Question

The FCS `tan_(fcs)(pix: pi) = tan (pix+ pi (tan (pix + pi tan (pix +...)`. How do you graph this FCS to reveal x-intercepts `0, +-1, +_2, +3|-...`?

Answer 1

See graph and explanation.

Explanation:

As #y = tan_(fcs)( pix; pi ) = tan ( pi ( x + y ) ), inversely,

`pi ( x + y ) = kpi + tan ^(-1) y, k = 0, +-1, +-2, +-3, ...`.

So, upon setting `y = 0, x = 0, +-1, +-2, +-3, ...`

See graph for these x-intercepts.

graph{y - tan ( pi ( x + y )) = 0[-10 10 -2 2] }