Cont only at a point
f(z) = |z| if |z| irrational and -|z| if |z| is irrational. (Rotate usual function.)
Note that if z_0 =/= 0 then lim_{z->z_0}f(z) DNE. Given \epsilon > 0 consider |f(z)-f(z_0)| for z\in B(\epsilon,z_0). If z_0 irrat then z->z_0 along rational approximations gives |f(z)-f(z_0)|=2|z_0|+\epsilon -/->0 as z->z_0. So f is not cts at z_0.
As z->0 |f(z)-f(0)|=|z_0|->0. So cts at 0.
Diff only at a pt. Example in lectures: |x| I think.
Cts *and* diffable only at a pt? f(z)=|z|^1/2 (|z| rational), -|z|^1/2 (|z| irrational). Cts at zero. Not cts away from zero. Let z_0 have |z_0| rational.
|f(z)-f(z_0)/z-z_0|=| +/- |z_0+\epsilon|^1/2- |z_0|^1/2 | / |epsilon|
= (same top line)/(|(z_0+\epsilon)|^1/2 - |z_0|^1/2)(|(z_0+\epsilon)|^1/2 + |z_0|^1/2
Nope, this is not the case! more work necessary...
lim_z_0->0 |\epsilon|^1/2 / |epsilon| which ->0 as \epsilon -> 0.
Sunday, March 18, 2007
Subscribe to:
Posts (Atom)