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    Calculus Challenge, Spring 2005 Solutions: LN (LN N) LN (LN N) (LN (LN N) )

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    Taghi Khaje
    1. The integral of (ln n)ln(ln n) from 1 to infinity diverges, similar to the integral of 1/n^2, which also diverges. 2. The absolute value of a function f(b) - f(a) is bounded above by the integral of the absolute value of the derivative of f from a to b. This can be used to bound the absolute value of f(b). 3. By applying the intermediate value theorem to the function g(x) = f(x) - x, there exists a value c where g(c) = 0, implying f(c) = c. 4. Making a substitution of x

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    Calculus Challenge, Spring 2005 Solutions: LN (LN N) LN (LN N) (LN (LN N) )

    Uploaded by

    Taghi Khaje
    17 views2 pages
    1. The integral of (ln n)ln(ln n) from 1 to infinity diverges, similar to the integral of 1/n^2, which also diverges. 2. The absolute value of a function f(b) - f(a) is bounded above by the integral of the absolute value of the derivative of f from a to b. This can be used to bound the absolute value of f(b). 3. By applying the intermediate value theorem to the function g(x) = f(x) - x, there exists a value c where g(c) = 0, implying f(c) = c. 4. Making a substitution of x

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    1. The integral of (ln n)ln(ln n) from 1 to infinity diverges, similar to the integral of 1/n^2, which also diverges. 2. The absolute value of a function f(b) - f(a) is bounded above by the integral of the absolute value of the derivative of f from a to b. This can be used to bound the absolute value of f(b). 3. By applying the intermediate value theorem to the function g(x) = f(x) - x, there exists a value c where g(c) = 0, implying f(c) = c. 4. Making a substitution of x

    Copyright:

    © All Rights Reserved
    Download as PDF, TXT or read online from Scribd
    Download as pdf or txt
    17 views2 pages

    Calculus Challenge, Spring 2005 Solutions: LN (LN N) LN (LN N) (LN (LN N) )

    Uploaded by

    Taghi Khaje
    1. The integral of (ln n)ln(ln n) from 1 to infinity diverges, similar to the integral of 1/n^2, which also diverges. 2. The absolute value of a function f(b) - f(a) is bounded above by the integral of the absolute value of the derivative of f from a to b. This can be used to bound the absolute value of f(b). 3. By applying the intermediate value theorem to the function g(x) = f(x) - x, there exists a value c where g(c) = 0, implying f(c) = c. 4. Making a substitution of x

    Copyright:

    © All Rights Reserved
    Download as PDF, TXT or read online from Scribd
    Download as pdf or txt
    of 2

    Calculus Challenge, Spring 2005

    Solutions
    1. Observe
    ln(ln n) 2
    (ln n)ln(ln n) = eln(ln n) = e[ln(ln n)]
    √
    and by the hint, ln(ln n) < ln n implies
    1 1 1
    > =
    e[ln(ln n)]2 eln n n
    P∞ 1
    P∞ 1
    Since 2 n diverges, so does 2 (ln n)ln(ln n) .

    2. (a)
    Z b Z b
    ′
    |f (b)| − |f (a)| ≤ |f (b) − f (a)| = | f (x)dx| ≤ |f ′ (x)|dx
    a a
    Z 1
    ≤ |f ′ (x)|dx
    0

    (b) Choose a ∈ (0, 1) such that f (a) ≤ f (x) for all x ∈ (0, 1) then
    Z 1 Z 1
    |f (a)| ≤ | f (x)dx| ≤ |f (x)|dx ⇒
    0 0
    Z 1 Z 1 Z 1
    ′ ′
    |f (b)| ≤ |f (x)|dx + |f (a)| ≤ |f (x)|dx + |f (x)|dx
    0 0 0
    Z 1
    = |f ′ (x)|dx + |f (x)|dx
    0

    3. Let g(x) = f (x) − x. Then g(0) ≥ 0 and g(1) ≤ 0. By intermediate value


    theotem there exists c ∈ [0, 1] such that g(c) = 0, whence the conclusion.
    4. Substitute x = π − z. Then, dx = −dz, x = 0 implies z = π, and x = π
    implies z = 0, so
    Z π Z 0 Z π Z π
    x sin x (π − z) sin(π − z) π sin z z sin z
    2
    =− 2
    dz = 2
    dz− 2
    dz
    0 1 + cos x π 1 + cos (π − z) 0 1 + cos z 0 1 + cos z

    Therefore,
    Z π Z Z 1
    x sin x π π sin x du π π π π 2
    2
    dx = dx = = ( + ) = ( )
    0 1 + cos x 2 0 1 + cos2 x −1 1 + u
    2 2 4 4 2

    1
    5. Let ε > 0, and let δ = ε. Then, for any |x| < ε, |f (x)| = 0 if x is rational and
    |f (x)| = |x| < ε if x is irrational. Thus, f is continuous at 0. For a 6= 0 and
    rational, |f (x)| = 0 or |x|. For ε < |a|, any δ > 0, there is an |x − a| < δ such
    that |x| > |a|, so f is not continuous at a. For a irrational, |f (x)−f (a)| = |a|
    or |x − a|, so again we fail for ε < |a|.

    6. Let limx→a f (x) = L. We note f is continuous at −a if the limit limx→−a f (x)


    exists and equals f (−a) = L if f is even, −L if f is odd. This limit is
    equivalent to limx→a f (−x) =: (∗). If f is even, f (−x) = f (x), so (∗) = L.
    If f is odd, f (−x) = −f (x), so (∗) = −L. We conclude f is continuous at
    −a.

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