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About arguments of functions 
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#1
Apr1614, 03:43 PM

P: 686

Commonly I see ##\sin(\omega t + k x)## I ask myself, if ωt is equal to θ thus kx is equal to what??
Also, the kernel of the laplace transform is ##\exp(st)##, very similar to kernel of the fourier, that is ##\exp(i \omega t)##, and and same question again, if ωt is equal to θ thus st is equal to what?? Finally, I see sometimes a connection between s and ω, that is ##s = \sigma + i \omega##, but, again, what this equation means and what is σ? 


#2
Apr1614, 04:16 PM

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#3
Apr1614, 04:40 PM

P: 686

https://en.wikipedia.org/wiki/Sinusoidal#General_form
What exactly you don't understand? k is the wave number, x is the coordinate spatial, ω is the angular velocity and t is the time. If the product ωt results the angle θ so the product kx results which physical/mathematical quantity? 


#4
Apr1614, 05:51 PM

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P: 3,288

About arguments of functions
$$\left\int_{\infty}^{\infty} f(t) e^{st} dt\right \leq \int_{\infty}^{\infty} f(t) e^{\sigma t} dt$$ So if ##f## decays fast enough on one of the intervals ##(0,\infty)## or ##(\infty, 0)## (for example, if its support on one of the intervals is compact) and grows no faster than exponentially on the other interval, the Laplace integral can converge for some values of ##\sigma##. Putting it another way, the Laplace transform of ##f(t)## is the same as the Fourier transform of ##f(t)e^{\sigma t}##. 


#5
Apr1614, 05:53 PM

P: 446

Reading the wiki article, I don't see where your confusion is coming from. I do not see ##\theta=\omega t## in that article. 


#6
Apr1614, 07:48 PM

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#7
Apr1714, 06:51 AM

P: 686

2) I was thought about what is ##\sigma##, so, I realized that exist a similarity between ##\exp(st)## and ##\exp(\frac{d\log(z)}{dt}t)##. If ##\frac{d \log(z)}{dt}## is constant, thus ##\frac{d \log(z)}{dt} \times t## is the antiderivative of ##\frac{d \log(z)}{dt}##, ie, ##\log(z)##. ##z = \exp(\log(z)) = \exp( \log(r) + i \theta) = \exp(\int \frac{d}{dt}(\log(r) + i \theta) dt )## ## = \exp(\int \frac{v_r}{r} + i \omega\;dt) = \exp((\frac{v_r}{r} + i \omega) t) = \exp((\sigma + i \omega)t) = \exp(st)##. All this means that the kernel of the laplace transform is the variable complex ##z## and that the laplace transform leads from time t to complex plan ##\frac{d \log(z)}{dt}##. 3) If the laplace transform has ##\sigma \neq 0## and ##\omega \neq 0## and the fourier transform has ##\sigma = 0## and ##\omega \neq 0##, for the party be complete, shouldn't there is a transform* that has ##\sigma \neq 0## and ##\omega = 0##? * ##\int f(t) \exp(\sigma t) dt## 


#8
Apr1714, 01:24 PM

P: 446

I have never seen ##\theta=\omega t## in a situation where ##\sin\theta## was a function of both time and position. Usually ##\theta## (except in geometry) is just an imaginary angle (not as in ##i##) that stands in for whatever argument is being used. If you write ##\sin(\omega t+k x)## then the argument is ##\omega t+kx##. If you want to, you could define ##\phi=kx##, but I have not ever seen this done unless ##x## is constant and then ##\phi## is just a phase shift. This isn't really a matter of right and wrong. I just don't think it is normally done the way you show it and I can't think of a reason to do it that way. 


#9
Apr1714, 04:17 PM

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$$\tilde{f}(\sigma) = \int f(t) \exp(\sigma t) dt$$ but would you be able to recover ##f(t)## given ##\tilde{f}(\sigma)##? If not, then it would not be a very useful transform. 


#10
Apr1714, 05:53 PM

P: 686

Actually, this transform $$\tilde{f}(\sigma) = \int_{\infty}^{+\infty} f(t) \exp(\sigma t) dt$$ or $$\tilde{f}(\sigma) = \int_{0}^{\infty} f(t) \exp(\sigma t) dt$$ has the face of the laplace transform. So this transform isn't a repeated ideia? Of course wouldn't be a repeated ideia if this new transform leads from time ##t## to radial velocity ##v_r## (like fourier that leads from time ##t## to angular velocity ##\omega##). So the transform would have a new face: $$\tilde{f}(v_r) = \int_{\infty}^{+\infty} f(t) \exp\left(\frac{v_r}{r}t\right) dt$$ But of course this analogy didn't go unnoticed for none good mathematical, however, I never see nobody mention this analogy. So, this make sense? 


#11
Apr1814, 11:02 PM

P: 686

For my surprise, I discovered that ##\theta = \theta (x, t)##!! This because ##\theta## is only an argument of a function and not a spatial coordinate, as I thought that was, (after all, how could a independent spatial coordinate be function of another spatial coordinate and of the time too, this no make sense). So, ##\frac{\partial \theta}{\partial t}=\omega## and ##\frac{\partial \theta}{\partial x} = k##, according with (https://en.wikipedia.org/wiki/Freque...s_of_frequency), from this follows 2 line of development:
$$f(\theta) = f(\omega t) = f(2\pi \nu t) = f\left(\frac{2\pi t}{T} \right)$$ $$f(\theta) = f(k x) = f(2\pi \xi x) = f\left(\frac{2\pi x}{\lambda} \right)$$ Being: ##\omega## = angular temporal frequency ##\nu## = temporal frequency ##k## = angular spatial frequency ##\xi## = spatial frequency If the laplace transform is for problems with temporal dependence and the fourier transform is for problems with spatial dependence, so is correct say that the laplace transform maps between the time ##t## and the angular temporal frequency ##\omega##, while that the fourier transform maps between the time ##t## and the angular spatial frequency ##k## (or spatial frequency ##\xi##)? If yes, thus no exist a connection between laplace and fourier, same with ##\sigma = 0##, because the kernel of the laplace would be ##\exp( i \omega t)## while that the kernel of fourier would be ##\exp(i k t)##, right? 


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