# Plotting partial sums of Fourier sine series

How do I plot this on Mathematica version 5.2?

$\frac{4}{\pi} \sin{x} + \frac{4}{3 \pi} \sin{3 x} + \cdots + \frac{4}{(2 N+1) \pi} \sin{(2 N+1) x}$

over $x \in [-\pi,\pi]$ for $N= 3, 6, 12, 24$.

I tried and got this error: This seems to work fine:

f[x_, n_] := (4/Pi) Sum[Sin[(2 k + 1) x]/(2 k + 1), {k, 0, n}];
Plot[Table[f[x, n], {n, 5}], {x, -Pi, Pi}] • May be it does not work in version 5.2 of mathematica Oct 10, 2013 at 2:47
• How to get get single curve for n=5 only? Oct 10, 2013 at 3:08
• Plot[f[x, 5], {x, -Pi, Pi}] gives just the n=5 curve. Oct 10, 2013 at 3:34
• This one works for version 5.2 s[n_, x_] := 4/( π) Sum[Sin[(2 k + 1) x]/(2 k + 1), {k, 0, n}] partialsums = Table[s[5, x], {n, 6}]; Plot[Evaluate[partialsums], {x, -π, π}] Oct 10, 2013 at 3:59

This is a documented behavior of Plot in or before v5.2. Just click the More... button in the Plot::plnr warning, you'll see the following explanation:

If Plot is used to plot a list of functions, that list should appear explicitly as the first argument in Plot or should be introduced using Evaluate or other means.

together with a typical example: So one solution is, as you've noticed, using Evaluate:

f[x_, n_] := (4/Pi) Sum[Sin[(2 k + 1) x]/(2 k + 1), {k, 0, n}];
Plot[Table[f[x, n], {n, 5}] // Evaluate, {x, -Pi, Pi}]


It's not the only solution, of course. A few other solutions:

With[{table = Table[f[x, n], {n, 5}]}, Plot[table, {x, -Pi, Pi}]]
Plot[#, {x, -Pi, Pi}] &@Table[f[x, n], {n, 5}]
Plot @@ {Table[f[x, n], {n, 5}], {x, -Pi, Pi}}