Mathematics Quick Review Cheat Sheet, Cheat Sheet of Mathematics

Download Mathematics Quick Review Cheat Sheet and more Cheat Sheet Mathematics in PDF only on Docsity! Mathematics Cheat Sheet Definitions Series 𝑓(𝑛) = 𝑂(𝑔(𝑛)) iff ∃ positive 𝑐, 𝑛0 such that 0 ≤𝑓(𝑛) ≤ 𝑐𝑔(𝑛) ∀𝑛 ≥ 𝑛0. 𝑓(𝑛) = Ω(𝑔(𝑛)) iff ∃ positive 𝑐, 𝑛0 such that 𝑓(𝑛) ≥𝑐𝑔(𝑛) ≥ 0 ∀𝑛 ≥ 𝑛0. 𝑓(𝑛) = Θ(𝑔(𝑛)) iff 𝑓(𝑛) = 𝑂(𝑔(𝑛)) and 𝑓(𝑛) =Ω(𝑔(𝑛)). 𝑓(𝑛) = 𝑜(𝑔(𝑛)) iff lim𝑛→∞ 𝑓(𝑛)/𝑔(𝑛) = 0. lim𝑛→∞ 𝑎𝑛 = 𝑎 iff ∀𝜖 > 0, ∃𝑛0 such that |𝑎𝑛 − 𝑎| < 𝜖, ∀𝑛 ≥ 𝑛0. sup 𝑆 least 𝑏 ∈ ℝ such that 𝑏 ≥ 𝑠, ∀𝑠 ∈𝑆. inf 𝑆 greatest 𝑏 ∈ ℝ such that 𝑏 ≤ 𝑠,∀𝑠 ∈ 𝑆. lim inf𝑛→∞ 𝑎𝑛 lim𝑛→∞ inf{𝑎𝑖 ∣ 𝑖 ≥ 𝑛, 𝑖 ∈ ℕ}. lim sup 𝑛→∞ 𝑎𝑛 lim𝑛→∞ sup{𝑎𝑖 ∣ 𝑖 ≥ 𝑛, 𝑖 ∈ ℕ}. (𝑛𝑘 ) Combinations: Size 𝑘 subsets of a size 𝑛 set. 𝑛 ∑ 𝑖=1 𝑖 = 𝑛(𝑛 + 1)2 , 𝑛 ∑ 𝑖=1 𝑖2 = 𝑛(𝑛 + 1)(2𝑛 + 1)6 , 𝑛 ∑ 𝑖=1 𝑖3 = 𝑛 2(𝑛 + 1)2 4 . In general: 𝑛 ∑ 𝑖=1 𝑖𝑚 = 1𝑚 + 1 [(𝑛 + 1) 𝑚+1 − 1 − 𝑛 ∑ 𝑖=1 ((𝑖 + 1)𝑚+1 − 𝑖𝑚+1 − (𝑚 + 1)𝑖𝑚)] 𝑛−1 ∑ 𝑖=1 𝑖𝑚 = 1𝑚 + 1 𝑚 ∑ 𝑘=0 (𝑚 + 1𝑘 ) 𝐵𝑘𝑛 𝑚+1−𝑘. Geometric series: 𝑛 ∑ 𝑖=0 𝑐𝑖 = 𝑐 𝑛+1 − 1 𝑐 − 1 , 𝑐 ≠ 1, ∞ ∑ 𝑖=0 𝑐𝑖 = 11 − 𝑐 , ∞ ∑ 𝑖=1 𝑐𝑖 = 𝑐1 − 𝑐 , |𝑐| < 1, 𝑛 ∑ 𝑖=0 𝑖𝑐𝑖 = 𝑛𝑐 𝑛+2 − (𝑛 + 1)𝑐𝑛+1 + 𝑐 (𝑐 − 1)2 , 𝑐 ≠ 1, ∞ ∑ 𝑖=0 𝑖𝑐𝑖 = 𝑐(1 − 𝑐)2 , |𝑐| < 1. Harmonic series: 𝐻𝑛 = 𝑛 ∑ 𝑖=1 1 𝑖 , 𝑛 ∑ 𝑖=1 𝑖𝐻𝑖 = 𝑛(𝑛 + 1) 2 𝐻𝑛 − 𝑛(𝑛 − 1) 4 . 𝑛 ∑ 𝑖=1 𝐻𝑖 = (𝑛 + 1)𝐻𝑛 − 𝑛, 𝑛 ∑ 𝑖=1 ( 𝑖𝑚) 𝐻𝑖 = ( 𝑛 + 1 𝑚 + 1) (𝐻𝑛+1 − 1 𝑚 + 1) . [𝑛𝑘 ] Stirling numbers (st kind): Ar- rangements of an 𝑛 element set into 𝑘 cycles. [𝑛𝑘 ] Stirling numbers (nd kind): Par- titions of an 𝑛 element set into 𝑘 non-empty sets. ⟨𝑛𝑘 ⟩ st order Eulerian numbers: Permutations 𝜋1𝜋2 … 𝜋𝑛 on {1, 2, … , 𝑛} with 𝑘 ascents. ⟪𝑛𝑘 ⟫  nd order Eulerian numbers. 𝐶𝑛 Catalan Numbers: Binary treeswith 𝑛 + 1 vertices. 1. (𝑛𝑘 ) = 𝑛! (𝑛 − 𝑘)!𝑘! , 2. 𝑛 ∑ 𝑘=0 (𝑛𝑘 ) = 2 𝑛, 3. (𝑛𝑘 ) = ( 𝑛 𝑛 − 𝑘), 4. (𝑛𝑘 ) = 𝑛 𝑘 ( 𝑛 − 1 𝑘 − 1 ), 5. ( 𝑛 𝑘 ) = ( 𝑛 − 1 𝑘 ) + ( 𝑛 − 1 𝑘 − 1 ), 6. ( 𝑛𝑚) ( 𝑚 𝑘 ) = ( 𝑛 𝑘 ) ( 𝑛 − 𝑘 𝑚 − 𝑘), 7. 𝑛 ∑ 𝑘=0 (𝑟 + 𝑘𝑘 ) = ( 𝑟 + 𝑛 + 1 𝑛 ), 8. 𝑛 ∑ 𝑘=0 ( 𝑘𝑚) = ( 𝑛 + 1 𝑚 + 1), 9. 𝑛 ∑ 𝑘=0 ( 𝑟𝑘) ( 𝑠 𝑛 − 𝑘) = ( 𝑟 + 𝑠 𝑛 ), 10. (𝑛𝑘 ) = (−1) 𝑘 (𝑘 − 𝑛 − 1𝑘 ), 11. [ 𝑛 1 ] = [ 𝑛 𝑛] = 1, 12. [ 𝑛2] = 2 𝑛−1 − 1, 13. [𝑛𝑘 ] = 𝑘 [ 𝑛 − 1 𝑘 ] + [ 𝑛 − 1 𝑘 − 1 ], 14. [𝑛1 ] = (𝑛 − 1)!, 15. [ 𝑛 2] = (𝑛 − 1)!𝐻𝑛−1, 16. [ 𝑛 𝑛] = 1, 17. [ 𝑛 𝑘 ] ≥ [ 𝑛 𝑘 ], 18. [ 𝑛 𝑘 ] = (𝑛 − 1) [ 𝑛 − 1 𝑘 ] + [ 𝑛 − 1 𝑘 − 1 ], 19. [ 𝑛𝑛 − 1] = [ 𝑛 𝑛 − 1] = ( 𝑛 2), 20. 𝑛 ∑ 𝑘=0 [𝑛𝑘 ] = 𝑛!, 21. 𝐶𝑛 = 1 𝑛 + 1 ( 2𝑛 𝑛 ), 22. ⟨ 𝑛 0⟩ = ⟨ 𝑛 𝑛 − 1⟩ = 1, 23. ⟨𝑛𝑘 ⟩ = ⟨ 𝑛 𝑛 − 1 − 𝑘⟩, 24. ⟨ 𝑛 𝑘 ⟩ = (𝑘 + 1) ⟨ 𝑛 − 1 𝑘 ⟩ + (𝑛 − 𝑘) ⟨ 𝑛 − 1 𝑘 − 1 ⟩, 25. ⟨ 0 𝑘 ⟩ = { 1 if k=, 0 otherwise , 26. ⟨𝑛1 ⟩ = 2 𝑛 − 𝑛 − 1, 27. ⟨ 𝑛2⟩ = 3 𝑛 − (𝑛 + 1)2𝑛 + (𝑛 + 12 ), 28. 𝑥 𝑛 = 𝑛 ∑ 𝑘=0 ⟨𝑛𝑘 ⟩ ( 𝑥 + 𝑘 𝑛 ), 29. ⟨ 𝑛𝑚⟩ = 𝑚 ∑ 𝑘=0 (𝑛 + 1𝑘 ) (𝑚 + 1 − 𝑘) 𝑛(−1)𝑘 , 30. 𝑚! [ 𝑛𝑚] = 𝑛 ∑ 𝑘=0 ⟨𝑛𝑘 ⟩ ( 𝑘 𝑛 − 𝑚), 31. ⟨ 𝑛 𝑚⟩ = 𝑛 ∑ 𝑘=0 [𝑛𝑘 ] ( 𝑛 − 𝑘 𝑚 ) (−1) 𝑛−𝑘−𝑚𝑘!, 32. ⟪ 𝑛0⟫ = 1, 33. ⟪ 𝑛 𝑛⟫ = 0 for 𝑛 ≠ 0, 34. ⟪ 𝑛 𝑘 ⟫ = (𝑘 + 1) ⟪ 𝑛 − 1 𝑘 ⟫ + (2𝑛 − 1 − 𝑘) ⟪ 𝑛 − 1 𝑘 − 1 ⟫, 35. 𝑛 ∑ 𝑘=0 ⟪𝑛𝑘 ⟫ = (2𝑛)𝑛 2𝑛 , 36. [ 𝑥𝑥 − 𝑛] = 𝑛 ∑ 𝑘=0 ⟪𝑛𝑘 ⟫ ( 𝑥 + 𝑛 − 1 − 𝑘 2𝑛 ), 37. [ 𝑛 + 1 𝑚 + 1] = ∑𝑘 (𝑛𝑘 ) [ 𝑘 𝑚] = 𝑛 ∑ 𝑘=0 [ 𝑘𝑚] (𝑚 + 1) 𝑛−𝑘 ,  Mathematics Cheat Sheet Identities Cont. Trees 38. [ 𝑛 + 1𝑚 + 1] = ∑𝑘 [𝑛𝑘 ] ( 𝑘 𝑚) = 𝑛 ∑ 𝑘=0 [ 𝑘𝑚] 𝑛 𝑛−𝑘 = 𝑛! 𝑛 ∑ 𝑘=0 1 𝑘! [ 𝑘 𝑚], 39. [ 𝑥 𝑥 − 𝑛] = 𝑛 ∑ 𝑘=0 ⟪𝑛𝑘 ⟫ ( 𝑥 + 𝑘 2𝑛 ), 40. [ 𝑛𝑚] = ∑𝑘 (𝑛𝑘 ) [ 𝑘 + 1 𝑚 + 1] (−1) 𝑛−𝑘 , 41. [ 𝑛𝑚] = ∑𝑘 [𝑛 + 1𝑘 + 1 ] ( 𝑘 𝑚) (−1) 𝑚−𝑘 , 42. [𝑚 + 𝑛 + 1𝑚 ] = 𝑚 ∑ 𝑘=0 𝑘 [𝑛 + 𝑘𝑘 ], 43. [ 𝑚 + 𝑛 + 1 𝑚 ] = 𝑚 ∑ 𝑘=0 𝑘(𝑛 + 𝑘) [𝑛 + 𝑘𝑘 ], 44. ( 𝑛𝑚) = ∑𝑘 [𝑛 + 1𝑘 + 1 ] [ 𝑘 𝑚] (−1) 𝑚−𝑘 , 45. (𝑛 − 𝑚)! ( 𝑛𝑚) = ∑𝑘 [𝑛 + 1𝑘 + 1 ] [ 𝑘 𝑚] (−1) 𝑚−𝑘 , for 𝑛 ≥ 𝑚, 46. [ 𝑛𝑛 − 𝑚] = ∑𝑘 (𝑚 − 𝑛𝑚 + 𝑘 ) ( 𝑚 + 𝑛 𝑛 + 𝑘 ) [ 𝑚 + 𝑘 𝑘 ], 47. [ 𝑛 𝑛 − 𝑚] = ∑𝑘 (𝑚 − 𝑛𝑚 + 𝑘 ) ( 𝑚 + 𝑛 𝑛 + 𝑘 ) [ 𝑚 + 𝑘 𝑘 ], 48. [ 𝑛ℓ + 𝑚] ( ℓ + 𝑚 ℓ ) = ∑𝑘 [𝑘ℓ ] [ 𝑛 − 𝑘 𝑚 ] ( 𝑛 𝑘 ), 49. [ 𝑛 ℓ + 𝑚] ( ℓ + 𝑚 ℓ ) = ∑𝑘 [𝑘ℓ ] [ 𝑛 − 𝑘 𝑚 ] ( 𝑛 𝑘 ) . Every tree with 𝑛 vertices has 𝑛 − 1 edges. Kraft inequality: If the depths of the leaves of a binary tree are 𝑑1, … , 𝑑𝑛: 𝑛 ∑ 𝑖=1 2−𝑑𝑖 ≤ 1, and equality holds only if every internal node has  sons. Recurrences Master method: 𝑇(𝑛) = 𝑎𝑇(𝑛/𝑏)+𝑓(𝑛), 𝑎 ≥ 1, 𝑏 > 1 If ∃𝜖 > 0 such that 𝑓(𝑛) = 𝑂(𝑛log𝑏 𝑎−𝜖) then 𝑇(𝑛) = Θ(𝑛log𝑏 𝑎). If 𝑓(𝑛) = Θ(𝑛log𝑏 𝑎) then 𝑇(𝑛) = Θ(𝑛log𝑏 𝑎 log2 𝑛). If ∃𝜖 > 0 such that 𝑓(𝑛) = Ω(𝑛log𝑏 𝑎+𝜖), and ∃𝑐 < 1 such that 𝑎𝑓(𝑛/𝑏) ≤ 𝑐𝑓(𝑛) for large 𝑛, then 𝑇(𝑛) = Θ(𝑓(𝑛)). Substitution (example): Consider the fol- lowing recurrence 𝑇𝑖+1 = 22 𝑖 ⋅ 𝑇2𝑖 , 𝑇1 = 2. Note that 𝑇𝑖 is always a power of two. Let 𝑡𝑖 = log2 𝑇𝑖. Then we have 𝑡𝑖+1 = 2𝑖 + 2𝑡𝑖, 𝑡1 = 1. Let 𝑢𝑖 = 𝑡𝑖/2𝑖. Dividing both sides of the previous equation by 2𝑖+1 we get 𝑡𝑖+1 2𝑖+1 = 2𝑖 2𝑖+1 + 𝑡𝑖 2𝑖 . Substituting we find 𝑢𝑖+1 = 1 2 + 𝑢𝑖, 𝑢1 = 1 2 , which is simply 𝑢𝑖 = 𝑖/2. So we find that 𝑇𝑖 has the closed form 𝑇𝑖 = 2𝑖2 𝑖−1 . Summing factors (example): Consider the following recurrence 𝑇(𝑛) = 3𝑇(𝑛/2) + 𝑛, 𝑇(1) = 1. Rewrite so that all terms involving 𝑇 are on the left side 𝑇(𝑛) − 3𝑇(𝑛/2) = 𝑛. Now expand the recurrence, and choose a factor which makes the left side “telescope” 1(𝑇(𝑛) − 3𝑇(𝑛/2) = 𝑛) 3(𝑇(𝑛/2) − 3𝑇(𝑛/4) = 𝑛/2) ⋮ ⋮ ⋮ 3log2 𝑛−1(𝑇(2) − 3𝑇(1) = 2) Let 𝑚 = log2 𝑛. Summing the left side we get 𝑇(𝑛)−3𝑚𝑇(1) = 𝑇(𝑛)−3𝑚 = 𝑇(𝑛)−𝑛𝑘 where 𝑘 = log2 3 ≈ 1.58496. Summing the right side we get 𝑚−1 ∑ 𝑖=0 𝑛 2𝑖 3 𝑖 = 𝑛 𝑚−1 ∑ 𝑖=0 (32) 𝑖 . Let 𝑐 = 32 . Then we have 𝑛 𝑚−1 ∑ 𝑖=0 𝑐𝑖 = 𝑛 (𝑐 𝑚 − 1 𝑐 − 1 ) = 2𝑛(𝑐 log2 𝑛 − 1) = 2𝑛(𝑐(𝑘−1) log𝑐 𝑛 − 1) = 2𝑛𝑘 − 2𝑛, and so 𝑇(𝑛) = 3𝑛𝑘 − 2𝑛. Full history recurrences can often be changed to limited history ones (ex- ample): Consider 𝑇𝑖 = 1 + 𝑖−1 ∑ 𝑗=0 𝑇𝑗 , 𝑇0 = 1. Note that 𝑇𝑖+1 = 1 + 𝑖 ∑ 𝑗=0 𝑇𝑗. Subtracting we find 𝑇𝑖+1 − 𝑇𝑖 = 1 + 𝑖 ∑ 𝑗=0 𝑇𝑗 − 1 − 𝑖−1 ∑ 𝑗=0 𝑇𝑗 = 𝑇𝑖. And so 𝑇𝑖+1 = 2𝑇𝑖 = 2𝑖+1. Generating functions: . Multiply both sides of the equation by 𝑥𝑖. . Sum both sides over all 𝑖 for which the equation is valid. . Choose a generating function 𝐺(𝑥). Usu- ally 𝐺(𝑥) = ∑∞𝑖=0 𝑥𝑖𝑔𝑖. . Rewrite the equation in terms of the gen- erating function 𝐺(𝑥). . Solve for 𝐺(𝑥). . The coefficient of 𝑥𝑖 in 𝐺(𝑥) is 𝑔𝑖. Example: 𝑔𝑖+1 = 2𝑔𝑖 + 1, 𝑔0 = 0. Multiply and sum: ∑ 𝑖≥0 𝑔𝑖+1𝑥𝑖 = ∑ 𝑖≥0 2𝑔𝑖𝑥𝑖 + ∑ 𝑖≥0 𝑥𝑖. We choose 𝐺(𝑥) = ∑𝑖≥0 𝑥𝑖𝑔𝑖. Rewrite in terms of 𝐺(𝑥): 𝐺(𝑥) − 𝑔0 𝑥 = 2𝐺(𝑥) + ∑𝑖≥0 𝑥𝑖. Simplify: 𝐺(𝑥) 𝑥 = 2𝐺(𝑥) + 1 1 − 𝑥 . Solve for 𝐺(𝑥): 𝐺(𝑥) = 𝑥(1 − 𝑥)(1 − 2𝑥) . Expand this using partial fractions: 𝐺(𝑥) = 𝑥 ( 21 − 2𝑥 − 1 1 − 𝑥) = 𝑥 (2 ∑ 𝑖≥0 2𝑖𝑥𝑖 − ∑ 𝑖≥0 𝑥𝑖) = ∑ 𝑖≥0 (2𝑖+1 − 1)𝑥𝑖+1. So 𝑔𝑖 = 2𝑖 − 1.  Mathematics Cheat Sheet Number Theory Graph Theory The Chinese remainder theorem: There exists a num- ber 𝐶 such that: 𝐶 ≡ 𝑟1 mod 𝑚1 ⋮ ⋮ ⋮ 𝐶 ≡ 𝑟𝑛 mod 𝑚𝑛 if 𝑚𝑖 and 𝑚𝑗 are relatively prime for 𝑖 ≠ 𝑗. Euler’s function: 𝜙(𝑥) is the number of positive inte- gers less than 𝑥 relatively prime to 𝑥. If ∏𝑛𝑖=1 𝑝 𝑒𝑖𝑖 is the prime factorization of 𝑥 then 𝜙(𝑥) = 𝑛 ∏ 𝑖=1 𝑝𝑒𝑖−1𝑖 (𝑝𝑖 − 1). Euler’s theorem: If 𝑎 and 𝑏 are relatively prime then 1 ≡ 𝑎𝜙(𝑏) mod 𝑏. Fermat’s theorem: 1 ≡ 𝑎𝑝−1 mod 𝑝. The Euclidean algorithm: if 𝑎 > 𝑏 are integers then gcd(𝑎, 𝑏) = gcd(𝑎 mod 𝑏, 𝑏). If ∏𝑛𝑖=1 𝑝 𝑒𝑖𝑖 is the prime factorization of 𝑥 then 𝑆(𝑥) = ∑ 𝑑|𝑥 𝑑 = 𝑛 ∏ 𝑖=1 𝑝𝑒𝑖+1𝑖 − 1 𝑝𝑖 − 1 . Perfect Numbers: 𝑥 is an even perfect number iff 𝑥 = 2𝑛−1(2𝑛 − 1) and 2𝑛 − 1 is prime. Wilson’s theorem: 𝑛 is a prime iff (𝑛 − 1)! ≡ −1 mod 𝑛. Möbius inversion: 𝜇(𝑖) = ⎧{ ⎨ {⎩ 1 if 𝑖 = 1. 0 if 𝑖 is not square-free. (−1)𝑟 if 𝑖 is the product of 𝑖𝑟 distinct primes. If 𝐺(𝑎) = ∑ 𝑑|𝑎 𝐹(𝑑), then 𝐹(𝑎) = ∑ 𝑑|𝑎 𝜇(𝑑)𝐺( 𝑎𝑑 ). Prime numbers: 𝑝𝑛 = 𝑛 ln 𝑛 + 𝑛 ln ln 𝑛 − 𝑛 + 𝑛 ln ln 𝑛 ln 𝑛 + 𝑂( 𝑛ln 𝑛), 𝜋(𝑛) = 𝑛ln 𝑛 + 𝑛 (ln 𝑛)2 + 2!𝑛 (ln 𝑛)3 + 𝑂( 𝑛(ln 𝑛)4 ). Definitions: Loop An edge connecting a vertex to itself. Directed Each edge has a direction. Simple Graph with no loops or multi-edges. Walk A sequence 𝑣0𝑒1𝑣1 … 𝑒ℓ𝑣ℓ. Trail A walk with distinct edges. Path A trail with distinct vertices. Connected A graph where there exists a path be- tween any two vertices. Component A maximal connected subgraph. Tree A connected acyclic graph. Free tree A tree with no root. DAG Directed acyclic graph. Eulerian Graph with a trail visiting each edge ex- actly once. Hamiltonian Graph with a cycle visiting each vertex exactly once. Cut A set of edges whose removal increases the number of components. Cut-set A minimal cut. Cut edge A size  cut. k-Connected A graph connected with the removal of any 𝑘 − 1 vertices. k-Tough ∀𝑆 ⊆ 𝑉, 𝑆 ≠ ∅ we have 𝑘 ⋅ 𝑐(𝐺 − 𝑆) ≤ |𝑆|. k-Regular A graph where all vertices have degree 𝑘. k-Factor A 𝑘-regular spanning subgraph. Matching A set of edges, no two of which are adjacent. Clique A set of vertices, all of which are adjacent. Ind. set A set of vertices, none of which are adja- cent. Vertex cover A set of vertices which cover all edges. Planar graph A graph which can be embeded in the plane. Plane graph An embedding of a planar graph. ∑ 𝑣∈𝑉 deg(𝑣) = 2𝑚. If 𝐺 is planar then 𝑛 − 𝑚 + 𝑓 = 2, so 𝑓 ≤ 2𝑛 − 4, 𝑚 ≤ 3𝑛 − 6. Any planar graph has a vertex with degree ≤ 5. Notation: 𝐸(𝐺) Edge set 𝑉(𝐺) Vertex set 𝑐(𝐺) Number of components 𝐺[𝑆] Induced subgraph deg(𝑣) Degree of 𝑣 Δ(𝐺) Maximum degree 𝛿(𝐺) Minimum degree 𝜒(𝐺) Chromatic number 𝜒𝐸(𝐺) Edge chromatic number 𝐺𝑐 Complement graph 𝐾𝑛 Complete graph 𝐾𝑛1 ,𝑛2 Complete bipartite graphr(𝑘, ℓ) Ramsey number Geometry Projective coordinates: triples (𝑥, 𝑦, 𝑧), not all 𝑥, 𝑦 and 𝑧 zero. (𝑥, 𝑦, 𝑧) = (𝑐𝑥, 𝑐𝑦, 𝑐𝑧) ∀𝑐 ≠ 0. Cartesian Projective (𝑥, 𝑦) (𝑥, 𝑦, 1) 𝑦 = 𝑚𝑥 + 𝑏 (𝑚, −1, 𝑏) 𝑥 = 𝑐 (1, 0, −𝑐) Distance formula, 𝐿𝑝 and 𝐿∞ metric: √(𝑥1 − 𝑥0)2 + (𝑦1 − 𝑦0)2, [|𝑥1 − 𝑥0|𝑝 + |𝑦1 − 𝑦0|𝑝] 1/𝑝 , lim𝑝→∞ [|𝑥1 −𝑥0| 𝑝 +|𝑦1 −𝑦0|𝑝] 1/𝑝. Area of triangle (𝑥0, 𝑦0), (𝑥1, 𝑦1) and (𝑥2, 𝑦2): 1 2 abs ∣ 𝑥1 − 𝑥0 𝑦1 − 𝑦0 𝑥2 − 𝑥0 𝑦2 − 𝑦0∣ . Angle formed by three points: (0, 0) (𝑥1, 𝑦1) (𝑥2, 𝑦2) ℓ1 ℓ2 𝜃 ∘∘ ∘ cos 𝜃 = (𝑥1, 𝑦1) ⋅ (𝑥2, 𝑦2)ℓ1ℓ2 . Line through two points (𝑥0, 𝑦0) and (𝑥1, 𝑦1): ∣ 𝑥 𝑦 1 𝑥0 𝑦0 1 𝑥1 𝑦1 1 ∣ = 0. Area of circle, volume of sphere: 𝐴 = 𝜋𝑟2, 𝑉 = 43𝜋𝑟 3. If I have seen farther than others, it is because I have stood on the shoulders of giants. – Issac Newton  Mathematics Cheat Sheet 𝜋 Calculus Wallis’ identity: 𝜋 = 2 ⋅ 2 ⋅ 2 ⋅ 4 ⋅ 4 ⋅ 6 ⋅ 6 ⋯1 ⋅ 3 ⋅ 3 ⋅ 5 ⋅ 5 ⋅ 7 ⋯ Brouncker’s continued fraction expansion: 𝜋 4 = 1 + 12 2 + 3 2 2+ 5 2 2+ 7 2 2+⋯ Gregrory’s series: 𝜋 4 = 1 − 1 3 + 1 5 − 1 7 + 1 9 − ⋯ Newton’s series: 𝜋 6 = 1 2 + 1 2 ⋅ 3 ⋅ 23 + 1 ⋅ 3 2 ⋅ 4 ⋅ 5 ⋅ 25 + ⋯ Sharp’s series: 𝜋 6 = 1 √3(1 − 1 31 ⋅ 3 + 1 32 ⋅ 5 − 1 33 ⋅ 7 + ⋯ ) Euler’s series: 𝜋2 6 = 1 12 + 1 22 + 1 32 + 1 42 + 1 52 + ⋯ 𝜋2 8 = 1 12 + 1 32 + 1 52 + 1 72 + 1 92 + ⋯ 𝜋2 12 = 1 12 − 1 22 + 1 32 − 1 42 + 1 52 − ⋯ Partial Fractions Let 𝑁(𝑥) and 𝐷(𝑥) be polynomial functions of 𝑥. We can break down 𝑁(𝑥)/𝐷(𝑥) using partial fraction expansion. First, if the degree of 𝑁 is greater than or equal to the degree of 𝐷, divide 𝑁 by 𝐷, obtaining 𝑁(𝑥) 𝐷(𝑥) = 𝑄(𝑥) + 𝑁 ′(𝑥) 𝐷(𝑥) , where the degree of𝑁 ′ is less than that of 𝐷. Second, factor 𝐷(𝑥). Use the following rules: For a non- repeated factor: 𝑁(𝑥) (𝑥 − 𝑎)𝐷(𝑥) = 𝐴 𝑥 − 𝑎 + 𝑁 ′(𝑥) 𝐷(𝑥) , where 𝐴 = [𝑁(𝑥)𝐷(𝑥) ]𝑥=𝑎 . For a repeated factor: 𝑁(𝑥) (𝑥 − 𝑎)𝑚𝐷(𝑥) = 𝑚−1 ∑ 𝑘=0 𝐴𝑘 (𝑥 − 𝑎)𝑚−𝑘 + 𝑁 ′(𝑥) 𝐷(𝑥) , where 𝐴𝑘 = 1 𝑘! [ 𝑑𝑘 𝑑𝑥𝑘 ( 𝑁(𝑥) 𝐷(𝑥) )]𝑥=𝑎 . The reasonable man adapts himself to the world; the unreasonable persists in trying to adapt the world to himself. Therefore all progress depends on the un- reasonable. – George Bernard Shaw Derivatives: 1. 𝑑(𝑐𝑢) 𝑑𝑥 = 𝑐 𝑑𝑢 𝑑𝑥 , 2. 𝑑(𝑢 + 𝑣) 𝑑𝑥 = 𝑑𝑢 𝑑𝑥 + 𝑑𝑣 𝑑𝑥 , 3. 𝑑(𝑢𝑣) 𝑑𝑥 = 𝑢 𝑑𝑣 𝑑𝑥 + 𝑣 𝑑𝑢 𝑑𝑥 , 4. 𝑑(𝑢𝑛) 𝑑𝑥 = 𝑛𝑢 𝑛−1 𝑑𝑢 𝑑𝑥 , 5. 𝑑(𝑢/𝑣) 𝑑𝑥 = 𝑣( 𝑑𝑢𝑑𝑥 ) − 𝑢( 𝑑𝑣 𝑑𝑥 ) 𝑣2 , 6. 𝑑(𝑒𝑐𝑢) 𝑑𝑥 = 𝑐𝑒 𝑐𝑢 𝑑𝑢 𝑑𝑥 , 7. 𝑑(𝑐𝑢) 𝑑𝑥 = (ln 𝑐)𝑐 𝑢 𝑑𝑢 𝑑𝑥 , 8. 𝑑(ln 𝑢) 𝑑𝑥 = 1 𝑢 𝑑𝑢 𝑑𝑥 , 9. 𝑑(sin 𝑢) 𝑑𝑥 = cos 𝑢 𝑑𝑢 𝑑𝑥 , 10. 𝑑(cos 𝑢) 𝑑𝑥 = − sin 𝑢 𝑑𝑢 𝑑𝑥 , 11. 𝑑(tan 𝑢) 𝑑𝑥 = sec 2 𝑢𝑑𝑢𝑑𝑥 , 12. 𝑑(cot 𝑢) 𝑑𝑥 = csc 2 𝑢𝑑𝑢𝑑𝑥 , 13. 𝑑(sec 𝑢) 𝑑𝑥 = tan 𝑢 sec 𝑢 𝑑𝑢 𝑑𝑥 , 14. 𝑑(csc 𝑢) 𝑑𝑥 = − cot 𝑢 csc 𝑢 𝑑𝑢 𝑑𝑥 , 15. 𝑑(arcsin 𝑢) 𝑑𝑥 = 1 √1 − 𝑢2 𝑑𝑢 𝑑𝑥 , 16. 𝑑(arccos 𝑢) 𝑑𝑥 = −1 √1 − 𝑢2 𝑑𝑢 𝑑𝑥 , 17. 𝑑(arctan 𝑢) 𝑑𝑥 = 1 1 + 𝑢2 𝑑𝑢 𝑑𝑥 , 18. 𝑑(arccot 𝑢) 𝑑𝑥 = −1 1 + 𝑢2 𝑑𝑢 𝑑𝑥 , 19. 𝑑(arcsec 𝑢) 𝑑𝑥 = 1 𝑢√1 − 𝑢2 𝑑𝑢 𝑑𝑥 , 20. 𝑑(arccsc 𝑢) 𝑑𝑥 = −1 𝑢√1 − 𝑢2 𝑑𝑢 𝑑𝑥 , 21. 𝑑(sinh 𝑢) 𝑑𝑥 = cosh 𝑢 𝑑𝑢 𝑑𝑥 , 22. 𝑑(cosh 𝑢) 𝑑𝑥 = sinh 𝑢 𝑑𝑢 𝑑𝑥 , 23. 𝑑(tanh 𝑢) 𝑑𝑥 = sech 2 𝑢𝑑𝑢𝑑𝑥 , 24. 𝑑(coth 𝑢) 𝑑𝑥 = − csch 2 𝑢𝑑𝑢𝑑𝑥 , 25. 𝑑(sech 𝑢) 𝑑𝑥 = − sech 𝑢 tanh 𝑢 𝑑𝑢 𝑑𝑥 , 26. 𝑑(csch 𝑢) 𝑑𝑥 = − csch 𝑢 coth 𝑢 𝑑𝑢 𝑑𝑥 , 27. 𝑑(arcsinh 𝑢) 𝑑𝑥 = 1 √1 + 𝑢2 𝑑𝑢 𝑑𝑥 , 28. 𝑑(arccosh 𝑢) 𝑑𝑥 = 1 √𝑢2 − 1 𝑑𝑢 𝑑𝑥 , 29. 𝑑(arctanh 𝑢) 𝑑𝑥 = 1 1 − 𝑢2 𝑑𝑢 𝑑𝑥 , 30. 𝑑(arccoth 𝑢) 𝑑𝑥 = 1 𝑢2 − 1 𝑑𝑢 𝑑𝑥 , 31. 𝑑(arcsech 𝑢) 𝑑𝑥 = −1 𝑢√1 − 𝑢2 𝑑𝑢 𝑑𝑥 , 32. 𝑑(arccsch 𝑢) 𝑑𝑥 = −1 |𝑢|√1 + 𝑢2 𝑑𝑢 𝑑𝑥 . Integrals: 1. ∫ 𝑐𝑢 𝑑𝑥 = 𝑐 ∫ 𝑢 𝑑𝑥, 2. ∫(𝑢 + 𝑣) 𝑑𝑥 = ∫ 𝑢 𝑑𝑥 + ∫ 𝑣 𝑑𝑥, 3. ∫ 𝑥𝑛 𝑑𝑥 = 1𝑛 + 1𝑥 𝑛+1, 𝑛 ≠ −1, 4. ∫ 1𝑥𝑑𝑥 = ln 𝑥, 5. ∫ 𝑒 𝑥 𝑑𝑥 = 𝑒𝑥, 6. ∫ 𝑑𝑥1 + 𝑥2 = arctan 𝑥, 7. ∫ 𝑢 𝑑𝑣 𝑑𝑥𝑑𝑥 = 𝑢𝑣 − ∫ 𝑣 𝑑𝑢 𝑑𝑥 𝑑𝑥, 8. ∫ sin 𝑥 𝑑𝑥 = − cos 𝑥, 9. ∫ cos 𝑥 𝑑𝑥 = sin 𝑥, 10. ∫ tan 𝑥 𝑑𝑥 = − ln | cos 𝑥|, 11. ∫ cot 𝑥 𝑑𝑥 = ln | cos 𝑥|, 12. ∫ sec 𝑥 𝑑𝑥 = ln | sec 𝑥 + tan 𝑥|, 13. ∫ csc 𝑥 𝑑𝑥 = ln | csc 𝑥 + cot 𝑥|, 14. ∫ arcsin 𝑥𝑎𝑑𝑥 = arcsin 𝑥 𝑎 + √𝑎 2 − 𝑥2, 𝑎 > 0,  Mathematics Cheat Sheet Calculus Cont. 15. ∫ arccos 𝑥𝑎𝑑𝑥 = arccos 𝑥 𝑎 − √𝑎 2 − 𝑥2, 𝑎 > 0, 16. ∫ arctan 𝑥𝑎𝑑𝑥 = 𝑥 arctan 𝑥 𝑎 − 𝑎 2 ln(𝑎 2 + 𝑥2), 𝑎 > 0, 17. ∫ sin2(𝑎𝑥)𝑑𝑥 = 12𝑎(𝑎𝑥 − sin(𝑎𝑥) cos(𝑎𝑥)), 18. ∫ cos 2(𝑎𝑥)𝑑𝑥 = 12𝑎(𝑎𝑥 + sin(𝑎𝑥) cos(𝑎𝑥)), 19. ∫ sec 2 𝑥 𝑑𝑥 = tan 𝑥, 20. ∫ csc2 𝑥 𝑑𝑥 = − cot 𝑥, 21. ∫ sin𝑛 𝑥 𝑑𝑥 = −sin 𝑛−1 𝑥 cos 𝑥 𝑛 + 𝑛 − 1 𝑛 ∫ sin 𝑛−2 𝑥 𝑑𝑥, 22. ∫ cos𝑛 𝑥 𝑑𝑥 = cos 𝑛−1 𝑥 sin 𝑥 𝑛 + 𝑛 − 1 𝑛 ∫ cos 𝑛−2 𝑥 𝑑𝑥, 23. ∫ tan𝑛 𝑥 𝑑𝑥 = tan 𝑛−1 𝑥 𝑛 − 1 − ∫ tan 𝑛−2 𝑥 𝑑𝑥, 𝑛 ≠ 1, 24. ∫ cot𝑛 𝑥 𝑑𝑥 = −cot 𝑛−1 𝑥 𝑛 − 1 − ∫ cot 𝑛−2 𝑥 𝑑𝑥, 𝑛 ≠ 1, 25. ∫ sec𝑛 𝑥 𝑑𝑥 = tan 𝑥 sec 𝑛−1 𝑥 𝑛 − 1 + 𝑛 − 2 𝑛 − 1 ∫ sec 𝑛−2 𝑥 𝑑𝑥, 𝑛 ≠ 1, 26. ∫ csc𝑛 𝑥 𝑑𝑥 = −cot 𝑥 csc 𝑛−1 𝑥 𝑛 − 1 + 𝑛 − 2 𝑛 − 1 ∫ csc 𝑛−2 𝑥 𝑑𝑥, 𝑛 ≠ 1, 27. ∫ sinh 𝑥 𝑑𝑥 = cosh 𝑥, 28. ∫ cosh 𝑥 𝑑𝑥 = sinh 𝑥, 29. ∫ tanh 𝑥 𝑑𝑥 = ln | cosh 𝑥|, 30. ∫ coth 𝑥 𝑑𝑥 = ln | sinh 𝑥|, 31. ∫ sech 𝑥 𝑑𝑥 = arctan sinh 𝑥, 32. ∫ csch 𝑥 𝑑𝑥 = ln ∣tanh 𝑥2 ∣, 33. ∫ sinh2 𝑥 𝑑𝑥 = 14 sinh(2𝑥) − 1 2𝑥, 34. ∫ cosh 2 𝑥 𝑑𝑥 = 14 sinh(2𝑥) + 1 2𝑥, 35. ∫ sech 2 𝑥 𝑑𝑥 = tanh 𝑥, 36. ∫ arcsinh 𝑥𝑎𝑑𝑥 = 𝑥 arcsinh 𝑥 𝑎 − √𝑥 2 + 𝑎2, 𝑎 > 0, 37. ∫ arctanh 𝑥𝑎𝑑𝑥 = 𝑥 arctanh 𝑥 𝑎 + 𝑎 2 ln |𝑎 2 − 𝑥2|, 38. ∫ arccosh 𝑥𝑎𝑑𝑥 = ⎧{⎨ {⎩ 𝑥 arccosh 𝑥𝑎 − √𝑥 2 + 𝑎2, if arccosh 𝑥𝑎 > 0 and 𝑎 > 0, 𝑥 arccosh 𝑥𝑎 + √𝑥 2 + 𝑎2, if arccosh 𝑥𝑎 < 0 and 𝑎 > 0, , 39. ∫ 𝑑𝑥 √𝑎2 + 𝑥2 = ln (𝑥 + √𝑎2 + 𝑥2) , 𝑎 > 0, 40. ∫ 𝑑𝑥𝑎2 + 𝑥2 = 1 𝑎 arctan 𝑥 𝑎 , 𝑎 > 0, 41. ∫ √𝑎 2 − 𝑥2 𝑑𝑥 = 𝑥2√𝑎 2 − 𝑥2 + 𝑎 2 2 arcsin 𝑥 𝑎 , 𝑎 > 0, 42. ∫(𝑎2 − 𝑥2)3/2𝑑𝑥 = 𝑥8 (5𝑎 2 − 2𝑥2)√𝑎2 − 𝑥2 + 3𝑎 4 8 arcsin 𝑥 𝑎 , 𝑎 > 0, 43. ∫ 𝑑𝑥 √𝑎2 − 𝑥2 = arcsin 𝑥𝑎 , 𝑎 > 0, 44. ∫ 𝑑𝑥𝑎2 − 𝑥2 = 1 2𝑎 ln ∣ 𝑎 + 𝑥 𝑎 − 𝑥 ∣, 45. ∫ 𝑑𝑥 (𝑎2 − 𝑥2)3/2 = 𝑥 𝑎2√𝑎2 − 𝑥2 , 46. ∫ √𝑎2 ± 𝑥2 𝑑𝑥 = 𝑥2√𝑎 2 ± 𝑥2 ± 𝑎 2 2 ln ∣𝑥 + √𝑎 2 ± 𝑥2∣, 47. ∫ 𝑑𝑥 √𝑥2 − 𝑎2 = ln ∣𝑥 + √𝑥2 − 𝑎2∣ , 𝑎 > 0, 48. ∫ 𝑑𝑥𝑎𝑥2 + 𝑏𝑥 = 1 𝑎 ln ∣ 𝑥 𝑎 + 𝑏𝑥 ∣, 49. ∫ 𝑥√𝑎 + 𝑏𝑥 𝑑𝑥 = 2(3𝑏𝑥 − 2𝑎)(𝑎 + 𝑏𝑥)3/2 15𝑏2 , 50. ∫ √𝑎 + 𝑏𝑥𝑥 𝑑𝑥 = 2√𝑎 + 𝑏𝑥 + 𝑎 ∫ 1 𝑥√𝑎 + 𝑏𝑥𝑑𝑥, 51. ∫ 𝑥 √𝑎 + 𝑏𝑥 𝑑𝑥 = 1 √2 ln ∣ √𝑎 + 𝑏𝑥 − √𝑎 √𝑎 + 𝑏𝑥 + √𝑎 ∣ , 𝑎 > 0, 52. ∫ √𝑎 2 − 𝑥2 𝑥 𝑑𝑥 = √𝑎 2 − 𝑥2 − 𝑎 ln ∣ 𝑎 + √𝑎2 − 𝑥2 𝑥 ∣, 53. ∫ 𝑥√𝑎 2 − 𝑥2 𝑑𝑥 = − 13 (𝑎 2 − 𝑥2)3/2, 54. ∫ 𝑥2√𝑎2 − 𝑥2 𝑑𝑥 = 𝑥8 (2𝑥 2 − 𝑎2)√𝑎2 − 𝑥2 + 𝑎 4 8 arcsin 𝑥 𝑎 , 𝑎 > 0, 55. ∫ 𝑑𝑥 √𝑎2 − 𝑥2 = − 1𝑎 ln ∣ 𝑎 + √𝑎2 − 𝑥2 𝑥 ∣, 56. ∫ 𝑥 𝑑𝑥 √𝑎2 − 𝑥2 = −√𝑎2 − 𝑥2, 57. ∫ 𝑥 2 𝑑𝑥 √𝑎2 − 𝑥2 = − 𝑥2√𝑎 2 − 𝑥2 + 𝑎 2 2 arcsin 𝑥 𝑎 , 𝑎 > 0, 58. ∫ √𝑎 2 + 𝑥2 𝑥 𝑑𝑥 = √𝑎 2 + 𝑥2 − 𝑎 ln ∣ 𝑎 + √𝑎 2 + 𝑥2 𝑥 ∣, 59. ∫ √𝑥2 − 𝑎2 𝑥 𝑑𝑥 = √𝑥 2 − 𝑎2 − 𝑎 arccos 𝑎|𝑥| , 𝑎 > 0, 60. ∫ 𝑥√𝑥2 ± 𝑎2 𝑑𝑥 = 13 (𝑥 2 ± 𝑎2)3/2, 61. ∫ 𝑑𝑥 𝑥√𝑥2 + 𝑎2 = 1𝑎 ln ∣ 𝑥 𝑎 + √𝑎2 + 𝑥2 ∣ , 