f(x)=cos(3pix-(pi/2))
Answers
This function is continuous on all R, it is simply a transformation of cos(x) to cos(ax+b), where a=3pi and b=-pi/2. cos(x) is continuous on all R, so it is it's transformed version. If you want the proof, for each t in R, lim (x->t-) f(x) = lim (x->t+) f(x) = cos(3pi*t-pi/2)
30 January 2012
the proof is similar to the proof that cos(x) is continuous everywhere.
30 January 2012
just put the usual ε
30 January 2012
cos(t+δ)-cos(t)
30 January 2012
cos(t+δ)=cos(t)cos(δ)-sin(t)sin(δ)
30 January 2012
cos(t+δ)-cos(t)=cos(t)(cos(δ)-1)-sin(t)sin(δ)
30 January 2012
now, if δ is small, so is sin(δ)
30 January 2012
and for δ>0 and δ small, cos(δ)=(1-sin(δ)^2)^(1/2)
30 January 2012
just put α=sin(δ)
30 January 2012
|cos(t+δ)-cos(t)|
30 January 2012
(1-cos(δ))+sin(δ)=1-(1-α^2)^(1/2)+α
30 January 2012
1-(1-α^2)^(1/2)+α
30 January 2012
(1+α-ε )
30 January 2012
etc...
30 January 2012
oops sorry, let's take it back from |cos(t+δ)-cos(t)|=|cos(t)(cos(δ)-1)-sin(t)sin(δ)|
30 January 2012
|cos(t)|
30 January 2012
now |cos(δ)-1|=|1-cos(δ)|
30 January 2012
|cos(δ)-1|+|sin(δ)|
30 January 2012
|cos(δ)-1|=|sin(δ)|^2 / |1+cos(δ)|
30 January 2012
now, |δ| is very small, so cos(δ)>0, therefore |sin(δ)|^2 / |1+cos(δ)|
30 January 2012
and since |sin(δ)|
30 January 2012
therefore |cos(δ)-1|+|sin(δ)|
30 January 2012
|sin(δ)|
30 January 2012
|δ|
30 January 2012
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