Anonymous
Anonymous asked in Science & MathematicsMathematics · 7 months ago

calculus II - series comparison test true or false questions - help pls?

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  • 7 months ago
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    (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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  • ted s
    Lv 7
    7 months ago

    T ; T ; T...................

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