1 0 2 b hence show that 2 sin 1 0 2 solution a for

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? > 1 ??? ??? 0 < ? < 𝜋 2 (b) Hence, show that 2 𝜋 sin ? ? < 1 ??? ??? 0 < ? ≤ 𝜋 2 Solution (a) For any 0 < ? < 𝜋 2 , we can apply mean value theorem on ?(?) = tan ? over [0, ?] and deduce that there exists ? ∈ (0, ?) such that tan ? − tan 0 ? − 0 = sec 2 ? ? (?) = 1 + tan 2 ? >0 ??? ?∈(0,?) > 1. (b) We consider a function ?: [0, 𝜋 2 ] → ℝ defined by ?(?) = { sin ? ? 𝑖? 0 < ? ≤ 𝜋 2 lim ?→0 sin ? ? = 1 𝑖? ? = 0
30 MATH2033 Mathematical Analysis Lecture Note 7: Differentiability We observe that ?(?) is continuous on [0, 𝜋 2 ] and is differentiable on (0, 𝜋 2 ) (because ?(?) = sin ? ? over (0, 𝜋 2 ) ). On the other hand, one can use the result of (a) and deduce that ? (?) = ? cos ? − sin ? ? 2 = cos ? ? >0 (1 − tan ? ? ) <0 < 0 for any 0 < ? < 𝜋 2 . It follows that ?(?) is strictly decreasing over 0 < ? < 𝜋 2 . Then for any 0 < ? < ? < ? < 𝜋 2 ?(?) < ?(?) = sin ? ? < ?(?). By taking the limits ? → 𝜋 2 (left-hand limits) and ? → 0 + , we have 2 𝜋 = ? ( 𝜋 2 ) = lim ?→ 𝜋 2 ?(?) ?(?) 𝑖? ????𝑖????? ?? ?= 𝜋 2 < ?(?) = sin ? ? < lim ?→0 + ?(?) = ?(0) = 1 ?(?) 𝑖? ????𝑖????? ?? ?=0 (?? ?????????𝑖??) .
31 MATH2033 Mathematical Analysis Lecture Note 7: Differentiability Generalized mean value theorem (or Cauchy mean value theorem) Technically, one can extend the mean value theorem into the following form which is useful in deriving L’Hopital rule. Remark: Geometrically, this theorem can be interpreted as follows: Theorem 4 (Generalized mean value theorem) Suppose that a function ?(?), ?(?) is continuous on [?, ?] , is differentiable over (?, ?) , then there exists ? ∈ (?, ?) such that [?(?) − ?(?)]? (?) = [?(?) − ?(?)]? (?) ?? ?(?) − ?(?) ?(?) − ?(?) = ? (?) ? (?) (????𝑖??? ?ℎ?? ?(?) ≠ ?(?)). 𝒗 ⃗⃗ (? (?), ?(?)) (?(?), ?(?)) (?(?), ?(?)) Parametric curve (?(?), ?(?)), ? ∈ [?, ?]
32 MATH2033 Mathematical Analysis Lecture Note 7: Differentiability We consider a parametric curve defined by (?(?), ?(?)) , where ? ∈ [?, ?] . Here, one can treat the parameter ? as time. Then (?(?), ?(?)) represents the position at time ? . Starting from (?(?), ?(?)) , the point moves to (?(?), ?(?)) as ? moves from ? to ? . It traces out a curve in 2-D plane. One can observe from the above figure that there exists a time ? such that the tangent vector (? (?), ? (?)) at that point is parallel to the vector ? , which is a vector joining (?(?), ?(?)) and (?(?), ?(?)) . Note that ? = (?(?) − ?(?), ?(?) − ?(?)) , then we have (? (?), ? (?)) = 𝜆? ⇔ (? (?), ? (?)) = (𝜆(?(?) − ?(?)), 𝜆(?(?) − ?(?))). This implies that ? (?) = 𝜆(?(?) − ?(?)) ??? ? (?) = 𝜆(?(?) − ?(?)) By eliminating 𝜆 , we deduce that ?(?) − ?(?) ?(?) − ?(?) = ? (?) ? (?) , which is the generalized mean value theorem.
33 MATH2033 Mathematical Analysis Lecture Note 7: Differentiability Proof of generalized mean value theorem When ?(?) = ? , the theorem is reduced to standard mean value theorem. So we shall prove this theorem by mimicking the proof of mean value theorem. We consider the function ?(?) = ?(?) − [?(?) + ?(?) − ?(?) ?(?) ?(?) ( ?(?) ?(?) )] or more generally (to avoid the possibility of ?(?) − ?(?) = 0 ), ?(?) = (?(?) − ?(?))(?(?) − ?(?)) − (?(?) − ?(?))(?(?) − ?(?)). Note that Since ?, ? are continuous on [?, ?] and differentiable on (?, ?) , so does ?(?) . One can verify that ?(?) = ?(?) = 0 . ? (?) = ? (?)(?(?) − ?(?)) − ? (?)(?(?) − ?(?)) It follows from Rolle’s theorem that there exists ? ∈ (?, ?) such that ? (?) = 0 ⇔ ? (?)(?(?) − ?(?)) − ? (?)(?(?) − ?(?)) ⇔ ?(?) − ?(?) ?(?) − ?(?) = ? (?) ? (?) 𝑖? ?(?)−?(?)≠0 .
34 MATH2033 Mathematical Analysis Lecture Note 7: Differentiability L’Hopital’s Rule It is one of the powerful tools for computing the limits of the forms of lim ?→? ?(?) ?(?) , where ?(?) and ?(?) are some functions. Remark of Theorem 5 The statement holds if all “ ? → ? + ” are replaced by "? → ? " The statement alslo holds if all ? → ? + ” are replaced by "? → ?" ( ? ∈ (?, ?) ) and the interval (?, ?) in the theorem is replaced by (?, ?)\{?} .

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