# Tag Info

4

The function Txk[x,k,n] calculates the contribution of the k^th zero at position x. The parameter n governs how many terms in the sum are used. This corresponds to Havil's equation on the bottom of page 196 of his book Gamma. Note that ExpIntegralEi should be used as @Guesswhoitis suggests, and as discussed here. I think there is a typo in the book, hence ...

2

You can integrate as follows. Integrate[BesselI[-nu, k*x]/x, {x, r1, r}, Assumptions :> {k \[Element] Complexes, r1 \[Element] Reals, r \[Element] Reals, nu \[Element] Reals, r1 > 0, r > r1} The result is a complicated expression in terms of Gamma and HypergeometricPFQRegularized functions. Nevertheless, it can be ...

2

i = 2; j = 2; FixedPoint[Plus[#, f1[i++]/f2[j++]] &, f1[1.]/f2[1]]

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You can find when $f_{2}(x) = 10^{-15}$, and then calculate the summation: f1[x_] := Exp[-x]; f2[x_] := 1/x; mybound = 10^-15; maxx = (x /. Solve[f2[x] == mybound, x])[[1]]; mysum = Sum[f1[x]/f2[x], {x, 0, maxx}] N[mysum, 10] (* 0.9206735942 *)

3

the NestWhile approach f1[x_] := Exp[-x]; f2[x_] := 1/x NestWhile[ {#[[1]] + f1[#[[2]]]/f2[#[[2]]] , #[[2]] + 1} & , {0, 1} , f2[#[[2]]] > 1/1000 & ] // First // N 0.920674 ( NSum is most certainly the better approach unless you have some peculiar functions )

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Have you tried using NSum? In your question, it seems to me that you try to sum numeric values (instead of analytically evaluating a sum) and I think NSum is better for this. Simple example: f1[x_] := x; f2[x_] := Exp[x]; NSum[f1[x]/f2[x], {x, 1, Infinity}] (* 0.920674 *)

1

SolveAlways can find coordinates of a polynomial with respect to any given basis. You set up an equation, setting the given polynomial equal to a linear combination of your basis polynomials. This approach will work generally with any polynomial that is a linear combination of a given set of (linearly independent) polynomials. poly = 1 + x + 3 x^2 + 7 ...

3

Posting this so the question doesn't remain unanswered. Here Michael Trott & Victor Adamchik) posted all the necessary material to solve a quintic in Mathematica. I'm not copying the whole thing because (perhaps) copyrights issues may arise. You need to use two notebooks: First this one to perform three transformations and reduce the general equation ...

3

Part1. The ordering of the roots and consequently which is the fourth root depends on when a is given its value. Table[Root[-7 a^4 #1^2 - 2 a^2 #1^4 + #1^6 &, n] // ToRadicals, {n, 6}] /. a -> 1. {0, 0, 0. + 1.35219 I, 0. - 1.35219 I, 1.95664, -1.95664} Table[Root[-7 a^4 #1^2 - 2 a^2 #1^4 + #1^6 &, n] /. a -> 1. // ToRadicals, {n, ...

3

Because the precision necessary to represent 18154980120849865. slightly exceeds, $MachinePrecision (* 15.9546 *) RiemannSiegelZ gives an inaccurate answer, RiemannSiegelZ[18154980120849865.] (* -0.563204 *) as compared with RiemannSiegelZ[SetPrecision[18154980120849865, 30] (* 1.22954136 *) and the same is true of the entire ListLinePlot shown in ... 5 Not an answer, but reporting a failed approach. I looked for asymptotic expansions of the zeta function for large arguments, thinking their evaluation might be faster. For instance, here The following took 100 s on my machine. Plot[ Re[(Exp[-((I (4 z^2 + Pi Sqrt[z^2]))/(8 z))] (z^2)^((I z)/4) Zeta[I z + 1/2])/(4^((I z)/4) Pi^((I z)/2))], {z, ... 3 Thanks to ConstantArray[], not much effort is needed: mysum[n_Integer, z_] := Sum[HypergeometricPFQ[ConstantArray[1, k], ConstantArray[2, k], z], {k, 1, n}] Test: mysum[3, z] (-1 + E^z)/z + HypergeometricPFQ[{1, 1, 1}, {2, 2, 2}, z] + (-EulerGamma - Gamma[0, -z] - Log[-z])/z 5 Erfc[-30. + 10^-1 I] used to return the result shown in the documentation through version 7.0.1. The implementation changed for version 8.0 and it started giving a machine precision answer (which is correct, more consistent and still demonstrates the same possible issue by being very close to 2). The (documentation) bug is that this example did not get ... 4 Here is an example that shows that a generic LaguerreL function is not automatically computed using the confluent hypergeometric function as suggested by MathWorld. First I'll define the rule that would amount to just such a replacement. Then I'll show that this replacement allows a particular integral of the LaguerreL function to be computed without any ... 3 I don't know what's so unacceptable about LaguerreL that anything would be acceptable in its place, but from functions.wolfram.com, we have the following transformation: simp = HoldPattern[LaguerreL[n_, λ_, z_]] :> (Gamma[λ + n + 1] HypergeometricPFQRegularized[{-n}, {λ + 1}, z])/ Gamma[n + 1]; We can apply it to the solution Assuming[U > 0 ... 6 From Wolfram MathWorld, EDIT : added definition using Hypergeometric1F1Regularized based on comment by @ilian and his reference LaguerreL[n, a, x] == Pochhammer[a + 1, n]/n! Hypergeometric1F1[-n, a + 1, x] == Gamma[a + n + 1]/Gamma[n + 1] Hypergeometric1F1Regularized[-n, a + 1, x] // FullSimplify True One of these definitions may be how ... 2 The reason for this error is that NIntegrate uses fixed precision when computing the integration ranges, while EllipticK needs to raise the precision internally to obtain a good result. N[EllipticK[7/10], 20] (* 2.0753631352924691439 *) Block[{$MinPrecision = $MaxPrecision = 20}, N[EllipticK[7/10], 20]] (* Divide::infy: Infinite expression ... 4 As Daniel Lichtblau showed in his comment, use exact numbers (or Rationalize) for input values alpha = 639/100; beta = 369/100; a1 = (1 - alpha)/2; a2 = (2 - alpha)/2; a3 = (1 - beta)/2; a4 = (2 - beta)/2; a5 = 1; b1 = 0; b2 = 1/2; SNR = 0; SNR0 = 10^(SNR/10); z = 2*(SNR0/(alpha*beta))^2; p1 = 2^(alpha + beta - 3)/(Pi*Sqrt[Pi]*Gamma[alpha]*Gamma[beta]); ... 1 It might be better just to ask SeriesCoefficient for the answer. From what I can tell it should always find the answer for functions of hypergeometric type. expr = Hypergeometric2F1[1/3 (2 - I Sqrt[2]), 1/3 (2 + I Sqrt[2]), 1/3, x] + Hypergeometric2F1[1/3 (4 - I Sqrt[2]), 1/3 (4 + I Sqrt[2]), 5/3, x]; Expand[Refine[SeriesCoefficient[expr, {x, 0, ... 5 The the function y is the integral of a logarithmic singularity, so it is relatively easy to understand. The function JacobiCN[s, 7/10] is relatively flat near s == EllipticK[7/10], so that the integral of its reciprocal can be approximated by a logarithm: y[s]/JacobiCN[s, 7/10] ~= y0 / (jp (s - EllipticK[7/10])) where y0 ~= y[s] at s == EllipticK[7/10] ... 3 As I've previously noted, when computing functions that involve ratios of gamma functions, it is manifestly better to re-express in terms of LogGamma[] and then do a final exponentiation afterwards. This deftly sidesteps the issue of huge intermediate values being used to compute a modestly-sized result. Plot3D[Exp[LogGamma[1 + (n + m)/2] - (LogGamma[1 + n] ... 4 Very recently, I learned a useful procedure due to J. P. Boyd (see also this and this) that involves expanding a function as a Chebyshev polynomial series, forming the so-called "colleague matrix", and then finding the eigenvalues of this matrix, which are often good approximations to the roots of the original function. I shall present how to do a barebones ... 13 Since EllipticTheta[] is a built-in function, and since the Eisenstein series$E_4(q)$and$E_6(q)$are expressible in terms of theta functions (I use the nome$q$as the argument in this answer, but you can convert to your convention by using the relation with the period ratio$\tau$:$q=\exp(2\pi i \tau)$), and since the higher-order Eisenstein series ... 5 As this is a special-functions question, I feel justified in using a bit of heavy artillery. Here goes nothing... In effect, what the OP seems to want to do is to evaluate $$\sum_{n=1}^\infty \frac{(q^{n+1};q)_\infty}{(q^n;q)_\infty} q^{n-1}$$ (where$(a;q)_n$is the$q\$-Pochhammer symbol) by approximating it with its partial sums. However, there is a ...

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