Best Answer:
(1) True
(2) True
(3) True
Additional discussion on the " Monster of Math"
(Weierstrass), after reading the message.
The series
cos(xpi3)/2+cos(xpi3^2)/2^2+
cos(xpi3^3)/2^3+...+T(n)+...
where T(n)=cos(xpi3^n)/2^n
The ratio test:
|cos[xpi3^(n+1)/2^(n+1)]*
[2^n/cos(xpi3^n)]
=
|cos[3xpi(3^n)]/[2cos(xpi3^n)|
=
|[4cos^3(xpi3^n)-3cos(xpi3^n)]/
cos(xpi3^n)|/2
=
|4cos^2(xpi3^n)-3|/2
<=
|4-3|/2=1/2<1 (independent of x).
=> the series converges absolutely.
In fact, for n=1 to infinity,
SIGMA[|cos(xpi3^n)/2^n|]<=
SIGMA[1/2^n]. By the comparison test,
the latter is a G.P. with the ratio=1/2 is
convergent, =>
the series is uniformly convergent with x
over D(-inf., inf.), as n->infinity.
Thus, we can write
..........inf.
f(x)=SIGMA[cos(xpi3^n)/2^n]
.........n=1
But, f(x), strictly speaking is a function of x
but it is not an injection & contains the
"many to one" types of corresponding. For
any x1& x2 in D, if
cos[(x1)pi3^n]=cos[(x2)pi3^n]
=>
cos[(x1)pi3^n]-cos[(x2)pi3^n]=0
=>
-2sin[pi3^n(x1+x2)/2]*
sin[pi3^n(x1-x2)/2]=0
=>
sin[pi3^n(x1+x2)/2]=0 or
sin[pi3^n(x1-x2)/2]=0
=>
x1=2k/3^n-x2 or
x1=2k/3^n+x2
for some k & n, but not exactly x1=x2.
The graph of it for n=20, is very complicated
like the "sound wave" but is continuous over
there. It may expect that f(x) is continuous
over D too.
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