I have an example from Stewart's Calculus where the equation $y'' + y = 0$ is solved using power series. The equation

Subscript[c, 2 + n] == -(Subscript[c, n]/((n + 1) (n + 2)))

is used to determine the coefficients recursively and is straightforward to solve "by hand", but I cannot see how to do this in Mathematica. Searching for "Recursion Relation" in the Wolfram Language & System Documentation Center returns a tutorial Functions That Remember Values They Have Found, which uses the example of a Fibonacci function

f[x_] := f[x] = f[x - 1] + f[x - 2] 

with conditions

f[0] = f[1] = 1

I have followed this example to input a function to calculate the coefficients recursively, but whatever I try I get a $RecursionLimit error. I would be grateful if someone could explain how to enter such a function since I am clearly misunderstanding the basic concept behind this in some way.

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    – Michael E2
    Commented Jun 20, 2019 at 19:52
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    – Michael E2
    Commented Jun 20, 2019 at 19:52
  • $\begingroup$ Related: mathematica.stackexchange.com/questions/25363/… $\endgroup$
    – Michael E2
    Commented Jun 20, 2019 at 19:55

2 Answers 2


One way is to use the provided function for solving recurrences. The following example code

sol = c[n] /. RSolve[ {c[n + 2] == 
   -c[n]/((n + 1) (n + 2)),
   c[0] == c0, c[1] == c1}, c[n], n][[1, 1]];
sol /. n -> # & /@ Range[0, 5] // InputForm

returns the result

{c0, c1, -c0/2, -c1/6, c0/24, c1/120}

as expected. Another way is to define a recursive function. The example code

c[0] = c0; c[1] = c1;
c[m_] := c[m] = With[{n = m - 2},
   -c[n]/((n + 1) (n + 2))];
c /@ Range[0, 5] // InputForm

returns the same result as the previous example. Notice the memoization c[m_]:=c[m]= as well as the ClearAll[c].

  • $\begingroup$ Both of your solutions give the answer I was looking for. $\endgroup$
    – BenNevis
    Commented Jun 20, 2019 at 18:56
  • $\begingroup$ @BenNevis If you like my answer you can accept it. $\endgroup$
    – Somos
    Commented Jun 20, 2019 at 18:58

y'' + y = 0 is solved using power series

Is this what you are looking for?

ode = y''[x] + y[x] == 0;
sol = AsymptoticDSolveValue[ode, y[x], {x, 0, 5}];
sol /. {C[1] -> y[0], C[2] -> y'[0]}

Mathematica graphics

  • $\begingroup$ @BenNevis This is a new function. I looked docs now, and it said it was added in 11.3 $\endgroup$
    – Nasser
    Commented Jun 20, 2019 at 17:40
  • $\begingroup$ @BenNevis You need to have your ODE in there. I'll add it now. Mathematica will not show you the steps of how it obtained the series solution like the text book does. For this you need step-by-step solver. I do not know any one which will show all the steps like in the textbook. I do not think Wolfram Alpha pro even shows these steps but you could try it. $\endgroup$
    – Nasser
    Commented Jun 20, 2019 at 17:41

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