Evaluate using Mathematica or otherwise the value of $f'(0)$

Question

If we have$$f(a)=_3F_2\left(\frac{1}{2},\frac{a+1}{2},\frac{a+2}{2}; \frac{a+3}{2},\frac{a+3}{2};1\right)$$ where $a\geq 0$ and $_3F_2$ is the Hypergeometric function with unit argument.

Evaluate using Mathematica or otherwise the value of $f'(0)$ (the first derivative of $f(a)$ at $0$).

I am requesting a code with answer if possible in Wolfram Mathematica.

Sorry, I am new to Wolfram Mathematica.

Any help would be appreciated.

Answer 1

From definition of HypergeometricPFQ as Series representations we have:

$\underset{a\to 0}{\text{lim}}\frac{\partial }{\partial a}\, _3F_2\left(\frac{1}{2},\frac{1}{2}+\frac{a}{2},1+\frac{a}{2};\frac{3}{2}+\frac{a}{2},\frac{3}{2}+\frac{a}{2};1\right)=\\\underset{a\to 0}{\text{lim}}\frac{\partial }{\partial a}\left(\sum _{k=0}^{\infty } \frac{\left(\frac{1}{2}\right)_k \left(\frac{1}{2}+\frac{a}{2}\right)_k \left(1+\frac{a}{2}\right)_k}{k! \left(\left(\frac{3}{2}+\frac{a}{2}\right)_k\right){}^2}\right)=\\\sum _{k=0}^{\infty } \left(\underset{a\to 0}{\text{lim}}\frac{\partial }{\partial a}\frac{\left(\frac{1}{2}\right)_k \left(\frac{1}{2}+\frac{a}{2}\right)_k \left(1+\frac{a}{2}\right)_k}{k! \left(\left(\frac{3}{2}+\frac{a}{2}\right)_k\right){}^2}\right)=\\\sum _{k=0}^{\infty } \left(\underset{a\to 0}{\text{lim}}\frac{\partial }{\partial a}\frac{(1+a) \Gamma \left(\frac{3}{2}+\frac{a}{2}\right) \Gamma \left(\frac{1}{2}+k\right) \Gamma \left(1+\frac{a}{2}+k\right)}{(1+a+2 k) \sqrt{\pi } \Gamma \left(1+\frac{a}{2}\right) \Gamma (1+k) \Gamma \left(\frac{3}{2}+\frac{a}{2}+k\right)}\right)=\\\sum _{k=0}^{\infty } \left(\underset{a\to 0}{\text{lim}}\left(-\frac{(1+a) \Gamma \left(\frac{3}{2}+\frac{a}{2}\right) \Gamma \left(\frac{1}{2}+k\right) \Gamma \left(1+\frac{a}{2}+k\right)}{(1+a+2 k)^2 \sqrt{\pi } \Gamma \left(1+\frac{a}{2}\right) \Gamma (1+k) \Gamma \left(\frac{3}{2}+\frac{a}{2}+k\right)}+\frac{\Gamma \left(\frac{3}{2}+\frac{a}{2}\right) \Gamma \left(\frac{1}{2}+k\right) \Gamma \left(1+\frac{a}{2}+k\right)}{(1+a+2 k) \sqrt{\pi } \Gamma \left(1+\frac{a}{2}\right) \Gamma (1+k) \Gamma \left(\frac{3}{2}+\frac{a}{2}+k\right)}-\frac{(1+a) \Gamma \left(\frac{3}{2}+\frac{a}{2}\right) \Gamma \left(\frac{1}{2}+k\right) \Gamma \left(1+\frac{a}{2}+k\right) \psi ^{(0)}\left(1+\frac{a}{2}\right)}{2 (1+a+2 k) \sqrt{\pi } \Gamma \left(1+\frac{a}{2}\right) \Gamma (1+k) \Gamma \left(\frac{3}{2}+\frac{a}{2}+k\right)}+\frac{(1+a) \Gamma \left(\frac{3}{2}+\frac{a}{2}\right) \Gamma \left(\frac{1}{2}+k\right) \Gamma \left(1+\frac{a}{2}+k\right) \psi ^{(0)}\left(\frac{3}{2}+\frac{a}{2}\right)}{2 (1+a+2 k) \sqrt{\pi } \Gamma \left(1+\frac{a}{2}\right) \Gamma (1+k) \Gamma \left(\frac{3}{2}+\frac{a}{2}+k\right)}+\frac{(1+a) \Gamma \left(\frac{3}{2}+\frac{a}{2}\right) \Gamma \left(\frac{1}{2}+k\right) \Gamma \left(1+\frac{a}{2}+k\right) \psi ^{(0)}\left(1+\frac{a}{2}+k\right)}{2 (1+a+2 k) \sqrt{\pi } \Gamma \left(1+\frac{a}{2}\right) \Gamma (1+k) \Gamma \left(\frac{3}{2}+\frac{a}{2}+k\right)}-\frac{(1+a) \Gamma \left(\frac{3}{2}+\frac{a}{2}\right) \Gamma \left(\frac{1}{2}+k\right) \Gamma \left(1+\frac{a}{2}+k\right) \psi ^{(0)}\left(\frac{3}{2}+\frac{a}{2}+k\right)}{2 (1+a+2 k) \sqrt{\pi } \Gamma \left(1+\frac{a}{2}\right) \Gamma (1+k) \Gamma \left(\frac{3}{2}+\frac{a}{2}+k\right)}\right)\right)=\\\sum _{k=0}^{\infty } \left(-\frac{\Gamma \left(\frac{1}{2}+k\right)}{2 (1+2 k)^2 \Gamma \left(\frac{3}{2}+k\right)}+\frac{\Gamma \left(\frac{1}{2}+k\right)}{2 (1+2 k) \Gamma \left(\frac{3}{2}+k\right)}+\frac{\gamma \Gamma \left(\frac{1}{2}+k\right)}{4 (1+2 k) \Gamma \left(\frac{3}{2}+k\right)}+\frac{\Gamma \left(\frac{1}{2}+k\right) \psi ^{(0)}\left(\frac{3}{2}\right)}{4 (1+2 k) \Gamma \left(\frac{3}{2}+k\right)}+\frac{\Gamma \left(\frac{1}{2}+k\right) \psi ^{(0)}(1+k)}{4 (1+2 k) \Gamma \left(\frac{3}{2}+k\right)}-\frac{\Gamma \left(\frac{1}{2}+k\right) \psi ^{(0)}\left(\frac{3}{2}+k\right)}{4 (1+2 k) \Gamma \left(\frac{3}{2}+k\right)}\right)=\\\sum _{k=0}^{\infty } -\frac{-2+(-1-2 k) H_k+(1+2 k) H_{\frac{1}{2}+k}+2 k (-4+\ln (4))+\ln (4)}{2 (1+2 k)^3}=\\\int_0^1 \left(\sum _{k=0}^{\infty } -\frac{-2+\frac{(-1-2 k) \left(1-t^k\right)}{1-t}+\frac{(1+2 k) \left(1-t^{\frac{1}{2}+k}\right)}{1-t}+2 k (-4+\ln (4))+\ln (4)}{2 (1+2 k)^3}\right) \, dt=\int_0^1 \frac{1}{16} \left(-\frac{2 \Phi \left(t,2,\frac{1}{2}\right)}{1+\sqrt{t}}-\pi ^2 (-4+\ln (4))-14 \zeta (3)\right) \, dt=\\\frac{\pi ^2}{4}-\frac{1}{16} \pi ^2 \ln (4)-\frac{7 \zeta (3)}{8}-\int_0^1 \frac{\Phi \left(t,2,\frac{1}{2}\right)}{8 \left(1+\sqrt{t}\right)} \, dt=\\\frac{\pi ^2}{4}-\frac{1}{16} \pi ^2 \ln (4)-\frac{7 \zeta (3)}{8}-\sum _{k=0}^{\infty } \int_0^1 \frac{t^k}{\left(\frac{1}{2}+k\right)^2 \left(8 \left(1+\sqrt{t}\right)\right)} \, dt=\\\frac{\pi ^2}{4}-\frac{1}{16} \pi ^2 \ln (4)-\frac{7 \zeta (3)}{8}-\sum _{k=0}^{\infty } \frac{-H_k+H_{\frac{1}{2}+k}}{2 (1+2 k)^2}=\\\frac{\pi ^2}{4}-\frac{1}{4} \pi ^2 \ln (2)-\frac{1}{2} \sum _{k=0}^{\infty } \frac{H_{\frac{1}{2}+k}}{(1+2 k)^2}=\\\frac{\pi ^2}{4}-\frac{1}{4} \pi ^2 \ln (2)-\frac{1}{2} \sum _{j=1}^{\infty } \left(\sum _{k=0}^{\infty } \frac{\frac{1}{2}+k}{\left(j \left(\frac{1}{2}+j+k\right)\right) (1+2 k)^2}\right)=\\\frac{\pi ^2}{4}-\frac{1}{4} \pi ^2 \ln (2)-\frac{1}{2} \sum _{j=1}^{\infty } \left(\frac{\gamma }{4 j^2}+\frac{\ln (4)}{4 j^2}+\frac{\psi ^{(0)}\left(\frac{1}{2}+j\right)}{4 j^2}\right)=\\\frac{\pi ^2}{4}-\frac{1}{4} \pi ^2 \ln (2)-\frac{1}{48} \pi ^2 (\gamma +\ln (4))-\sum _{j=1}^{\infty } \frac{\psi ^{(0)}\left(\frac{1}{2}+j\right)}{4 j^2}=\\\frac{\pi ^2}{4}-\frac{1}{4} \pi ^2 \ln (2)-\frac{1}{48} \pi ^2 (\gamma +\ln (4))-\int_0^{\infty } \left(\sum _{j=1}^{\infty } \frac{\frac{e^{-t}}{t}-\frac{(1+t)^{-\frac{1}{2}-j}}{t}}{4 j^2}\right) \, dt=\\\frac{\pi ^2}{4}-\frac{1}{4} \pi ^2 \ln (2)-\frac{1}{48} \pi ^2 (\gamma +\ln (4))-\int_0^{\infty } \left(\frac{e^{-t} \pi ^2}{24 t}-\frac{\text{Li}_2\left(\frac{1}{1+t}\right)}{4 t \sqrt{1+t}}\right) \, dt=\\\frac{\pi ^2}{4}-\frac{1}{4} \pi ^2 \ln (2)-\frac{7 \zeta (3)}{16}$

Using definitions:

$H_x=\int_0^1 \frac{1-t^x}{1-t} \, dt$

$\Phi \left(t,2,\frac{1}{2}\right)=\sum _{k=0}^{\infty } \frac{t^k}{\left(\frac{1}{2}+k\right)^2}$

$H_k=\sum _{j=1}^{\infty } \frac{k}{j (j+k)}$

$\psi ^{(0)}\left(\frac{1}{2}+j\right)=\int_0^{\infty } \left(\frac{e^{-t}}{t}-\frac{(1+t)^{-\frac{1}{2}-j}}{t}\right) \, dt$

---

$$\underset{a\to 0}{\text{lim}}\frac{\partial }{\partial a}\, _3F_2\left(\frac{1}{2},\frac{a+1}{2},\frac{a+2}{2};\frac{a+3}{2},\frac{a+3}{2};1\right)=\color{red}{-\frac{7 \zeta (3)}{16}+\frac{\pi ^2}{4}-\frac{1}{4} \pi ^2 \ln (2)}$$

HoldForm[Limit[D[HypergeometricPFQ[{1/2, (a + 1)/2, (a + 2)/2}, {(a + 3)/2, (a + 3)/2}, 1], a], a -> 0] ==
\[Pi]^2/4 - 1/16 \[Pi]^2 Log[4] - (7 Zeta[3])/8 - 
Integrate[LerchPhi[t, 2, 1/2]/(8 (1 + Sqrt[t])), {t, 0, 1}] == \[Pi]^2/4 - 1/4 \[Pi]^2 Log[2] - 
1/2*Sum[HarmonicNumber[1/2 + k]/(1 + 2 k)^2, {k, 0, 
  Infinity}] == \[Pi]^2/4 - 1/4 \[Pi]^2 Log[2] - (7 Zeta[3])/16]

We can check:

 SOL1 = N[\[Pi]^2/4 - 1/4 \[Pi]^2 Log[2] - (7 Zeta[3])/16, 50]
 (* 0.23122908917573801863436751581925439504379858686425 *)
 Needs["NumericalCalculus`"];
 SOL2 = ND[f[a], a, 0, WorkingPrecision -> 50, Method -> NIntegrate]
 (* 0.23122908917573801863436751581925439504379858686430 - 2.*10^-50 I *)
 SOL1 - SOL2

 (*-5.*10^-50 + 2.*10^-50 I *)

Answer 2

Let's define

f[a_] := HypergeometricPFQ[{1/2, (a + 1)/2, (a + 2)/2}, {(a + 3)/2, (a + 3)/2}, 1]

HypergeometricPFQ[{1/2, (a + 1)/2, (a + 2)/2}, {(a + 3)/2, (a + 3)/2}, 1]//
TraditionalForm

HypergeometricPFQ[{a1,..., ap},{ b1,...,bq},z] is the generalized
hypergeometric function pFq(a;b;z). 

If we evaluate simply

Derivative[1][f][0] // TraditionalForm

we get a symbolic representation by introducing another special functions (derivatives of $_3 F_2$ with respect to its parameters $a_i$ and $b_j$) which can't be expanded into other well (better) known functions. Moreover the system seems to choke on numerical evaluation.

Table[ f[a], {a, Range[-2, 2]}]

 {1, 1, Pi^2/8, (4 (-2 + Pi))/Pi, 9/32 (-4 + Pi^2)}

Now we can estimate symbolically derivative of $f'(0)$ simply taking $\frac{f(1)-f(0)}{1}$ or $\frac{f(1)-f(-1)}{2}$

{(f[1] - f[0])/1, (f[1] - f[-1])/2} // Simplify
N @ %

{ 4 - 8/Pi - Pi^2/8, 3/2 - 4/Pi}
{ 0.21982, 0.22676}   

Comparing with the plot this appears to be a good approximation. Moreover evaluation at close points yields a similar, even more precise result (up to $10^{-5}$) without using external packages

(f'[0.00001] + f'[-0.00001])/2

 0.231229 

Plot[{f[a], (4 - 8/Pi - Pi^2/8) a + Pi^2/8}, {a, -5, 4}, 
  AxesOrigin -> {0, 0}, AspectRatio -> Automatic, PlotRange -> {0, 6.5}, 
  PlotStyle -> {Thick, Dashed}, PlotLegends -> Placed["Expressions", {0.8, 0.75}], 
  Epilog -> {Red, PointSize[0.02], Point[{0, Pi^2/8}]}]

Derivation of the exact result within Mathematica

Partial sums of geometric series can be calculated simply with

ForAll[ n, n ∈ Integers && n >= 2, 
       (1 - q^n)/(1 - q) == Sum[q^k, {k, 0, n - 1}]] // Resolve

True

This also yields with $n \to \infty$ the sum of geometric series as well as the starting point to the definition of the hypergeometric series and by the analytic continuation the hypergeometric function:

Sum[ Gamma[k + a] Gamma[k + b] Gamma[c]/(
       Gamma[a] Gamma[b] Gamma[k + c] k!) q^k, { k, 0,3}]

Hypergeometric2F1[a, b, c, q]

The above formula yields for $a=b=c=1$ the sum of geometric series, more precisely this should be a conditional expression. Analogously we can define $_3 F_2$ function.

The following yields a basic definition of the Euler gamma function

Integrate[ t^(x - 1) Exp[-t], {t, 0,  ∞}, Assumptions -> Re[x] > 0]

Gamma[x]

which by the analytic continuation provides a meromorphic function in the whole complex plane

{FunctionMeromorphic[#, x], FunctionDomain[#, x, Complexes]}&@Gamma[x]

{True, Re[x] > 0 || x \[NotElement] Integers}

We use also

FullSimplify[ Gamma[a1 + k]/Gamma[a1] == Pochhammer[a1, k]]

 True

Denoting by $d_0 = f'(0)$ we can write from the definition that wherever the series is absolutely convergent:

$$d_0=\lim_{a\to 0}\frac{d}{d a}\, _3F_2\left(\frac{1}{2},\frac{a+1}{2},\frac{a}{2}+1;\frac{a+3}{2},\frac{a+3}{2};1\right)=\sum _{k=0}^{\infty }\left(\underset{a\to 0}{\lim}\frac{d }{d a} \frac{\left(\frac{1}{2}\right)_k \left(\frac{1}{2}+\frac{a}{2}\right)_k \left(1+\frac{a}{2}\right)_k}{k! \left(\left(\frac{3}{2}+\frac{a}{2}\right)_k\right){}^2}\right)$$

which is the case here, e.g.

SumConvergence[ (Pochhammer[1/2, k] Pochhammer[(a + 1)/2, k]
                 Pochhammer[(a + 2)/2, k]/(Pochhammer[(a + 3)/2, k]^2 k!), k]

 True

The same holds also including the oprerator $\underset{a\to 0}{\lim}\frac{d}{d a}$ under the sum sign. Then differentiating with respect to $a$ we obtain

Assuming[k ∈ PositiveIntegers, 
  Limit[ D[ FunctionExpand[(
    Pochhammer[1/2, k] Pochhammer[(a + 1)/2, k] Pochhammer[(a + 2)/2, k])/(
    Pochhammer[(a + 3)/2, k]^2 k!)], a], a -> 0] // FullSimplify]

-((-2 + (-1 - 2 k) HarmonicNumber[k] + (1 + 2k) HarmonicNumber[1/2 + k]
    + 2 k (-4 + Log[4]) + Log[4])/(2 (1 + 2 k)^3))

The above limit yields $0$ for $k=0$. The only problematic term when summing from $k=1$ to $k=\infty$ is -HarmonicNumber[1/2 + k]/(2(1 + 2k)^2)

d0 = Sum[-((-2 + (-1 - 2 k) HarmonicNumber[k] + 2 k (-4 + Log[4])
         + Log[4])/(2 (1 + 2 k)^3)), {k, ∞}] 
   - Sum[ HarmonicNumber[1/2 + k]/(2 (1 + 2 k)^2), {k, ∞}]

 -1/8 ((-4 + Pi^2) (-2 + Log[4])) 
  - Sum[ HarmonicNumber[1/2 + k]/(2(1 + 2k)^2), {k, ∞}]

We denote:

 d01 = -1/8 ((-4 + Pi^2) (-2 + Log[4]));
 d02 = - Sum[ HarmonicNumber[1/2 + k]/(2(1 + 2k)^2), {k, ∞}];

Using the following identity

Sum[-((k + 1/2)/(2 m (2 k + 1)^2 (m + k + 1/2))), {m, ∞}]// FullSimplify

-HarmonicNumber[1/2 + k]/(2 (1 + 2 k)^2)

we can recast the hard term by summing it first with respect to $k$

Sum[-((k + 1/2)/(2 m (2 k + 1)^2 (m + k + 1/2))), {k, ∞}]//FunctionExpand

-((-2 + EulerGamma + Log[4] + PolyGamma[0, 3/2 + m])/(8 m^2))

Next we exploit two identities

-PolyGamma[0, 3/2 + m]/(8 m^2) == -(1/(
   8 m^2)) (PolyGamma[0, 1/2 + m] + 1/(1/2 + m)) // FullSimplify

 True

and

Integrate[ Exp[-t]/t - 1/((t + 1)^(m + 1/2) t), {t, 0, ∞},
  Assumptions -> m > 0]

PolyGamma[0, 1/2 + m]

Summing up

d02 = 1 - Log[2] - 1/48 Pi^2 (EulerGamma + Log[4]) + 
      Integrate[-Sum[ Exp[-t]/(8 m^2 t) - 1/(8 m^2 (t + 1)^(m + 1/2) t), 
                     {m, ∞}], {t, 0, ∞}] // FullSimplify

1 - Log[2] - (7 Zeta[3])/16

In mathematical notation we have $$ d_{02}= -\sum_{k=1}^{\infty}\frac{H_{k+\frac{1}{2}}}{2(2k+1)^2}=-\sum_{m=1}^{\infty}\frac{\gamma-2+\ln(4)+\psi^{(0)}(\frac{3}{2}+m)}{8m^2}$$ and $$-\sum_{m=1}^{\infty}\frac{\psi^{(0)}(\frac{3}{2}+m)}{8m^2}=-\sum_{m=1}^{\infty}\frac{1}{4m^2(2m+1)}-\sum_{m=1}^{\infty}\frac{\psi^{(0)}(\frac{1}{2}+m)}{8m^2}=\\=1-\ln(2)-\frac{\pi^2}{48}(\gamma+\ln(4))-\int_{0}^{\infty}\left( \sum_{m=1}^{\infty}\frac{\exp(-t)}{8m^2 t}-\sum_{m=1}^{\infty}\frac{1}{8m^2 (t+1)^{\frac{1}{2}+m} t}\right)=\\=1-\ln(2)-\frac{\pi^2}{48}(\gamma+\ln(4))-\int_{0}^{\infty}\left(\frac{\pi^2 \exp(-t)}{48t}-\frac{6\operatorname{Li}_{2}{\frac{1}{t+1}}}{48\sqrt{t+1}\; t} \right) $$ and finally

d0 = d01 + d02 // FullSimplify
N @ %

 1/16 ( -4 Pi^2 (-1 + Log[2]) - 7 Zeta[3])
 0.231229

$$d_0=d_{01}+d_{02}= \frac{4\pi^2\left(1-\ln(2)\right)- 7\zeta(3)}{16}$$

Answer 3

$Version

(* "13.2.1 for Mac OS X ARM (64-bit) (January 27, 2023)" *)

Clear["Global`*"]

f[a_] := HypergeometricPFQ[{1/2, (a + 1)/2, (a + 2)/2}, 
  {(a + 3)/2, (a + 3)/2}, 1]

This can be evaluated exactly for integers, a > -3

pts1 = {#, f[#]} & /@ Range[-2, 5] // Simplify

(* {{-2, 1}, {-1, 1}, {0, π^2/8}, {1, 4 - 8/π}, {2, 
  9/32 (-4 + π^2)}, {3, (64 (-7 + 3 π))/(27 π)}, {4, 
  75/512 (-16 + 3 π^2)}, {5, (64 (-149 + 60 π))/(375 π)}} *)

Or numerically for reals

SeedRandom[1234];

val = RandomReal[{-5, 5}, 10];

pts2 = {#, f[#]} & /@ val;

Plot[f[a], {a, -5, 5},
 Epilog -> {AbsolutePointSize[5],
   Red, Point[pts1],
   Purple, Point[pts2]}]

EDIT: For the derivative, f'[a]

Needs["NumericalCalculus`"]

ND[f[a], a, 0, WorkingPrecision -> 15] // N

(* 0.231229 *)

ListLinePlot[Table[{a0, ND[f[a], a, a0, WorkingPrecision -> 15]},
  {a0, 0, 5, 1/4}]]

Answer 4

Mathematica evaluates a closed form as a ConditionalExpression

Limit[(f[a] - f[0])/a, a -> 0]
  

which is probably not useful because it contains unevaluated expressions of HypergeometricPFQ !