# Tag Info

1

It is nothing profound. "The Lapace transform of the probability density function for a random variable" is way too many words to keep saying and writing. "The Lapace transform of the random variable" is a bit quicker to say. Mathematicians are lazy. That is okay as long as the true meaning is understood.   Although things like this do cause ...

1

The Laplace transform that you are looking for is given, of course, by the formula $$\mathcal L (\omega) (s) = \int \limits _0 ^\infty \omega (u) \ \Bbb e ^{-su} \ \Bbb d u .$$ The problem is that the definition of Buchstab's function (the series in Tao's Ex. 28.i) is almost useless for computations. Fortunately, there comes Ex. 28.iii which gives the ...

0

No: Recalling the quotient rule $$\frac{d}{dx}\dfrac{f(x)}{g(x)}=\dfrac{\frac{df}{dx}g(x)-f(x)\frac{dg}{dx}}{g(x)^2}=\dfrac{1}{g(x)}\dfrac{df}{dx}-\frac{f(x)g'(x)}{g(x)^2},$$ we conclude that $\dfrac{1}{g(x)}\dfrac{d}{dx}f(x) \neq \dfrac{d}{dx}\dfrac{f(x)}{g(x)}$ unless $g(x)=$ const.

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Let's suppose that $f:[0,\infty)\to\Bbb R$ is convex and of slower than exponential growth, so that it has a Laplace transform. Then $f'_+$ (the right-hand derivative of $f$) is right-continuous and non-decreasing. As such $f'_+$ is the "distribution function" of a measure $\mu$ on $[0,\infty)$: $\mu((a,b]) =f'_+(b)-f'_+(a)$ for $0\le a<b$. Integrate by ...

0

The equation has no solution. Let $f(t) = \int_0^1 \cos(t-\tau ) x(\tau) d \tau$. Then we have $f'(t) = -\int_0^1 \sin(t-\tau ) x(\tau) d \tau$ and hence $f''(t) = - f(t)$. It follows that $f(t) = a \cos t + b \sin t$ for some $a,b$ and hence can not be equal to the right hand side above. Alternatively: If we let $g(t) = t \cos t$, we see that $g'' \... 2$\newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{\mathrm{i}} \...

0

If I properly understand, you look for the solution of $$h'(x)=\frac k{(1+x^n)^{1/n}}$$ The solution exists but it involves the hypergeometric function $$h(x)= k x \, _2F_1\left(\frac{1}{n},\frac{1}{n};1+\frac{1}{n};-x^n\right)+C$$ The only simple forms are $$n=1 \implies h(x)=k \,\log (1+x)+C$$ $$n=2 \implies h(x)=k\, \sinh ^{-1}(x)+C$$ Don't be afraid ...

3

Oh, I hate the Laplace transform! I have yet to find a differential equation that cannot be solved more easily using simpler methods. Here, the differential equation is $y''+ y= x^2$. The associated homogeneous equation is $y''+ y= 0$. Its characteristic equation is $r^2+ 1= 0$ which has roots $r= \pm i$ so the general solution to the associated ...

3

As for the Laplace solution you asked for, you can split the fraction like this: $$\frac 1{s^3(s^2+2)}=\frac As+\frac B{s^2}+\frac C{s^3}+\frac{Ds+E}{s^2+1}$$ $$=\frac{As^4+As^2+Bs^3+Bs+Cs^2+2C+Ds^4+Es^3}{s^3(s^2+1)}$$ $$=\frac{(A+D)s^4+(B+E)s^3+(A+C)s^2+Bs+C}{s^3(s^2+1)}$$ By identification, you find $B=0,E=0,C=1,A=-1,D=1$

1

My question is how do I find this inverse Laplace transform of $\dfrac{1}{s^3(s^2+1)}$? Hint. If one wants to proceed on your route, by a partial fraction decomposition, one has $$\frac{1}{s^3(s^2+1)}=-\frac{1}{s}+\frac{1}{s^3}+\frac{s}{1+s^2}$$ giving $$\mathcal{L}^{-1}\left(\frac{1}{s^3(s^2+1)}\right)(t)=-1+\frac{t^2}2+\cos t$$ using standard ...

1

$\newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{\mathrm{i}} \... 2 First, we should find the inverse Laplace transform of$\frac{1}{(s+1)^4}$. If you look at a table of Laplace transforms, you'll see that $$L(t^ne^{at}) = \frac{n!}{(s-a)^{n+1}}$$ (this formula can be shown by induction & integration by parts). So we can see that $$L(\frac{t^3e^{-t}}{3!}) = \frac{1}{(s+1)^4}$$ Most tables will also mention that $$L(f'(... 1 Hint. From standard properties (see this table) and from the given result one gets$$ \mathscr{L}\left(t \sin bt \right)(s)=-\left(\mathscr{L}\left( \sin bt \right)\right)'(s)=\frac{2bs}{(s^2+b^2)^2} $$then using$$ \mathscr{L}\left(e^{at}f(t) \right)(s)=\left(\mathscr{L}f\right)(s-a) $$gives finally$$ \mathscr{L}\left(t e^{at}\sin (bt) \right)(s)=\... 2 The laplace transformations of$coshbtis the following $$\int_0^\infty cosh({bt})e^{-st} dt$$ $$= \int_0^\infty \frac{(e^bt + e^{-bt})e^{-st}}{2} dt$$ $$= \frac{1}{2} \int_0^\infty e^{-st + bt} + e^{-st - bt} dt$$ $$= \frac{1}{2} \int_0^\infty e^{(-s + b)t} + e^{(-s - b)t} dt$$ $$=\frac{1}{2} \begin{bmatrix} \frac {e^{(-s+ b)t}}{-s+b} + \frac{e^{-st - bt}}... 0 I found an example of such a bell-shaped Schwartz function whose Laplace transform has two zeros at the same imaginary part in its domain of convergence :$$\varphi(x) = e^{x/4}\left(\frac{1}{e^{e^x}-1}-\frac{c}{e^{c e^x}-1}\right)\mathcal{L}[\varphi](1/4-s) = \Gamma(s) (1-c^{1-s}) \zeta(s)$$choosing c = e^{2 \pi /T} \approx 2.43 where T is the ... 2 Hint: As you wrote, by taking Laplace from your equation, we obtained$$Y(s)=\frac{1}{s(s^4+1)}$$Then, using partial fractions, we have$$Y(s)=\dfrac{1}{s} - \dfrac{s^3}{s^4+1}$$Can you continue by taking inverse Laplace from this to find y(t)? 0 First you must know how to rewrite a piecewise defined function in terms of unit step functions. I will illustrate with a function defined in three pieces. Let f(x)=\begin{cases}y_0(t)\text{ for }0<t\le a\\ y_1(t)\text{ for }a<t\le b\\ y_2(t)\text{ for }t\ge b \end{cases} Then f(t)=y_0(t)\,u_0(t)+[y_1(t)-y_0(t)]\,u_a(t)+[y_2(t)-y_1(t)]\,u_b(t) ... 0 Hint$$y''+6y'+9y=5t-5t\,\,{{u}_{3}}(t)s^2F(s)-sf(0)-f'(0)+6F(s)-6f(0)+9F(s)=\frac{5}{s^2}-\frac{5e^{-3s}}{s^2}-\frac{15e^{-3s}}{s}0 s' is a dummy variable, not the derivative of s with respect to anything. In any case, it can sometimes be helpful to start at the far side of the equals sign. We have \begin{align} \int_s^{\infty}F(s')ds' & = \int_s^{\infty}\int_0^{\infty}e^{-s' t}f(t)dtds'\\ & = \int_0^{\infty}\int_s^{\infty}e^{-s't}f(t)ds'dt\\ & = \int_0^{\infty}\... 1\sin(t)\frac{\partial^2y}{\partial t^2}+\cos(t)\frac{\partial y}{\partial t}+\cos(t)\frac{\partial ^2y}{\partial x^2}=0$$You tried a solution like y=y(x,t)=Ae^{\alpha t}sin(Ct)+Be^{\beta t}sin(Dt)+Fe^{\zeta x}sin(Ex)+Ge^{\gamma x}sin(Hx) via method of undetermined coefficients. This can succeed only if each of the terms is solution of the PDE. Putting ... 1 Explicit computation yiels$$\int_{0}^{\infty} y''(t) \> \sin t \> e^{-st} \> dt=\sin t \> e^{-st} \> y'(t) \> |_{0}^{\infty} + \int_{0}^{\infty} y'(t) (se^{-st} \sin t - e^{-st} \cos t) \> dt $$We see the first term is 0 if y is of exponential order. Assuming this we continue$$= se^{-st} \sin t \> y \> |_0^{\infty} - \... 1 I assumey$and its derivatives are exponentially bounded. Then using integration by parts twice, I get (for$\text{Re}(s)$sufficiently large) $$y(0) + (s^2-1) F(s) - 2 s G(s)$$ where$F(s)$and$G(s)$are the Laplace transforms of$y(t) \sin(t)$and$y(t) \cos(t)$respectively. 0 There are expressions which are not Laplace transforms of functions of exponential order. A necessary condition for$F(s)$to be the Laplace transform of a function of exponential order is that$|s F(s)|$is bounded as$s \to +\infty$. Your function has$|F(s)| \sim -2 \ln(s)$as$s \to +\infty$. However, your$F$is the Laplace transform of a ... 0 Hola podrían ayudarme con esta ecuación Por favor y''-4y'+4y=t^3*e^- 3t se los agradecería mucho por favor ayudenme 0 Assuming that$u_{2\pi}(t)$is the function that equals$1$for$t\in[0,2\pi]$and zero otherwise, by assuming$g(s)=\int_{0}^{+\infty}y(t)e^{-st}\,dt$and applying the properties of the Laplace transform we get: $$(4+s^2) g(s) = \frac{1-e^{-2\pi s}}{1+s^2}$$ hence$y$is given by the inverse Laplace transform of$\frac{1-e^{-2\pi s}}{(1+s^2)(4+s^2)}$, ... 1 Let$y(x)=z(x)$(otherwise, after the Laplace transform they look like the same functions): $$\begin{cases} 3\cdot\frac{\text{d}^2y(t)}{\text{d}t^2}+\frac{\text{d}z(x)}{\text{d}x}=0\\ 5\cdot\frac{\text{d}^2z(x)}{\text{d}x^2}-\frac{\text{d}y(t)}{\text{d}t}=0 \end{cases}$$ Take the Laplace transform of both sides: -First ODE: $$3\cdot\left(s^2\text{Y}(s)-... 0 Question 1 Observing that$$\mathcal L \{t^2u(t)\}=\frac{2}{s^3}$$where u(t) is the Heaviside's function, and using the shifting property we have$$\mathcal L \{g(t)\}=\mathcal L \{(t+4)^2u(t+4)\}=\frac{2}{s^3}\mathrm e^{4s}$$Question 2 Observing that$$\mathcal L \{\mathrm e^{-at}f(t)\}=F(s+a)$$and that$$\mathcal L \{h(t-b)u(t-b)\}=\mathrm e^{-bs}H(... 2 I've been able to prove simple inverse laplace transforms of$\dfrac {1}{(s+a)} $quite easily, but what about$\dfrac {1}{(s+a)^3+b^2} $? Hint. One may just use a partial fraction decomposition over$\mathbb{C}(X)$, giving $$\frac {1}{(s+a)^3+b^3}=\frac {A}{s+a+b}+\frac {B}{s+a-b\tau}+\frac {C}{s+a-b\bar{\tau}}$$ where$\tau=\frac12+i\frac{\sqrt{3}...

2

$$y''(t)-y'(t)-2y(t)=2t+1\Longleftrightarrow$$ $$\mathcal{L}_t\left[y''(t)-y'(t)-2y(t)\right]_{(s)}=\mathcal{L}_t\left[2t+1\right]_{(s)}\Longleftrightarrow$$ $$\mathcal{L}_t\left[y''(t)\right]_{(s)}-\mathcal{L}_t\left[y'(t)\right]_{(s)}-\mathcal{L}_t\left[2y(t)\right]_{(s)}=\mathcal{L}_t\left[2t\right]_{(s)}+\mathcal{L}_t\left[1\right]_{(s)}\... 0 Notice: For the -3 dB points, you know that \left|\text{H}(\omega)\right|=\frac{1}{\sqrt{2}} and we can say that:$$\color{red}{20\log\left(\left|\text{H}(\omega)\right|\right)=20\log\left(\frac{1}{\sqrt{2}}\right)=-10\log(2)\approx-3.0103}$$Where \log is the base 10 logarithm. First, when z\in\mathbb{C}:$$|z|=\left|\Re[z]+\Im[...

1

I think you have a sign error in your function, it should be: $$\tag 1 f(t) = 8 ~\sin(t)~u(t-\pi)$$ We want to write the function in $(1)$ without the use of the unit step function. For $(1)$, the unit step function is equal to zero for all times less than $\pi$ and is then just the sine function for times greater than $\pi$. If we were to plot $(1)$, we ...

0

$\newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{\mathrm{i}} \... 1 My free interpretation is: how to use the Laplace transform to compute the integral $$I=\int_{0}^{+\infty}\frac{e^{-t}\sin^2(t)}{t}\,dt$$ ? Well, since: $$\mathcal{L}\left(e^{-t}\sin^2(t)\right)=\frac{2}{(1+s)(5+2s+s^2)}\tag{1}$$ we have:$\$ I = 2\int_{0}^{+\infty}\frac{ds}{(1+s)(5+2s+s^2)}=2\int_{0}^{+\infty}\frac{du}{4+e^{2u}}=\int_{0}^{+\infty}\frac{dv}{...

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