# About arguments of functions

1. Apr 16, 2014

### Jhenrique

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. Apr 16, 2014

### micromass

Staff Emeritus
What do you mean?? I'm sorry, but this question makes no sense at all.

3. Apr 16, 2014

### Jhenrique

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. Apr 16, 2014

### jbunniii

I'm not sure whether $\sigma$ has a physical meaning or not, but mathematically it is a "decay" factor which can allow the Laplace transform to converge even if the Fourier transform (which is the special case with $\sigma = 0$) does not. Since $e^{-st} = e^{-(\sigma + i \omega)t} = e^{-\sigma t}e^{-i \omega t}$, the absolute value $|e^{-st}|$ is $e^{-\sigma t}$. Thus if we compute the Laplace transform of some function $f$, then
$$\left|\int_{-\infty}^{\infty} f(t) e^{-st} dt\right| \leq \int_{-\infty}^{\infty} |f(t)| e^{-\sigma t} dt$$
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 $f(t)e^{-\sigma t}$.

Last edited: Apr 16, 2014
5. Apr 16, 2014

### DrewD

If $\sin(\omega t+kx)=\sin\theta$, then $\theta=\omega t+kx$ not $\omega t$. I am guessing this is what you mean, but I may be misinterpreting. If $x$ is constant, then $\theta$ changes as $\omega t$ or if $t$ is held constant $\theta=kx$.

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. Apr 16, 2014

### FactChecker

$\sigma$is in there to study exponential responses. Not everything is sinusoidal. An interesting youtube series that covers both Laplace and Fourier transforms is at https://www.youtube.com/watch?v=ZGPtPkTft8g&list=PLUMWjy5jgHK3j74Z5Tq6Tso1fSfVWZC8L

7. Apr 17, 2014

### Jhenrique

1) Are you sure? IMO, $\sin(\theta) = \sin(\omega t) \neq \sin(kx)$. Is sure that exist a relationship between $\theta$, $t$ and $\omega$, that is $\frac{d\theta}{dt}=\omega$, if $\omega$ is constant, so $\theta = \omega t$. The only way of connect $\theta$ and $x$ is saying that the wave number $k$ is the derivative of $\theta$ wrt $x$. Is this true?

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 $\omega = 0$?

* $\int f(t) \exp(-\sigma t) dt$

Last edited by a moderator: Apr 17, 2014
8. Apr 17, 2014

### DrewD

If you want to define $\theta=\omega t$, then sure. But why would you do that if there is also a dependance on position.

Sure, if $\theta=\omega t+kx$ then $\frac{\partial\theta}{\partial t}=\omega$ and $\frac{\partial\theta}{\partial x}=k$. The first implies $\theta=\omega t+f(x)$ and the second implies $\theta=kx+g(t)$.

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. Apr 17, 2014

### jbunniii

You could define such a transform, say
$$\tilde{f}(\sigma) = \int f(t) \exp(-\sigma t) dt$$
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. Apr 17, 2014

### Jhenrique

Yeah, I'm assuming that exist the inverse. I didn't think in the superior and inferior limits nor in the inverse, I just did think in the ideia of transform a function from time to sigma.

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. Apr 18, 2014

### Jhenrique

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/Frequency#Other_types_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?

Last edited: Apr 18, 2014