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SolveAlways[(-s^2 + 40 s + 50)/(s (s + 1) (s + 5)^2) == A/s + B/(s + 1) + (C*s + D)/(s + 5)^2, s]

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Apart does what you need to get done but to make this example work you need == between the two expressions and get rid of the last "{s}". Addendum: 1 should be just solve without the Apart. 2 is wrong usage. 3 is incomplete and hence wrong. 4 is correct but it didn't give you what you expected because you setup incorrectly. Since you have $(s+5)^2$ in ...

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Try Apart[]: Apart[(-s^2 + 40 s + 50)/(s (s + 1) (s + 5)^2)] 2/s - 9/(16 (1 + s)) - 35/(4 (5 + s)^2) - 23/(16 (5 + s))

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Mathematica: Assuming[{ta > 0, ta > te, te > 0}, Block[{f, igrand, lamtbl}, f[ta_, lambda_] := lambda/(4*(Sqrt[Pi ta^3])) Exp[-lambda^2/(4 ta)]; igrand[ta_, te_, lambda_] := f[ta, lambda] Erf[lambda/(2*Sqrt[ta - te])]; lamtbl = {0.4, 0.8, 1.2, 2, 2.4}; Table[ NIntegrate[igrand[ta, te, lamtbl[[which]]], {ta, te, Infinity}], ...

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II. SOLVE THE PROBLEM Why doesn't a simple mean give the position of a centroid in a polygon? leads to formulas. Frown. I want to know how to set up and solve the problem. A solution can always be converted to a formula. The reverse --figuring out how a problem was solved by looking at the formula-- is very hard. So I will solve problem from ...

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I. CONSIDERATIONS. (In 2 parts. The formulas are at the end of each.) Case: Point List I have sixteen marbles of equal mass. Let us agree they all lie in one plane. Q1: What is the center of mass of the set of marbles? The center of mass of a sphere is its center. So let the coordinates of the marble centers be: $\;\;\;\;\;Pts = A_1, A_2.... A_{16}. ... 2 $$\sinh(l-i n \pi)=\tfrac12(e^{l-i n \pi}-e^{-l+i n \pi})=(-1)^n\sinh(l)$$ since$e^{i\pi}=-1$, and $$\frac1{l-i n \pi}=\frac{l+i n \pi}{(l-i n \pi)(l+i n \pi)}=\frac{l+i n \pi}{l^2+n^2 \pi^2}$$ so all your quoted expressions are actually equal. 1 This seems to be what you're looking for: x = 2; y = 3; a = {{1, 0},{0, 1}}; b ={{I, 0},{0, I}}; m = a x + b y; Edit: here's what you are probably actually looking for: Solve[a x + b y == m && Re[x] == 0 && Re[y] == 0, {x, y}] With$a,b$and$m$as above. 0 The function you want in matlab is the quiver function. The following will produce the required phase portrait, as I understand them. % no domain is given, so I will use [-5,5] x [-5,5] % with 50 subintervals in each direction xdom = linspace(-5,5,51); ydom = linspace(-5,5,51); [X,Y] = meshgrid(xdom,ydom); % generate mesh of domain U = X.*(7 - X - ... 0 The question concerns the light path in a vertically inhomogeneous medium with the velocity$v(y)$supposed linearly dependent on the height$y(x)$. Here $$v=v_0\left(1-\frac y\alpha \right)$$From Fermat's principle one obtains the ode $$\frac 1{v\sqrt{1+y'^2}}=\text{const}$$Here one solves the "boundary value problem" $$\begin {cases}y'=\dfrac{\sqrt ... 2 One point that seems to have been largely ignored in the other answers is that the typesetting system in Mathematica deals with more than appearance - it deals with semantics as well and this is a primary consideration in its design. Consider the following two inputs and outputs. In the first, I've typed in the integral in Mathematica's typeset input ... 2 If a and b are integers, with a composite, then the number of primes in the interval [a,b] is the number of primes less than or equal to b minus the number of primes less than or equal to a. If a itself is prime, we should account for this by adding that one prime back on. To determine the number of primes less than or equal to a number in ... 4 Basically, Mathematica is designed for calculations, and can do some typesetting too. \LaTeX is designed for typesetting, and can do some calculations too. So the main strength of \LaTeX here is that if and when you want to add some other common document feature to your math papers, like cross-referencing, bibliography, numbered lists, etc., you'll ... 3 Yes. In Mathemtica use the function DSolve. But you can do it by hand. We have$$ \frac{\dot W}{A}=\frac{\dot M}{B}\implies W=\frac{A}{B}(M-m_0)+w_0. $$Substitution in the second equation yields$$ \dot M=B\Bigl(k-\frac{B\,M}{A(M-m_0)+B\,w_0)}\Bigr).$$This is a first order ODE in separated variables that can be integrated explicitly. 10 OP asks: is there something significant I can do in LATEX/TEX that I can't do in Mathematica? LaTeX far better at handling print page layout … for example, 2 column layout, or page breaks, footnotes etc some journals/conferences require submissions in LaTeX format I am sure there are other benefits! — but much steeper learning curve … though I believe ... 3 Luckily, in the open community world GNU TeXmacs can work as standard Office word in WYSIWYG way. BTW, it can be used as front-end for many open alternative scientific computing packs, such as Maxima, a symbolic computing software. You can have it a try. 32 I use both Mathematica and$\LaTeX$extensively. While it is possible to make nice looking documents in each, they really have two different aims. Mathematica is mainly about mathematical computing and interactive content.$\LaTeX$is about typesetting and publication-ready articles with fine language-based (as opposed to WYSIWYG) control of layout. As ... 34$\LaTeX$is not only free like in beer, but free like in speech. There are enormous advantages to using open source standards. Right now you like Mathematica and are happy with its software. Suppose that five years from now the company decides to put out a new version and break backwards-compatibility. Or the company goes out of business. And suddenly ... 12 Sharing documents may not be easy - LaTeX is a standard that almost everybody is expected to be able to use, and is free. Mathematica requires a purchase, and lots of people never saw it. 4 The following inputs will plot the following 6 hearts in the picture below respectively. ContourPlot[(x^2 + y^2 - 1)^3 - x^2 y^3 == 0, {x, -1.5, 1.5}, {y, -1.5, 1.5}, MaxRecursion -> 5] ContourPlot[x^2 + (y - (2 (x^2 + Abs[x] - 6))/(3 (x^2 + Abs[x] + 2)))^2 == 36, {x, -9, 9}, {y, -9, 9}, MaxRecursion -> 5] ContourPlot[x^2 + (5/4 y - Sqrt[Abs[x]])^2 ... 0 Here is a screen shot from this equation on Wolfram Alpha. I don't have a license for Mathematica. (x^2+y^2-1)^3 = x^2 1 It is true. Since Wikipedia does a good job of explaining it in the general case (here$\Bbb C$isn't much different from$\Bbb R^2\$), I'll contend myself to linking to the relevant part on the Mean Value theorem page (see equation (**)).

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I see what you've done now, you need to put a space in-between the I3.1, so it should be either I 3.1 or I*3.1 . If you don't do that Mathematica looks at it as a variable and not as the number 3.1 multiplied by I but as the variable "I3.1". When it highlights something as blue that means it's an undefined variable, so you know something hasn't been input ...

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