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2

For any square matrix M which is the sum of two similar matrices M = A + B the determinant can be written as a sum of determinants as follows (example for two dimensions): det(M) = det( ( A11 + B11, A12 + B12), (A21 + B21, A22 + B22) ) = det( ( A11 + 0, A12 + 0), (A21 + B21, A22 + B22) ) + det( ( 0 + B11, 0 + B12), (A21 + B21, A22 + B22) ) and, ...


1

How about Table[First@Timing[N[Sum[(MatrixPower[A, j] t^j)/j!, {j, 0, i}]]], {i, 0, 200}] ListLinePlot[%] If you want the iteration number side by side, just change the body of the Table to {i, First@...}.


5

There are a few problems with your code. Allow me to highlight them and guide you to a solution. First of all, the proper way to define a function in Mathematica is using :=. So your code should read F[x_] := NSum[Exp[-x BesselJZero[0, a]^2]/BesselJZero[0, a]^2, {a, 1, Infinity}] Furthermore, you should note that the zeroes of the Bessel function are ...


1

Manipulate[ Plot[{Sin@x, Normal@Series[Sin@u, {u, x0, n}] /. u -> x}, {x, -2 Pi, 2 Pi}, PlotRange -> {Automatic, {-2, 2}}, Epilog -> {PointSize[Medium], Point@{x0, Sin@x0}}], {n, 0, 10, 1}, {x0, -Pi, Pi}]


0

This is not an answer but a general note. for the case of this function: E^x^(1/2) you have to note that the derivative of E^x^(1/2) at x=0 is ComplexInfinity. to show you that, the general series of a function is as follows: Clear[f] s = Series[f[x], {x, 0, 1}] // Normal if you set f[x] to your function it will result in ComplexInfinity. f[x_] ...


0

Series[1/2 (e1 + e2 - Sqrt[e1^2 - 2 e1 e2 + e2^2 + 4 V12^2]), {V12, 0, 2}] // Simplify[#, e1 > e2] &


1

If we remove the subscripts from the converted TeX, ToExpression[ "\\frac{1}{2} \\left(e_1+e_2-\\sqrt{e_1^2-2e_1e_2+e_2^2+4V_{12}^2} \\right)", TeXForm] /. Subscript[b_, e_] :> b[e] (* 1/2 (e[1] + e[2] - Sqrt[e[1]^2 - 2 e[1] e[2] + e[2]^2 + 4 V[12]^2]) *) we can simply use PowerExpand: Series[ 1/2 (e[1] + e[2] - Sqrt[e[1]^2 - 2 e[1] e[2] + e[2]^2 ...


1

Assumptions given to Series are not used by FullSimplify that you have in your code. To pass Asumptions to all functions in given expression, wrap said expression with Assuming: Assuming[e1 > e2, Series[1/2 (e1 + e2 - Sqrt[e1^2 - 2 e1 e2 + e2^2 + 4 V12^2] ), {V12, 0, 2}] // Simplify ] (* e2 + V12^2/(-e1+e2) + O[V12]^3*) Or, if you want to always ...



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