# How to make a discontinuous function a continuous function?

I want to make function a[t] a continuous which has indeterminate form at some points.

tT = 0.0042643923240938165;

y[t_] := (1625.0688606154426*Cos[1473.4069545336129*t]*Cos[0.9403762801519631*Tanh[2.4*Sin[1.2636183784438946 + 1473.4069545336129*t]]])/
Sqrt[1 - 0.48999999999999994*Sin[1473.4069545336129*t]^2];

z[t_] := -((1625.0688606154426*Cos[1473.4069545336129*t]*Sin[0.9403762801519631*Tanh[2.4*Sin[1.2636183784438946 + 1473.4069545336129*t]]])/
Sqrt[1 - 0.48999999999999994*Sin[1473.4069545336129*t]^2]);

NSolve[{y[t] == 0, z[t] == 0}, t]

a[t_] = If[y[t] == 0 && z[t] == 0, 0, Sin[2*ArcTan[y[t], z[t]]]];

Plot[a[t], {t, 0, tT}, ImageSize -> Large]


At some point t=0.0010661 both y[t] and z[t] becomes zero. So i gave If condition when ever y[t] and z[t] both are zero take the value to be zero. It is not working.
Any suggestion how to make this function continuous. Thank you.

• I cannot reproduce your discontinuity in version 10.0 for linux. Which version are you using? – mattiav27 Nov 29 '19 at 9:27
• @mattiav27 Probably because of the improvement of singularity detection in v11. Adding Exclusions->None helps. Related: mathematica.stackexchange.com/a/143728/1871 – xzczd Nov 29 '19 at 9:48

Try this one

p = 0.9403762801519631;
q = 2.4;
r = 1.2636183784438946;
s = 1473.4069545336129;
Clear[f]
f[t_] := -Sin[2 p Tanh[q Sin[r + s t]]]


instead of a[t].

• Thanks a lot. now i'm trying to use this /. ArcTan[Tan[x_]] -> x at the end of my equation. Not sure whether it will cause any error. Though to tell you. – Gummala Navneeth Nov 29 '19 at 12:06