I have the following generating functions:

$l_{2i-1}=l_{1}-(i-2)(w+s)$ with $i\geq 2$ and $l_{2i}=l_{2}-(i-1)(w+s)$ with $i\geq 2$, so the first one is for odd index and the second for even index. Then I need to use the two for a function, let´s say a sum, So i'm going to need $l_{1},l_2,l_3,...$ etc but these are generated from the two different functions, so how can I call the $l_{odd}$ using the first function and $l_{even}$ using the second function in the same sum ?

I tried simply defining my first function like f[2*i_-1] but it doesn´t work :(

Many thanks,


One way is to use OddQ and EvenQ. For example:

m[x_] := x^2 /; OddQ[x]
m[x_] := 5 x^3 /; EvenQ[x]

defines m differently for odd and even arguments. For your f, maybe something like

f[x_] := l1 - (x - 2) (w + s) /; OddQ[x]
f[x_] := l2 - (x - 1) (w + s) /; EvenQ[x]
  • 1
    $\begingroup$ g[x_?OddQ] := l1 - (x - 2) (w + s);g[x_?EvenQ] := l2 - (x - 1) (w + s); is also an alternative and maybe even a bit faster. $\endgroup$ – Henrik Schumacher Sep 27 '18 at 22:08
  • $\begingroup$ To actually match the mathematical definition given by the OP, you need to do g[x_?OddQ] := l1 - Quotient[x - 3, 2] (w + s) and g[x_?EvenQ] := l2 - Quotient[x - 2, 2] (w + s). $\endgroup$ – J. M.'s ennui Sep 28 '18 at 0:12
  • $\begingroup$ But is any way to add the condition of Odd or Even to the subindex ? Because the subindex "i" in the two functions take Odd and Even values starting from 2, but because of the definition (2i-1) generate the Even indexes and (2i) the odd indexes, $\endgroup$ – Felipe Gonzalez Sep 28 '18 at 19:47

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