the limit formula for the derivative, and then splitting and rearranging the resulting expression into pieces that can be approximated well.

It's one of those things that, if you understand the ideas behind using limits and approximations, will be very straightforward. And if you don't find it straightforward, then it's really worth studying as an exercise.