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1

The title of your question states that you will be satisfied with an instance of {p, n} that satisfies Sum[p^k/k!, {k, 0, n}] > 0. Here is a function to find any number of such instances: instances[m_Integer?Positive] := Block[{p, n, k}, FindInstance[Sum[p^k/k!, {k, 0, n}] > 0 && p ∈ Reals && n ∈ Integers, {n, p}, m]] And ...


2

My best guess as to what you want: f[m_, z_] := (1/m)*Sum[Exp[z*Exp[2*Pi*I/m]^k], {k, 0, m - 1}] g[t_] := 1/(2 - f[5, t^(1/5)]); Series[ExpToTrig @ g[t], {t, 0, 10}] 1 + t/120 + (253 t^2)/3628800 + (762763 t^3)/1307674368000 + ( 43173223 t^4)/8846916393369600 + (633287284180541 t^5)/15511210043330985984000000 + ( 633594892177711781 ...


6

They should not be equal. Note that $$ \lim_{x\to0}\frac{d^n}{dx^n} e^{-\sqrt{x}}=\infty $$ for every integer $n$. What Series does is computing the fractional power series, a.k.a Puiseux Series, which cannot be obtained by evaluating the derivative. You can see this problem with a simpler example: try expanding the function $f=\frac{1}{x}$ or $g=\sqrt{x}$ ...


3

Interesting, doing nearly the same thing as @BobHanlon: clist = CoefficientList[ Normal@Series[Log[1/2 (1 + Sqrt[1 - x])], {x, 0, 12}], x]; FindSequenceFunction[clist, n] yields a (Correct) DifferenceRoot expression: DifferenceRoot[ Function[{[FormalY], [FormalN]}, {(-1 + 3 [FormalN] - 2 [FormalN]^2) [FormalY][[FormalN]] + ...


9

Here is a workaround: f[x_] = Log[1/2 (1 + Sqrt[1 - x])]; Generate a sequence from several of the coefficients seq = Table[ SeriesCoefficient[f[x], {x, 0, n}], {n, 8}]; Find the function that generates that sequence coef[n_] = FindSequenceFunction[seq, n] // FullSimplify -(Pochhammer[3/2, -1 + n]/ (4*n*Gamma[1 + n])) For an ...


3

I have found a simple method to mitigate the limit bug: using an intermediate Series[], the Dirichlet coefficients of a function f[x] can be calculated correctly as a Limit[]. The formulas for the coefficients are cD := Module[{}, a[1] = Limit[f[x], x -> ∞]; a[n_] := a[n] = Limit[ Series[n^x (f[x] - Sum[a[k] k^-x, {k, 1, n - 1}]), {x, ∞, ...


2

If everything else fails, you can always do Total[SeriesCoefficient[f@x, {x, 0, #}] x^# & /@ Range[0, 10]]


2

Why not to subtract two expansions like in t1 = Series[1/Sin[x], {x, 0, 10}] t2 = Series[1/Sin[x], {x, 0, 0}] Then Normal[t1] - Normal[t2] Out[3]:= x/6 + (7 x^3)/360 + (31 x^5)/15120 + (127 x^7)/604800 + ( 73 x^9)/3421440


2

I'm not sure this approach is applicable to all series, but from a quick test it seems to work for rational exponents: Looking at the FullForm of ser = Series[Exp[x]/x^(2/3), {x, 0, 5}] (* x^(-2/3) + x^(1/3) + x^(4/3)/2 + x^(7/3)/6 + x^(10/3)/24 + x^(13/3)/120 + O[x]^(16/3) *) gives FullForm[ser] (* ...


0

Is this as simple as DeleteCases[s1, _*x^c_ /; c<0] That is going to find all the terms in your series with negative exponents and simply delete them.


3

You can sum your infinite series by shifting the sum over n along by 1/2, which then simplifies the summand to a form that Mathematica can handle. (1/(-1)^(1/2)) Sum[((-1)^n Log[2 n])/(2 n), {n, 1/2, Infinity, 1}] (* (1/4) (Pi Log[4] + StieltjesGamma[1, 1/4] - StieltjesGamma[1, 3/4]) *) The overall factor 1/(-1)^(1/2) cancels the (-1)^(1/2) factor that ...


4

Shown is the actual output visible on the screen. z = Series[-q (c[x1] - c[x2]), {x2, x1, 1}] (* q c'[x1] (x2 - x1) + O[x2 - x1]^2 *) Exp[z] // Normal // Simplify (* 1 + q (-x1 + x2) c'[x1] *) (Exp[z // Normal]) // Simplify (* E^(q (-x1 + x2) c'[x1]) *) The third expression Exp[z // Normal] converts the Series to normal linear expression and then ...


2

== is needed instead of = in eqn, and a space needs to be inserted in ut. With these changes, sol = DSolve[eqn, u, t] returns unevaluated. This is not surprising, because DSolve generally can solve only equations that are known to be solvable analytically. Instead, try sol = NDSolveValue[{(eqn /. {p -> 1, n -> 0}), u[1] == 1, u'[1] == 0}, u, {t, ...



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