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0

DEGREE = 2; x0 = 1; y0 = \[Pi]; f = x^2*Sin[(x*y)/2]; g = Normal@Series[f, {x, x0, DEGREE}, {y, y0, DEGREE}]; $$ g(x,y)=(x-1)^2 \left(\left(\frac{\pi ^2}{64}-\frac{3}{4}\right) (y-\pi )^2-\frac{3}{4} \pi (y-\pi )-\frac{\pi ^2}{8}+1\right)+(x-1) \left(-\frac{1}{2} (y-\pi )^2-\frac{1}{4} \pi (y-\pi )+2\right)-\frac{1}{8} (y-\pi )^2+1 $$ Plot3D[#, ...


7

It seems, that findExponentialGeneratingFunction would be equivalent to FindGeneratingFunction of the sequence {a1/0!,a2/1!,a3/2!,...}. Therefore findExponentialGeneratingFunction = FindGeneratingFunction[#1/(Factorial /@ Range[0, Length@#1 - 1]), #2] & findExponentialGeneratingFunction[{1,1,1,1,1,1,1,1},x] (* Exp[x] *)


2

Log[2] + Sum[(-1)^(k + 1) (x - 2)^k/(k 2^k), {k, 1, Infinity}] Log[x] Since Log is complex or unbounded for non-positive argument, the maximum difference is undefined for x <= 0 Log /@ {0, -1/4} // InputForm {-Infinity, I*Pi - Log[4]} The maximum difference will be arbitrarily large as x gets arbitrarily close to 0+. Clear[f] f[x_, n_] := ...


8

Your formula should be either 4 + Sum[(-1)^x*4/(2*x + 1), {x, 1, Infinity}] Pi or Sum[(-1)^x*4/(2*x + 1), {x, 0, Infinity}] Pi The finite series approximation is then Clear[f] f[n_] := Sum[(-1)^x*4/(2*x + 1), {x, 0, n}] As n increases, f[n] converges slowly to Pi. With[{nmax = 150}, DiscretePlot[f[n], {n, 1, nmax}, Epilog -> {Red, ...


0

What you probably mean you can easily get exchanging integration and series expansion. And could also do without Normal[] but this way it looks better after Simplify[]. Here it is up to the order eps^2 Integrate[Series[ Exp[1/2 (I*Exp[I*x[l]] + \[Epsilon]*M2[l])], {\[Epsilon], 0, 2}] // Normal, {l, 0, L}] // Simplify $\int_0^L \frac{1}{8} ...


4

Borrowing from Chip Hurst: f = Inactive[Sum][ Assuming[n > 0, SeriesCoefficient[Sin[x], {x, 0, n}]] x^n, {n, 0, Infinity}] ExpToTrig[Activate[f]] // Simplify (*Sin[x]*)


4

Try this FullSimplify[SeriesCoefficient[Sin[x], {x, 0, n}], n >= 0 && n ∈ Integers] ((-1)^(1 + n) Sin[n π/2])/n!


3

The slowness is caused by unnecessary use of SetDelayed particularly for functions which include derivatives. AA0[q_] = (3/(16 q^5) - 1/(4 q^3) + 1/q) // Simplify; q[rr_] = rr^-4 (q0 + q1 rr^(1/3)); a0p2[rr_] = (AA0[q[rr]] Sqrt[1 + rr^2 D[q[rr], rr]^2])/(2 Sqrt[ 1 + rr^4 + AA0[q[rr]]^2]) // Simplify; NUM[rr_] = rr^2 D[q[rr], rr] Sqrt[bb^2 + rr^4] // ...


2

As to the second question: you could change ConditionalExpression itself. Unprotect[ConditionalExpression] ConditionalExpression[a_, ___] := a but this is potentially dangerous. Who knows where else ConditionalExpression is being used? So, a safer alternative is just change the way it's being output Unprotect[ConditionalExpression] ...


1

Alternatively, and perhaps more generally, conditions can be eliminated by prepending First, as in First@Integrate[1/(x^2 + y^2), {y, -Infinity, Infinity}]


1

Use Normal with Series f[x_] = Series[1/Sqrt[x], {x, 1, 2}] // Normal 1 + (1 - x)/2 + (3/8)*(-1 + x)^2 Use GenerateConditions -> False with Integrate Integrate[1/(x^2 + y^2), {y, -Infinity, Infinity}, GenerateConditions -> False] Pi*Sqrt[1/x^2]


3

Appreciating the nice answers of Daniel Lichtblau and glance I would like to note that they go in the classical mathematical direction. Let me add here a different, Mathematica-oriented view on the problem at hand. Let us make a table with the structure {n, Integral} where I substitute x=10^n as follows: lst = Table[{n, NIntegrate[Exp[-10^n t^3 ...


6

First of all, for a mathematical explanation on how to compute the first terms of the series with pen&paper check out this math.SE question. That said, here is what I ended up doing with Mathematica 10.0 to obtain an arbitrary number of terms of the asymptotic expansion. I warn you that I'm not very proficient with Mathematica so any advice on how to ...


10

Chances are it will behave similarly to Integrate[Exp[-x t^3 ], {t, 0, π/2}]. I show a numerical verification below. We'll use the dominant series term, which we can compute, for the above integral, and compare to numerical integrations of the one we cannot compute analytically. ee1[x_] := Exp[-x*t^3] ee2[x_] := Exp[-x*t^3*Cos[t]]; Let's get the candidate ...



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