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Feynman's Calculus 
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#19
Jul705, 06:23 PM

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[tex]\frac{d}{dt}F(t)=F'(t)=2F(t)[/tex] while [tex]\frac{d}{dx}F(t)=0[/tex] 


#20
Jul805, 07:11 AM

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[tex]\frac{d}{dt}\int_0^{\infty}\exp(x^2t^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). 


#21
Jul805, 07:33 AM

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#22
Jul805, 07:48 AM

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#23
Jul1005, 05:10 PM

P: 240

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. 


#24
Jul1005, 05:11 PM

P: 240

Sorry for that post.
Castilla. 


#25
Jan808, 07:54 AM

P: 89

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


#26
Jan808, 01:14 PM

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P: 1,391

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). 


#27
Jan808, 01:34 PM

P: 89

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


#28
Jan808, 07:20 PM

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P: 1,391

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)) 


#29
Mar1108, 10:14 PM

P: 1

[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.... 


#30
Mar1108, 10:39 PM

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PF Gold
P: 16,098

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. 


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