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Results tagged with Search options user 148

Questions related to the calculus and analysis branches of Mathematica, including, but not limited to, limits, derivatives, integrals, series, and residues.

0
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David Parks's Presentations add-on application includes a number of functions for doing indefinite integration step-by-step. For example: << Presentations` integrate[Sin[x] Exp[Cos[x]], x] (* o …
answered Aug 15 by murray
7
votes
Explicitly: Integrate[Boole[x^2/a^2 + y^2/b^2 <= 1], {x, -a, a}, {y, -b, b}, Assumptions -> {a > 0, b > 0}] Function Boole delineates the region; the Assumptions option is needed for sym …
answered Feb 10 '13 by murray
3
votes
The questions say to use the integral remainder estimate. tailUpper[n_] := Integrate[1/(x^2 + 1), {x, n, ∞}] Reduce[tailUpper[n] <= Rationalize[0.0005], n, Integers] (* n ∈ Integers && n >= 2000 *) …
answered Feb 27 '17 by murray
5
votes
At least with Mathematica version 10.0.1, using option Assumptions does provide a correct answer: Integrate[Cos[m*x]*Cos[n*x], {x, 0, 2 Pi}, Assumptions -> Element[{m, n}, Integers]] (* (Sin[2* …
answered Nov 17 '14 by murray
4
votes
Here's a more down-to-earth solution in Mathematica that's probably closer to your current level of mathematics study. Just directly use the usual one-dimensional integrals involved in finding plane a …
answered Jan 2 '13 by murray
6
votes
Mathematica 10 introduced a new set of functions that can help with this: Inactive[Integrate][r^3, {t, 0, 2 Pi}, {r, 0, 2}, {z, r^2, 4}] Or, obtaining same output: Inactivate[Integrate[r …
answered Dec 1 '14 by murray
5
votes
In Mathematica: Series[1/(z^2 - 3 z + 2), {z, 1, 10}, Assumptions -> {0 < Abs[z - 1 ]< 1}] (or whatever the top-order term is that you want). Mathematical method: Step (i): expand the function i …
answered Mar 21 '14 by murray
4
votes
2answers
(1) I want to insert into a Text cell conventional in-line mathematical notation like $\lim_{x\to a} f(x)$. I know I can do it by typing the StandardForm Limit[f[x], x -> a] in an Input cell, conver …
asked Sep 8 '15 by murray
1
vote
You'll find some notebooks about complex analysis, and especially about visualizing complex functions, on my web page: http://blogs.umass.edu/math421-murray/files/ Disclaimer" These notebooks are se …
answered Jan 7 by murray
3
votes
Note that you can also use the new (as of Version 12) ComplexPlot function, too: ComplexPlot[1 - z Log[(1 + z)/z], {z, -2 - 2 I, 2 + 2 I}, Mesh -> 10, MeshFunctions -> {Re[#2] &, Im[#2] &}] O …
answered yesterday by murray