# Why is this Do-Loop taking so much time?

I have this simple Do Loop that, for some reason, takes a really long time for $$h<\frac{1}{15}$$. Why is this so? Is there a more efficient way in programming such a recursion?

h = 1/20;
v1[0] = 1;
v2[0] = 0;
Do[v1[j + h] = h v2[j] + v1[j];
v2[j + h] = h (v2[j] + 1/v1[j]) + v2[j], {j, 0, 2, h}]

• Yes, there is a simple solution. Replace exact numbers with approximate numbers, and it will be significantly faster. This means: replace v1[0] = 1; v2[0] = 0; with v1[0] = 1.; v2[0] = 0.; Commented Aug 4, 2022 at 12:37
• To elaborate on @Domen's comment, your code uses exact numbers, and with say h = 1/15, your v2[2] is a rational number where numerator and denominator have millions of digits. Commented Aug 4, 2022 at 12:41
• Are you trying to solve a differential equation? If yes, then a built-in method will be much more effective: NDSolve[{v1'[j] == v2[j], v2'[j] == v2[j] + 1/v1[j], v1[0] == 1, v2[0] == 0}, {v1, v2}, {j, 0, 2}] or even just NDSolve[{v1''[j] == v1'[j] + 1/v1[j], v1[0] == 1, v1'[0] == 0}, v1, {j, 0, 2}] Commented Aug 4, 2022 at 13:17

As @Domen says, use machine-precision numbers instead of exact rationals. Apart from this, I'd comment that there is no need to pre-compute the numbers in a Do loop: it is more Mathematica-style to define a memoizing recursion and then access the functions $$v_1(j)$$ and $$v_2(j)$$ randomly (without having to be explicit about the order of their evaluation):

h = 1/20;
Clear[v1, v2];
v1[0] = 1.;
v2[0] = 0.;
v1[j_ /; Divisible[j, h] && j > 0] := v1[j] = h v2[j - h] + v1[j - h];
v2[j_ /; Divisible[j, h] && j > 0] := v2[j] = h (v2[j - h] + 1/v1[j - h]) + v2[j - h];

ListPlot[Transpose@Table[{{j, v1[j]}, {j, v2[j]}}, {j, 0, 2, h}], PlotLegends -> {"v1", "v2"}]


The NDSolve curves in the above plot give the limit $$h\to0^+$$ and were calculated with

NDSolve[{v1''[j] == v1'[j] + 1/v1[j], v1[0] == 1, v1'[0] == 0}, v1, {j, 0, 2}]


Random access without explicit in-order precomputation:

v1[3]
(*    12.2335    *)

v2[7]
(*    629.188    *)