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Unformatted text preview: Harold’s Calculus Notes Cheat Sheet 1 October 2015 AP Calculus Limits Definition of Limit Let f be a function defined on an open interval containing c and let L be a real number. The statement: lim () = → means that for each > 0 there exists a > 0 such that if 0 < | − | < , then |() − | < Tip : Direct substitution: Plug in f(a) and see if it provides a legal answer. If so then L = f(a). The Existence of a Limit The limit of f(x) as x approaches a is L if and only if: lim () = → − lim () = → + Prove that () = − is a continuous function. Definition of Continuity A function f is continuous at c if for every > 0 there exists a > 0 such that | − | < and |() − ()| < . Tip: Rearrange |() − ()| to have |( − )| as a factor. Since | − | < we can find an equation that relates both and together. Two Special Trig Limits |() − ()| = |( 2 − 1) − ( 2 − 1)| = | 2 − 1 − 2 + 1| = | 2 − 2 | = |( + )( − )| = |( + )| |( − )| Since |( + )| ≤ |2| |() − ()| ≤ |2||( − )| < So given > 0, we can choose = | | > in the Definition of Continuity. So substituting the chosen for |( − )| we get: 1 |() − ()| ≤ |2| (| | ) = 2 Since both conditions are met, the function is continuous. =1 →0 1 − =0 →0 Copyright © 2015 by Harold Toomey, WyzAnt Tutor 1 Derivatives Definition of a Derivative of a Function Slope Notation for Derivatives The Constant Rule The Power Rule The General Power Rule The Constant Multiple Rule The Sum and Difference Rule Position Function Velocity Function Acceleration Function Jerk Function The Product Rule The Quotient Rule The Chain Rule Exponentials ( , ) Logorithms ( , ) Sine Cosine Tangent Secent Cosecent Copyright © 2015 by Harold Toomey, WyzAnt Tutor (See Larson’s 1-pager of common derivatives) ( + ∆) − () ′ () = lim ∆→0 ∆ () − () ′ () = lim → − ′ ′ (), () (), , , [()], [] [] = 0 [ ] = −1 [] = 1 (think = 1 0 = 1) [ ] = −1 ′ [()] = [()] −1 [()] = ′ () [() ± ()] = ′ () ± ′ () 1 () = 2 + 0 + 0 2 () = ′ () = + 0 () = ′ () = ′′ () () = ′ () = ′′ () = (3) () [] = ′ + ′ ′ − ′ = [ ] 2 = (), = () ≠ 0 = · [(())] = ′ (())′ () [ ] = , [ ] = (ln ) 1 1 [ln ] = , [log ] = (ln ) [()] = cos() [()] = −() [()] = 2() [()] = () () [()] = − () () 2 [()] = − 2 () Cotangent Applications of Differentiation Rolle’s Theorem f is continuous on the closed interval [a,b], and f is differentiable on the open interval (a,b). If f(a) = f(b), then there exists at least one number c in (a,b) such that f’(c) = 0. () − () − () = () + ( − )′() Find ‘c’. () lim () = lim = → → () Mean Value Theorem If f meets the conditions of Rolle’s Theorem, then L’Hôpital’s Rule ′ () = 0 ∞ { , , 0 • ∞, 1∞ , 00 , ∞0 , ∞ − ∞} , {0∞ }, 0 ∞ () ′ () ′′ () ℎ lim = lim ′ = lim ′′ =⋯ → () → () → () Graphing with Derivatives Test for Increasing and Decreasing Functions The First Derivative Test The Second Deriviative Test Let f’(c)=0, and f”(x) exists, then Test for Concavity Points of Inflection Change in concavity Analyzing the Graph of a Function x-Intercepts (Zeros or Roots) y-Intercept Domain Range Continuity Vertical Asymptotes (VA) Horizontal Asymptotes (HA) Infinite Limits at Infinity Copyright © 2015 by Harold Toomey, WyzAnt Tutor 1. If f’(x) > 0, then f is increasing (slope up) 2. If f’(x) < 0, then f is decreasing (slope down) 3. If f’(x) = 0, then f is constant (zero slope) 1. If f’(x) changes from – to + at c, then f has a relative minimum at (c, f(c)) 2. If f’(x) changes from + to - at c, then f has a relative maximum at (c, f(c)) 3. If f’(x), is + c + or - c -, then f(c) is neither 1. If f”(x) > 0, then f has a relative minimum at (c, f(c)) 2. If f”(x) < 0, then f has a relative maximum at (c, f(c)) 3. If f’(x) = 0, then the test fails (See 1st derivative test) 1. If f”(x) > 0 for all x, then the graph is concave upward 2. If f”(x) < 0 for all x, then the graph is concave downward If (c, f(c)) is a point of inflection of f, then either 1. f”(c) = 0 or 2. f” does not exist at x = c. (See Harold’s Illegals and Graphing Rationals Cheat Sheet) f(x) = 0 f(0) = y Valid x values Valid y values No division by 0, no negative square roots or logs x = division by 0 or undefined lim− () → and lim+ () → →∞ →∞ lim− () → ∞ and lim+ () → ∞ →∞ →∞ 3 Differentiability Relative Extrema Concavity Points of Inflection Limit from both directions arrives at the same slope Create a table with domains, f(x), f’(x), and f”(x) If ”() → +, then cup up ∪ If ”() → −, then cup down ∩ f”(x) = 0 (concavity changes) Approximating with Differentials Newton’s Method Finds zeros of f, or finds c if f(c) = 0. Tangent Line Approximations Function Approximations with Differentials Related Rates Copyright © 2015 by Harold Toomey, WyzAnt Tutor ( ) ′( ) = + = ′ ()( − ) + () +1 = − ( + ∆) ≈ () + = () + ′ () 1. Identify the known variables and rates of change. ( = 2 ; = −3 ; ′ = 4 ; ′ = ? ) 2. Construct an equation relating these quantities. ( 2 + 2 = 2 ) 3. Differentiate both sides of the equation. (2 ′ + 2 ′ = 0) 4. Solve for the desired rate of change. ( ′ = − ′ ) 5. Substitute the known rates of change and quantities into the equation. 2 8 ( ′ = − ⦁4= ) −3 3 4 Integration Basic Integration Rules Integration is the “inverse” of differentiation. Differentiation is the “inverse” of integration. ∫ ′ () = () + () = 0 ∫ 0 = () = = 0 ∫ = + [∫ () ] = () The Constant Multiple Rule The Sum and Difference Rule The Power Rule ∫ () = ∫ () ∫[() ± ()] = ∫ () ± ∫ () ∫ = () = +1 + , ℎ ≠ −1 +1 If = −1 then ∫ −1 = ln|| + The General Power Rule If = (), ′ = () then +1 ∫ ′ = + , ℎ ≠ −1 +1 Reimann Sum ∑ ( ) ∆ , ℎ −1 ≤ ≤ =1 ‖∆‖ = ∆ = Definition of a Definite Integral Area under curve ‖∆‖→0 The Second Fundamental Theorem of Calculus =1 ∫ () = − ∫ () ∫ () = ∫ () + ∫ () The Fundamental Theorem of Calculus lim ∑ ( ) ∆ = ∫ () Swap Bounds Additive Interval Property − ∫ () = () − () (See Harold’s Fundamental Theorem of Calculus Cheat Sheet) = [∫ () ] = () () [∫ () ] = () ′() ′ () = (See Harold’s Fundamental Theorem of Calculus Cheat Sheet) Mean Value Theorem for Integrals Copyright © 2015 by Harold Toomey, WyzAnt Tutor ∫ () = ()( − ) Find ‘c’. 5 1 ∫ () − The Average Value for a Function ∑ = =1 ( + 1) 2 ∑= = + 2 2 2 =1 ∑ 2 = =1 ∑ 3 = =1 ∑ 4 = =1 ( + 1)(2 + 1) 3 2 = + + 6 3 2 6 2 ( + 1)2 4 3 2 = + + 4 4 2 4 ( + 1)(2 + 1)(32 + 3 − 1) 30 5 4 3 + + − 5 2 3 30 2 ( + 1)2 (22 + 2 − 1) ∑ 5 = 12 = =1 ∑ 6 = =1 Summation Formulas for Sum of Powers ∑ 7 = =1 2 ( + 1)2 (34 + 63 − 2 − 4 + 2) 24 + ∑ =1 ( + 1)(2 + 1)(34 + 63 − 3 + 1) 42 = +1 | +1 ℎ ≠ −1 1 ∑ = ln() =1 ∑ ( + 1) = =1 ∑ =1 ( + 1)( + 2) 3 1 = ( + 1) +1 ∑ ( + 1)( + 2) = =1 ∑ =1 ( + 1)( + 2)( + 3) 4 1 ( + 3) = ( + 1)( + 2) 4( + 1)( + 2) ∑(2 + 1) = ( + 2) =1 ∑(2 − 1) = 2 =1 Copyright © 2015 by Harold Toomey, WyzAnt Tutor 6 Integration Methods 1. Memorized See Larson’s 1-pager of common integrals ∫ (())′ () = (()) + Set = (), then = ′ () 2. U-Substitution ∫ () = () + = _____ = _____ ∫ = − ∫ = ____ = _____ = _____ = _____ Pick ‘’ using the LIATED Rule: L – Logarithmic : ln , log , . I – Inverse Trig.: tan−1 , sec −1 , . A – Algebraic: 2 , 3 60 , . T – Trigonometric: sin , tan , . E – Exponential: , 19 , . 3. Integration by Parts D – Derivative of: () () where () () are polynomials ∫ 4. Partial Fractions Case 1: If degree of () ≥ () then do long division first Case 2: If degree of () < () then do partial fraction expansion ∫ √ 2 − 2 5. Trig Substitution for √ 6. Trig Substitution for √ − − + Substutution: = sin Identity: 1 − 2 = 2 ∫ √ 2 − 2 Substutution: = sec Identity: 2 − 1 = 2 ∫ √ 2 + 2 7. Trig Substitution for √ 8. Table of Integrals 9. Computer Algebra Systems (CAS) 10. Numerical Methods 11. WolframAlpha Copyright © 2015 by Harold Toomey, WyzAnt Tutor Substutution: = tan Identity: 2 + 1 = 2 CRC Standard Mathematical Tables book TI-Nspire CX CAS Graphing Calculator TI –Nspire CAS iPad app Riemann Sum, Midpoint Rule, Trapezoidal Rule, Simpson’s Rule Google of mathematics. Shows steps. Free. 7 Partial Fractions Condition Case I: Simple linear ( degree) Case II: Multiple degree linear ( degree) Case III: Simple quadratic ( degree) Case IV: Multiple degree quadratic ( degree) Typical Solution for Cases I & II Typical Solution for Cases III & IV Numerical Methods ( mposition) () () = () where () () are polynomials and degree of () < () ( + ) + + 1 2 ( + ) ( + ) ( + )3 + 2 + + ) ( + + + + + ( 2 + + ) ( 2 + + )2 ( 2 + + )3 ∫ = | + | + + ∫ 2 = −1 ( ) + 2 + ∫ () = lim ∑ ( ∗ ) ∆ ‖‖→0 Riemann Sum =1 where = 0 < 1 < 2 < ⋯ < = and ∆ = − −1 and ‖‖ = {∆ } Types: Left Sum, Middle Sum, Right Sum ∫ () ≈ ∑ (̅ ) ∆ = Midpoint Rule =1 ∆[(̅1 ) + (̅2 ) + (̅3 ) + ⋯ + (̅ )] − where ∆ = 1 2 and ̅ = ( −1 + ) = [ −1 , ] Error Bounds: | | ≤ (−)3 242 ∫ () ≈ Trapezoidal Rule ∆ [(0 ) + 2(1 ) + 2(3 ) + ⋯ + 2( −1 ) 2 + ( )] − where ∆ = and = + ∆ Error Bounds: | | ≤ Copyright © 2015 by Harold Toomey, WyzAnt Tutor (−)3 122 8 ∫ () ≈ ∆ [(0 ) + 4(1 ) + 2(2 ) + 4(3 ) + ⋯ 3 + 2( −2 ) + 4( −1 ) + ( )] Where n is even − and ∆ = and = + ∆ Simpson’s Rule Error Bounds: | | ≤ Infinite Sequences and Series (−)5 1804 (See Harold’s Series Cheat Sheet) lim = (Limit) Sequence →∞ ∞ Arithmetic Series (infinite) Example: ( , +1 , +2 , …) ∑ = = 1 + 2 + 3 + ⋯ + + ⋯ =1 Geometric Series (finite) ∑ −1 = = + + 2 + ⋯ + −1 =1 (1 − ) () 1− (1 − ) = lim = →∞ 1− 1− only if || < 1 where is the radius or interval of convergence = Geometric Series (infinite) Partial Sum (finite) = ∑ = 1 + 2 + 3 + ⋯ + Similar to an arithmetic series Misc. Euler’s Equation Copyright © 2015 by Harold Toomey, WyzAnt Tutor + 1 = 0 = cos() + sin() 9 Convergence Tests (See Harold’s Series Convergence Tests Cheat Sheet) 1. Term Test 2. Geometric Series Test ∞ ∑( − +1 ) =1 3. Telescoping Series Converges if lim = → ∞ Diverges if N/A Sum: = 1 − 4. p-Series Test 5. Alternating Series Test 6. Integral Test 7. Ratio Test 8. Root Test 9. Direct Comparison Test 10. Limit Comparison Test Taylor Series +∞ Power Series ∑ ( − ) = 0 + 1 ( − ) + 2 ( − )2 + ⋯ =0 Power Series About Zero +∞ ∑ = 0 + 1 + 2 2 + ⋯ =0 Taylor Series +∞ () () () ≈ () = ∑ ( − ) ! =0 () = () + () +∞ =∑ =0 Taylor Series with Remainder () () ( − ) + () ! (+1) ( ∗ ) () = ( − ) +1 ( + 1)! where ≤ ∗ ≤ and lim () = 0 →+∞ Maclaurin Series Taylor series about zero +∞ () (0) () ≈ () = ∑ ! =0 () = () + () +∞ =∑ =0 Maclaurin Series with Remainder () (0) + () ! (+1) ( ∗ ) +1 () = ( + 1)! where ≤ ∗ ≤ and lim () = 0 →+∞ Copyright © 2015 by Harold Toomey, WyzAnt Tutor 10 Common Series Exponential Function ∞ =∑ =0 ∞ ! = ln() = ∑ =0 Natural Logarithm ( ln()) ! ∞ ln (1 − ) = ∑ =1 ∞ ∞ ln (1 + ) = ∑ =1 1 + ln() + || < 1 (−1) ( − 1) ln () = ∑ || < 1 =1 1+ + + ( − 1) + (−1) −1 || < 1 2 3 4 + + +⋯ 2! 3! 4! ( ln())2 ( ln())3 + +⋯ 2! 3! 2 3 4 5 + + + +⋯ 2 3 4 5 ( − 1)2 ( − 1)3 ( − 1)4 + + +⋯ 2 3 4 − 2 3 4 5 + − + −⋯ 2 3 4 5 Geometric Series ∞ 1 = ∑(−1) ( − 1) 0 < < 2 =0 1 − ( − 1) + ( − 1)2 − ( − 1)3 + ( − 1)4 + ⋯ ∞ 1 = ∑(−1) || < 1 1+ =0 1 − + 2 − 3 + 4 − ⋯ ∞ 1 = ∑ || < 1 1− 1 + + 2 + 3 + 4 + ⋯ 1 = ∑ −1 || < 1 (1 − )2 1 + 2 + 3 2 + 4 3 + 5 4 + ⋯ =0 ∞ =1 ∞ 1 ( − 1) −2 =∑ || < 1 3 (1 − ) 2 1 + 3 + 6 2 + 10 3 + 15 4 + ⋯ =2 Binomial Series +∞ (1 + ) = 1 + ∑ ( ) =1 || < 1 and all complex r where − +1 ( )=∏ =1 ( − 1)( − 2) … ( − + 1) = ! Trigonometric Functions ∞ sin () = ∑ =0 (−1) 2+1 (2 + 1)! ∞ cos () = ∑ =0 (−1) 2 (2)! Copyright © 2015 by Harold Toomey, WyzAnt Tutor 1 + + ( − 1) 2 ( − 1)( − 2) 3 + +⋯ 2! 3! − 3 5 7 9 + − + −⋯ 3! 5! 7! 9! 1− 2 4 6 8 + − + −⋯ 2! 4! 6! 8! 11 ∞ 2 (−4) (1 − 4 ) 2−1 (2)! =1 || < 2 Bernoulli Numbers: 1 1 1 1 0 = 1, 1 = , 2 = , 4 = , 6 = , 2 6 30 42 1 5 691 7 8 = , 10 = , 12 = , 14 = 30 66 2730 6 ∞ (−1) 2 2 sec () = ∑ (2)! =0 || < 2 Euler Numbers: 0 = 1, 2 = −1, 4 = 5, 6 = −61, 8 = 1,385, 10 = −50,521, 12 = 2,702,765 ∞ (2)! arcsin () = ∑ 2 2+1 (2 !) (2 + 1) tan () = ∑ =0 || ≤ 1 arccos () = − arcsin () 2 || ≤ 1 ∞ (−1) 2+1 arctan () = ∑ (2 + 1) +2 + 1+ 3 5 7 9 + 16 + 272 + 7936 − ⋯ 3! 5! 7! 9! 1 3 2 5 17 7 2 9 + + + −⋯ 3 15 315 945 2 4 6 8 10 + 5 + 61 + 1385 + 50521 +⋯ 2! 4! 6! 8! 10! + 3 1 ∙ 3 5 1 ∙ 3 ∙ 5 7 + + +⋯ 2∙3 2∙4∙5 2∙4∙6∙7 3 1 ∙ 3 5 1 ∙ 3 ∙ 5 7 − − − − −⋯ 2 2∙3 2∙4∙5 2∙4∙6∙7 3 5 7 9 − + − + −⋯ 3 5 7 9 =0 || < 1, ≠ ± Hyperbolic Functions ∞ − − 2+1 sinh () = =∑ (2 + 1)! 2 + 3 5 7 9 + + + +⋯ 3! 5! 7! 9! + − 2 cosh () = =∑ (2)! 2 1+ 2 4 6 8 + + + +⋯ 2! 4! 6! 8! =0 ∞ =0 ∞ 2 4 (4 − 1) 2−1 tanh () = ∑ (2)! =1 || < 2 ∞ (−1) (2)! arcsinh () = ∑ 2 2+1 (2 !) (2 + 1) =0 || ≤ 1 3 5 7 9 − 2 + 16 − 272 + 7936 − ⋯ 3! 5! 7! 9! 1 3 2 5 17 7 2 9 − + − + −⋯ 3 15 315 945 − 3 1 ∙ 3 5 1 ∙ 3 ∙ 5 7 + − +⋯ 2∙3 2∙4∙5 2∙4∙6∙7 ∞ 2− arccosh () = − ∑ 2+1 2 ! (2 + 1) =0 || ≤ 1 ∞ arctanh () = ∑ =0 2+1 (2 + 1) || < 1, ≠ ±1 Copyright © 2015 by Harold Toomey, WyzAnt Tutor 3 ∙ 1 ∙ 3 5 ∙ 1 ∙ 3 ∙ 5 7 − − − − −⋯ 2 2∙3 2∙4∙5 2∙4∙6∙7 + 3 5 7 9 + + + +⋯ 3 5 7 9 12 ...
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