# Tag Info

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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 ...

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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 ...

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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 ...

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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 ...

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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 ...

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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 $B=0,E=0,C=1,A=-1,D=1$

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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 ...

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$$y''(t)-y'(t)-2y(t)=2t+1\Longleftrightarrow$$ $$\mathcal{L}_t\left[y''(t)-y'(t)-2y(t)\right]_{(s)}=\mathcal{L}_t\left[2t+1\right]_{(s)}\Longleftrightarrow$$ $$\mathcal{L}_t\left[y''(t)\right]_{(s)}-\mathcal{L}_t\left[y'(t)\right]_{(s)}-\mathcal{L}_t\left[2y(t)\right]_{(s)}=\mathcal{L}_t\left[2t\right]_{(s)}+\mathcal{L}_t\left[1\right]_{(s)}\... 0 Notice: For the -3 dB points, you know that \left|\text{H}(\omega)\right|=\frac{1}{\sqrt{2}} and we can say that:$$\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[...

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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 ...

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