Lecture (5) Analytical Geometry & Calculus I PDF

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This document is a lecture on analytical geometry and calculus, specifically focusing on hyperbolic functions, their derivatives, and Taylor/Maclaurin series. It contains definitions, formulas, and examples.

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Lecture (5) 1 Analytical Geometry & Calculus I MATH 111i Outline 2  Inverse Hyperbolic Functions  Derivatives of Inverse Hyperbolic Functions  Taylor and Maclaurin series Revision: Definition of Hyperbo...

Lecture (5) 1 Analytical Geometry & Calculus I MATH 111i Outline 2  Inverse Hyperbolic Functions  Derivatives of Inverse Hyperbolic Functions  Taylor and Maclaurin series Revision: Definition of Hyperbolic Functions 3 −x e −e x Hyperbolic Sine: sinh ( x ) = 2 −x e +e x Hyperbolic Cosine: cosh ( x ) = 2 sinh ( x ) e x − e− x Hyperbolic Tan: tanh ( x ) = = x −x cosh ( x ) e + e Revision: Definition of Hyperbolic Functions 4 cosh ( x ) e x + e− x Hyperbolic Cot: coth ( x ) = = x ,x 0 sinh ( x ) e − e − x 1 2 Hyperbolic Sec: sech ( x ) = = x −x cosh ( x ) e + e 1 2 Hyperbolic Cosec: csch ( x ) = = x ,x 0 sinh ( x ) e − e − x Revision: Identities of Hyperbolic Functions 5 𝐜𝐨𝐬𝐡 𝒙 + 𝐬𝐢𝐧𝐡 𝒙 = 𝒆𝒙 𝐜𝐨𝐬𝐡 𝒙 − 𝐬𝐢𝐧𝐡 𝒙 = 𝒆−𝒙 𝐜𝐨𝐬𝐡𝟐 𝒙 − 𝐬𝐢𝐧𝐡𝟐 𝒙 = 𝟏 𝟏 − 𝐭𝐚𝐧𝐡𝟐 𝒙 = 𝐬𝐞𝐜𝐡𝟐 𝒙 𝐜𝐨𝐭𝐡𝟐 𝒙 − 𝟏 = 𝐜𝐬𝐜𝐡𝟐 𝒙 𝐬𝐢𝐧𝐡 𝒙 ± 𝒚 = 𝐬𝐢𝐧𝐡 𝒙 𝐜𝐨𝐬𝐡 𝒚 ± 𝐜𝐨𝐬𝐡 𝒙 𝐬𝐢𝐧𝐡 𝒚 𝐜𝐨𝐬𝐡 𝒙 ± 𝒚 = 𝐜𝐨𝐬𝐡 𝒙 𝐜𝐨𝐬𝐡 𝒚 ± 𝐬𝐢𝐧𝐡 𝒙 𝐬𝐢𝐧𝐡 𝒚 To prove the Identities use the definition Derivatives of Hyperbolic Functions 6 d d sinh x = cosh x cosh x = sinh x dx dx d d tanh x = sec h x 2 coth x = −cs c h x 2 dx dx d d sec hx = −sec h x tanh x cs c hx = −cs c h x coth x dx dx Revision: Graphs of Hyperbolic Functions 7 𝐷: ℜ 𝐷: ℜ 𝐷: ℜ 𝑅: ℜ 𝑅: ሾ1, ∞) 𝑅: −1,1 𝐷: ℜ − 0 𝐷: ℜ 𝐷: ℜ − 0 𝑅: ℜ − 0 𝑅: (0, 1ሿ 𝑅: ℜ − −1,1 Inverse Hyperbolic Functions 8  𝑦 = sinh−1 𝑥 ⇔ 𝑥 = sinh 𝑦  𝑦 = cosh−1 𝑥 ⇔ 𝑥 = cosh 𝑦 , 0≤𝑦  𝑦 = tanh−1 𝑥 ⇔ 𝑥 = tanh 𝑦  𝑦 = coth−1 𝑥 ⇔ 𝑥 = coth 𝑦  𝑦 = sech−1 𝑥 ⇔ 𝑥 = sech 𝑦 , 0≤𝑦  𝑦 = csch−1 𝑥 ⇔ 𝑥 = csch 𝑦 Graphs of Inverse Hyperbolic Functions 9 one to one Graphs of Inverse Hyperbolic Functions 10 Not one to one Graphs of Inverse Hyperbolic Functions 11 one to one Graphs of Inverse Hyperbolic Functions one to one Graphs of Inverse Hyperbolic Functions Not one to one Graphs of Inverse Hyperbolic Functions one to one Graphs of Inverse Hyperbolic Functions 15 𝐷: ℜ 𝐷: ሾ1, ∞) 𝐷: −1,1 𝑅: ℜ 𝑅: ሾ0, ∞) 𝑅: ℜ 𝐷: ℜ − 0 𝐷: (0, 1ሿ 𝐷: ℜ − −1,1 𝑅: ℜ − 0 𝑅: ሾ0, ∞) 𝑅: ℜ − 0 Inverse Hyperbolic Identities 16 sinh−1 −𝑥 = − sinh−1 𝑥 tanh−1 −𝑥 = − tanh−1 𝑥 coth−1 −𝑥 = − coth−1 𝑥 csch−1 −𝑥 = − csch−1 𝑥 Inverse Hyperbolic Identities 17 −1 −1 1 csch 𝑥 = sinh 𝑥 −1 −1 1 sech 𝑥 = cosh 𝑥 −1 −1 1 coth 𝑥 = tanh 𝑥 Proof −1 1 1 𝑦 = csch 𝑥 ⇒ csch 𝑦 = 𝑥 ⇒ = csch 𝑦 𝑥 1 −1 1 sinh 𝑦 = ⇒ 𝑦= sinh 𝑥 𝑥 1 ⇒ csch−1 𝑥 = sinh −1 𝑥 Derivatives of Inverse Hyperbolic Functions d −1 1 d 1 sinh x = −1 cosh x = dx 1+ x 2 dx x2 −1 d −1 1 d 1 tanh x = −1 coth x = dx 1− x 2 dx 1 − x2 d −1 1 d −1 1 sec h x = − csc h x = − dx x 1− x 2 dx x 1 + x2 18 Derivatives of Inverse Hyperbolic Functions 19 Proofs (i) 𝑦 = sinh−1 𝑥 ⇒ sinh 𝑦 = 𝑥 cosh 𝑦 𝑦 ′ = 1 ′ 1 1 1 𝑦 = = = cosh 𝑦 1+sinh2 𝑦 1+𝑥 2 Derivatives of Inverse Hyperbolic Functions 20 Proofs (ii) 𝑦 = tanh−1 𝑥 ⇒ tanh 𝑦 = 𝑥 sech2 𝑦 𝑦 ′ = 1 ′ 1 1 1 𝑦 = = = sech2 𝑦 1−tanh2 𝑦 1−𝑥 2 Derivatives of Inverse Hyperbolic Functions 21 Proofs (iii) 𝑦 = sech−1 𝑥 ⇒ sech 𝑦 = 𝑥 − sech 𝑦 tanh 𝑦 𝑦 ′ = 1 ′ −1 −1 𝑦 = = sech 𝑦 tanh 𝑦 sech 𝑦 1−sech2 𝑦 −1 = 𝑥 1−𝑥 2 Derivatives of Inverse Hyperbolic Functions 22 Example ( ) Find 𝑦 ′ if 𝑦 = 𝑒 𝑥 tanh−1 𝑥 + 3𝑥 cosh−1 𝑥 𝑥 𝑒 𝑦 ′ = 𝑒 𝑥 tanh−1 𝑥 + 1 − 𝑥2 𝑥 −1 𝑥 1 +3 ln 3 cosh 𝑥 + 3 𝑥2 − 1 Derivatives of Inverse Hyperbolic Functions 23 Example () Find 𝑦 ′ if 𝑦 = sin−1 𝑥 + 𝑥 sech−1 𝑥 + 1 − 𝑥 2 ′ 1 −1 −1 −2𝑥 𝑦 = + sech 𝑥+𝑥∗ + 1−𝑥 2 𝑥 1−𝑥 2 2 1−𝑥 2 −1 𝑥 = sech 𝑥− 1−𝑥 2 Example () show that 𝑦 ′′ = csc 𝑥 cot 𝑥 − sech 𝑥 tanh 𝑥 if 𝑦 = tan−1 sinh 𝑥 + sech−1 sin 𝑥 24 ′ 1 −1 𝑦 = 2 cosh 𝑥 + cos 𝑥 1 + sinh 𝑥 sin 𝑥 1 − sin2 𝑥 cosh 𝑥 cos 𝑥 = 2 − cosh 𝑥 sin 𝑥 cos 𝑥 1 1 = − = sech 𝑥 − csc 𝑥 cosh 𝑥 sin 𝑥 𝑦 ′′ = −sech 𝑥 tanh 𝑥 − − csc 𝑥 cot 𝑥 = csc 𝑥 cot 𝑥 − sech 𝑥 tanh 𝑥 Brook Taylor Colin Maclaurin (1685-1731) (1698–1746) Taylor and Maclaurin series How to represent the function in power series(polynomials)? ? MACLAURIN SERIES  If a function f(x) has derivatives of all orders at x = 0, then the series f "(0) 2 f "'(0) 3 f ( x ) = f (0) + f '(0)( x ) + ( x) + ( x ) +... 2! 3! is called the Maclaurin series for f(x). MACLAURIN SERIES Example: Find the Maclaurin series of the function f(x) = ex f ( x) = e x f (0) = 1 f ( x) = e ' x f (0) = 1 ' f ( x) = e '' x f (0) = 1 '' f (n) ( x) = e x f (n) (0) = 1 f "(0) 2 f "'(0) 3 f ( x ) = f (0) + f '(0)( x ) + ( x) + ( x ) +... 2! 3! MACLAURIN SERIES Example: Find the Maclaurin series of the function f(x) = ex f "(0) 2 f "'(0) 3 f ( x ) = f (0) + f '(0)( x ) + ( x) + ( x ) +... 2! 3! 1 1 e = 1 + ( x ) + ( x ) + ( x ) +... x 2 3 2! 3!  n x = 2 3 x x e = 1+ x + x + +... 2! 3! n =0 n ! MACLAURIN SERIES Guidelines for Finding Maclaurin Series 1. Differentiate f(x) several times (n) f (0), f "(0), f '''(0),..., f (0) Evaluate each derivative at x=0 f ( n ) (0) 2. Find the Maclaurin coefficients Cn = n! MACLAURIN SERIES Example: Find the Maclaurin series of the function f(x) = cos(x) f ( x ) = cos( x ) f (0) = 1 f ( x ) = − sin x ' f (0) = 0 ' f ( x ) = − cos x '' f (0) = −1 '' f '''( x ) = sin x f '''(0) = 0 f (4) ( x ) = cos x f (4) (0) = 1 repetition: MACLAURIN SERIES Example: Find the Maclaurin series of the function f(x) = cos(x) f "(0) 2 f "'(0) 3 f ( x ) = f (0) + f '(0)( x ) + ( x) + ( x ) +... 2! 3! −1 1 cos x = 1 + 0( x ) + ( x ) + 0( x ) + ( x ) 4... 2 3 2! 4! x2 x4 x6 cos x = 1 − + −... 2! 4! 6! MACLAURIN SERIES Similarly you can proof the following. x3 x5 x7 sin x = x − + −... 3! 5! 7! x3 x5 x7 sinh x = x + + +... 3! 5! 7! x2 x4 x6 cosh x = 1 + + +... 2! 4! 6! 1 = 1 + x + x 2 + x 3 +... 1− x x2 x3 x4 ln(1 − x) = − x − − − −... 2 3 4 Maclaurin Series for Composite Function  What about when f(x) = cos(x2)? The series for cos x is x2 x4 x6 cos x = 1 − + − +... 2! 4! 6! Now substitute by x2 (replace each x by x2) 4 8 12 x x x cos x = 1 − + − 2 +... 2! 4! 6! MACLAURIN SERIES Example: Find the first three non zero terms of the Maclaurin series of the function f ( x ) = x 2 e x 2 First find the Maclaurin series of ex (as before) 2 3 x x e = 1+ x + x + +... 2! 3! Replace each x by x2 4 6 x x e = 1+ x + + +... 2 x 2 2! 3! Multiply by x 2 6 8 x x x2ex = x2 + x4 + + +... 2 2! 3! MACLAURIN SERIES Examples (try): Find the first three non zero terms of the Maclaurin series of the following f ( x) = x cosh x f ( x ) = x 3 sinh( x 2 ) f ( x ) = ln(1 − x 2 ) Taylor Series(more general)  If a function f(x) has derivatives of all orders at x = c, then the series f "( c ) f ( x ) = f ( c ) + f '( c )( x − c ) + ( x − c ) 2 +... 2!  f ( n ) (c) = ( x − c) n n =0 n! is called the Taylor series for f(x) at c.  If c = 0, the series is the Maclaurin series for f(x). Taylor Series(more general) Example: Find the Taylor series of the function  f ( x ) = sin x at x= 2  f ( x ) = sin( x ) f( ) =1 2  f '( x ) = cos x f '( ) =0 2  f ''( x ) = − sin x f ''( ) = −1 2  f '''( x ) = − cos x f '''( )=0 2  f (4) ( x ) = sin x f (4) ( ) =1 2 repetition: Taylor Series(more general) Example: Find the Taylor series of the function  f ( x ) = sin x at x= 2 f "( c ) f ( x ) = f ( c ) + f '( c )( x − c ) + ( x − c ) 2 +... 2!  −1  2 0  3 1  4 sin x = 1 + 0( x − ) + ( x − ) + ( x − ) + ( x − ) +... 2 2! 2 3! 2 4! 2 1  2 1  4 sin x = 1 − ( x − ) + ( x − ) +... 2! 2 4! 2 Finish See you next week

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