# Calculating the n-th term of the series expansion of a special function [closed]

I am trying to calculate the $n^{\text{th}}$ term of the following polynomial:

$$\, _2F_1\left(-n,n+3;\frac{3}{2};x\right)$$

To do this I calculate:

c[k_] = SeriesCoefficient[
Hypergeometric2F1[-n, n+3, 3/2, x], {x, 0, k},
Assumptions -> k >= 0
];


and get:

c[k] //TeXForm


$\frac{\sqrt{\pi } (k-n-1)! (k+n+2)!}{2 k! \left(k+\frac{1}{2}\right)! (-n-1)! (n+2)!}$

and the problem is when I try to calculate for $n=10$:

Block[{n = 10}, Sum[c[k] x^k, {k, 0, 10}]]


Infinity::indet: Indeterminate expression (0 2 Sqrt[π] ComplexInfinity)/Sqrt[π] encountered.

Infinity::indet: Indeterminate expression (0 4 Sqrt[π] ComplexInfinity)/(3 Sqrt[π]) encountered.

Infinity::indet: Indeterminate expression (0 8 Sqrt[π] ComplexInfinity)/(15 Sqrt[π]) encountered.

General::stop: Further output of Infinity::indet will be suppressed during this calculation.

Indeterminate

I get error messages and an incorrect answer. The correct result is:

Hypergeometric2F1[-n, n+3, 3/2, x] /. n->10


1/33 (33 - 2860 x + 72072 x^2 - 823680 x^3 + 5125120 x^4 - 19009536 x^5 + 43868160 x^6 - 63504384 x^7 + 56033280 x^8 - 27525120 x^9 + 5767168 x^10)

Other manifestations of problems with c:

Block[{n=10}, c[k]]
Block[{n=10}, c[5]]


0

Infinity::indet: Indeterminate expression -((128 0 64 316234143225 Sqrt[π] Sqrt[π] ComplexInfinity)/(135135 10395 4096 Sqrt[π] Sqrt[π])) encountered.

Indeterminate

c does not give a useful symbolic result for the $k^{\text{th}}$ term of the series.

I try to use assumptions but it does not help.

• It will be impossible to determine why you are getting an error without code. Aug 30, 2018 at 12:30
• please how to paste the code i copy from mathematica +crtl K but do not work Aug 30, 2018 at 12:47
• The code corresponding to the question gives me the correct answer with no error messages. I suggest saving your notebook (without any extraneous material), restarting Mathematica, and then executing the notebook. If that does not work, check your code for errors. Aug 30, 2018 at 13:49
• Do the Sum first (with n left undefined, then substitute n->10 in the result. This will avoid the indeterminate forms. Aug 30, 2018 at 17:04
• If the comment by Daniel ( %/.n->10 ) doesn't solve your problem, when using a fresh kernel as suggested by bbgodfrey, then it's not clear what you are asking. Please edit your question to clarify. Aug 31, 2018 at 7:51

It seems that the output generated by SeriesCoefficient is difficult for Mathematica to simplify into a version that can evaluate properly for integer n. So, I recommend using the new symbolic order derivatives introduced in M11.1:

c[k_] = Assuming[
n>=1,
Simplify @ D[Hypergeometric2F1[-n, n+3, 3/2, x], {x, k}]/k! /. x->0
]


(Pochhammer[-n, k] Pochhammer[3 + n, k])/(k! Pochhammer[3/2, k])

Note that this version of the symbolic coefficient of the series evaluates correctly for explicit values of n and k:

Block[{n=10}, c[5]]


-6336512/11

Let's check:

r1 = Block[{n=10}, Sum[c[k] x^k, {k, 0, n}]];
r1 // TeXForm


$\frac{524288 x^{10}}{3}-\frac{9175040 x^9}{11}+\frac{18677760 x^8}{11}-\frac{21168128 x^7}{11}+\frac{14622720 x^6}{11}-\frac{6336512 x^5}{11}+\frac{465920 x^4}{3}-24960 x^3+2184 x^2-\frac{260 x}{3}+1$

r2 = Block[{n=10}, Expand @ Hypergeometric2F1[-n,n+3,3/2,x]];
r2 // TeXForm


$\frac{524288 x^{10}}{3}-\frac{9175040 x^9}{11}+\frac{18677760 x^8}{11}-\frac{21168128 x^7}{11}+\frac{14622720 x^6}{11}-\frac{6336512 x^5}{11}+\frac{465920 x^4}{3}-24960 x^3+2184 x^2-\frac{260 x}{3}+1$

They are the same:

r1 === r2


True

The OP asks in a comment about a different hypergeometric function argument:

c[k_]=Assuming[
n>=1,
Simplify @ D[Hypergeometric2F1[3/2+n, -(3/2)-n, 3/2, x], {x, k}]/k! /. x->0
];

r1 = Block[{n=10}, Sum[c[k] x^k, {k, 0, n}]];
r1 //TeXForm


$\frac{515830463005 x^{10}}{262144}-\frac{264205846905 x^9}{65536}+\frac{165491574435 x^8}{32768}-\frac{8448518815 x^7}{2048}+\frac{2304141495 x^6}{1024}-\frac{2304141495 x^5}{2816}+\frac{24775715 x^4}{128}-\frac{452295 x^3}{16}+\frac{18515 x^2}{8}-\frac{529 x}{6}+1$

r2 = Block[{n=10}, Hypergeometric2F1[3/2+n, -(3/2)-n, 3/2, x] //Expand];
r2 //TeXForm


$-\frac{524288}{3} \sqrt{1-x} x^{11}+\frac{33292288}{33} \sqrt{1-x} x^{10}-\frac{27852800}{11} \sqrt{1-x} x^9+\frac{39845888}{11} \sqrt{1-x} x^8-\frac{35790848}{11} \sqrt{1-x} x^7+\frac{20959232}{11} \sqrt{1-x} x^6-\frac{24134656}{33} \sqrt{1-x} x^5+\frac{540800}{3} \sqrt{1-x} x^4-27144 \sqrt{1-x} x^3+\frac{6812}{3} \sqrt{1-x} x^2-\frac{263}{3} \sqrt{1-x} x+\sqrt{1-x}$

The difference between them is that r1 is a series approximation of r2. When r2 is not a degree 10 polynomial, than the two expressions will not be the same. Instead compare r1 with the series approximation of r2:

r2 + O[x]^11 //TeXForm


$1-\frac{529 x}{6}+\frac{18515 x^2}{8}-\frac{452295 x^3}{16}+\frac{24775715 x^4}{128}-\frac{2304141495 x^5}{2816}+\frac{2304141495 x^6}{1024}-\frac{8448518815 x^7}{2048}+\frac{165491574435 x^8}{32768}-\frac{264205846905 x^9}{65536}+\frac{515830463005 x^{10}}{262144}+O\left(x^{11}\right)$

• Thanks Carlk for your help it is works when n is neative interger but i try with n=3/2-n and do not work for example c[k_] = Assuming[n >= 1, Simplify@D[Hypergeometric2F1[3/2 + n, -(3/2) - n, 3/2, x], {x, k}]/ k! /. x -> 0] r1 = Block[{n = 10}, Sum[c[k] x^k, {k, 0, n}]]; r1 r2 = Block[{n = 10}, Hypergeometric2F1[3/2 + n, -(3/2) - n, 3/2, x] // Expand]; r2 gives different result check it if you will ,(sorry i could paste de code cause i paste from mathematica +Crtl K but do not work for me thanks anyway Aug 31, 2018 at 9:59
• Thanks Carlk now it is very clear Aug 31, 2018 at 13:42

Instead = write := at function c[k_]

$Version (* "11.3.0 for Microsoft Windows (64-bit) (March 7, 2018)"*) ClearAll["Global*"]; Remove["Global*"];(* Clears the kernel *) c[k_] := SeriesCoefficient[Hypergeometric2F1[-n, n + 3, 3/2, x], {x, 0, k}] Block[{n = 10}, Sum[c[k] x^k, {k, 0, 10}]] `$\frac{524288 x^{10}}{3}-\frac{9175040 x^9}{11}+\frac{18677760 x^8}{11}-\frac{21168128 x^7}{11}+\frac{14622720 x^6}{11}-\frac{6336512 x^5}{11}+\frac{465920 x^4}{3}-24960 x^3+2184 x^2-\frac{260 x}{3}+1\$