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Given $d\in\mathbb{N}_0$, the Taylor series about $i/2^d$ is a polynomial of degree at most $d$ for all $i\in\mathbb{Z}$. Let $S_d$ be the set of such Taylor series. There exist unique polynomials $pol_0,pol_1,\ ...\ ,pol_d$ of degree $0,1,\ ...\ ,d$ and a function $c:\mathbb{N}\times\mathbb{R}\mapsto\{-1,0,1\}$ such that for all $x\in\mathbb{R}$, the ...

6

Using the conditions in the book ClearAll[a, x]; expr = Log[1 + 2 a*Cos[2 x] + a^2]*Sin[x]^2; r = Integrate[expr, {x, 0, Pi/2}, Assumptions -> a^2 > 1] Book result You used $a>1$ but the book says to use $a^2>1$. These are not the same. Update: I asked about this on another forum. Experts opinions says that result should be valid for \$...

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Define your rule as a function: ruleDerF = (# /. Module[{x}, Derivative[n_][f][var_] :> (D[x f[x], {x, n - 1}] /. x -> var) ])& f''[x] // ruleDerF f[x] + x Derivative[1][f][x] FixedPoint[ruleDerF, f''[x]] f[x] + x^2 f[x] FixedPoint[ruleDerF, f''[1]] 2 f[1] (havent tested heavily)

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Thanks to @J. M. this solution gives the answer: Sum[a^i, {i, ∞}, GenerateConditions -> True] It returns a ConditionalExpression, in the above case: ConditionalExpression[-(a/(-1 + a)), Abs[a] < 1]

1

When you sum it over there is no free variable. If you use the same index, Mathematica is converting the derivative of a sum to a sum of derivatives which give you n. Choose a number j such that 1<j<n and you will see your answer. For example D[Sum[q[i], {i, 10}], q[3]] 1 I am not sure if there is a straightforward way to do it symbolically,

1

I found an error with the syntax in the model where a curly brace was used rather than parenthesis Exp[-{Log[w/δ]}^2/(2 σ^2)] -> Exp[-(Log[w/δ])^2/(2 σ^2)] With this change model is defined as: model[Ms_?NumericQ, δ_?NumericQ, σ_?NumericQ, x_?NumericQ] := NIntegrate[ Ms (Coth[(Ms x (1/6 Pi w^3))/(k T)] - ((k T)/(Ms x (1/6 Pi w^3)))) 1/(Sqrt[2 ...

1

I'm surprised that nobody followed up on QuantumDot's suggestion to use Dt[]: With[{n = 9}, (Table[Dt[y, {x, k}], {k, n}] /. First @ Solve[Table[Dt[Sin[x] + Sin[x + y] - y, {x, k}] == 0, {k, n}], Table[Dt[y, {x, k}], {k, n}]]) /. {x -> π, y -> 0}] {-1, 0, 1/2, 0, -1/2, 0, 1/2, 0, 17}

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