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and our terms of service . home questions tags users unanswered profile activity network profile reputation user apparently, this user prefers to keep an air of mystery about them. answers questions ~ people reached member for days profile views last seen mins ago communities ( ) mathematics bronze badges
answers | votes newest prove $\sin( )\sin( )\sin( )\cdots\sin(n)$ converges as $n o\infty$ [duplicate] dec ' the series $\sum_{n= }^\infty\sin(n)\cdot\sin( /n)$ [duplicate] dec ' convergence of series $\sum_{n= }^\infty\sin(n)\cdot\sin( /n)$ [closed] dec ' polynomials over q. irreducible hours ago badges...
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respect a laser-like focus on one topic. top tags ( ) limits score posts sequences-and-series score posts epsilon-delta score posts top posts ( ) all questions answers | votes newest epsilon n proof of limit of a trigonometric sequence hours ago epsilon n proof with n-th powers for a sequence hours ago badges...
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. $$ fourier-transform d convolution spherical-geometry fast-fourier-transform share | cite | improve this question edited hour ago moritz asked jan at : moritz moritz bronze badges $\endgroup$ add a comment | answer active oldest votes $egingroup$ indeed you can use the discrete sine transform. just
powers of $r$, so this symmetric extension is then also smooth.) whether this yields a good approximation of course strongly depends on the range and resolution of the $r$ values compared to the range and structure of $ ho$. share | cite | improve this answer answered feb at : joriki joriki k gold badges...
https://math.stackexchange.com/questions/3527070/how-to-compute-the-dft-of-a-radially-symmetric-function-in-3d
notation for the last example. i should also note that for this example, a notation like $(a imes a)+b imes b+ imes a imes b$ is not allowed as it contains constants. thank you. combinatorics arithmetic expression-simplification share | cite | improve this question asked hour ago robert robert bronze badges
$\endgroup$ add a comment | answer active oldest votes $egingroup$ there are more: $$a+b+c=(a+b)+c=a+(b+c)=(b+a)+c=b+(a+c)=(a+c)+b=$$ $$=a+(c+b)=(c+a)+b=c+(a+b)=(b+c)+a=$$ $$=b+(c+a)=(c+b)+a=c+(b+a).$$ share | cite | improve this answer answered hour ago michael rozenberg michael rozenberg k gold badges...
https://math.stackexchange.com/questions/3545105/all-different-notations-of-an-arithmetic-expression
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