1 / 75

Adding Spice to A level Maths Lessons Graham Winter 2007

Adding Spice to A level Maths Lessons Graham Winter 2007. 5% interest on ¼ d since 1066. 1  960 × 1.05 2007 – 1066 = £90 54 3 898 922 419 141.99 Total GDP for world in 2003 = £25 000 000 000 000. Fold a piece of paper in half. Then fold it in half again.

gareth-hayden
Télécharger la présentation

Adding Spice to A level Maths Lessons Graham Winter 2007

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Adding Spice to A level Maths LessonsGraham Winter 2007

  2. 5% interest on ¼ d since 1066 1  960 × 1.05 2007 – 1066 = £90 543 898 922 419 141.99 Total GDP for world in 2003 = £25 000 000 000 000

  3. Fold a piece of paper in half. Then fold it in half again. And again, fifty times in all. It now has a thickness of 78 000 000 miles, which is 4/5 of the distance to the sun – a 7½ year trip on Concorde.

  4. Average Point Scores Mathematics A2 point average: Althon College 2560 points from 10 students: 256 average Basing College 3600 points from 20 students: 180 average Advanced FSM point average: Althon College 2340 points from 60 students: 39 average Basing College 1200 points from 40 students: 30 average Total Maths point average: Althon College 4900 points from 70 students: 70 average Basing College 4800 points from 60 students: 80 average

  5. Obtaining a formula for π

  6. Obtaining a formula for π

  7. Obtaining a formula for π

  8. Obtaining a formula for π

  9. Obtaining a formula for π

  10. Rearranging:

  11. Rearranging: This formula converges veryslowly.A computer performing 10 12 calculations per second, which began calculating this formula at the Big Bang 4.4 billion years ago, would have just established the 29th decimal place.

  12. A graphics calculator can be simply programmed to calculate  using this formula. : Clrhome : 4  A : 3  B : Repeat 0 : A – 4/B + 4/(B + 2)  A : Disp A : B + 4  B : End The calculator would have to run the program for 8½ years to establish the 9th decimal place.

  13.  has been calculated to 206 billion decimal places. The diameter of the universe is 40 billion light years. Hence just 30 decimal places of  are needed to find the circumference of the universe correct to the nearest mm.

  14. Let S = 1 + 2 + 4 + 8 + 16 + 32 + 64 + . . .  S = 1 + 2( 1 + 2 + 4 + 8 + 16 + 32 + . . . )  S = 1 + 2S  S – 2S = 1  –S = 1  S = –1

  15. To prove 1 = 2 Let x = y x2 = xy  x2– y 2 = xy – y 2  (x + y)(x – y) = y(x – y)  x + y = y  y + y = y  2y = y  2 = 1

  16. Solve: 2 cos x sin x = cos x, 0  x < 360 2 cos x sin x = cos x  2 sin x = 1  sin x = ½  x = 30 o or 150 o

  17. A formula for the Fibonacci sequence 1 , 1 , 2 , 3 , 5 , 8 , 13 , 21 , 34 , . . . . . . . u1 = 1 , u2 = 1 u n + 2 = u n + 1 + u n

  18. A formula for the Fibonacci sequence 1 , 1 , 2 , 3 , 5 , 8 , 13 , 21 , 34 , . . . . . . . u1 = 1 , u2 = 1 u n + 2 = u n + 1 + u n

  19. is the Golden ratio. This was widely used in architecture and art.

  20. A formula for any sequence e.g. 2 , 4 , 8 , 30 ,  , . . . . . .

  21. A formula for any sequence e.g. 2 , 4 , 8 , 30 ,  , . . . . . .

  22. A formula for any sequence e.g. 2 , 4 , 8 , 30 ,  , . . . . . .

  23. A formula for any sequence e.g. 2 , 4 , 8 , 30 ,  , . . . . . .

  24. “Student” cancelling

  25. “Student” cancelling works here

  26. Algebraic symbols Before the 17th century, algebraic manipulation was very cumbersome. The following slide is a copy of part of Cardan’s work on solving cubic equations, published in 1545, together with a translation. Note that the translation uses “modern” symbols e.g. +, not present in the original.

  27. Cardan’s solution of a cubic equation, 1545

  28. Cardan was professor of science at Milan university. He divided his time equally between mechanics, astrology and debauchery. One of his sons was executed for poisoning his wife, and he cut off the ears of his other in a fit of rage after some offence had been committed . He was imprisoned for heresy, became the astrologer to the Pope, and felt obliged to commit suicide after predicting the date of his own death. In his Ars Magna he found a general solution for cubic equations, introducing negative and imaginary numbers in the process.

  29. Roman numerals were still used extensively for accounting until 1600. One of the first appearances of decimal notation was in a work by Pitiscus in 1608. The unknown in an equation was called rei (Latin for thing) and its square called zensus, so for example x 2 + 3x – 2 was written Z p3R m 2by Pacioli in 1500. In 1553 Stifel used AA for A 2.

  30. The German mathematician Jordanus first used letters for unknowns c. 1200, but there were no symbols for + or –. His work Algorithmus was not printed until 1534. The + and – symbols were first consistently used by the French mathematician Vieta in 1591. The × symbol was invented by the English mathematician William Oughtred in 1631. The = symbol was invented by the Welsh mathematician Robert Record in 1557.

  31. RSA Coding and Decoding as a Function and its Inverse For RSA coding , two numbers are chosen: • a product of 2 primes e.g. 1189 = 29  41 • a number coprime to1189 e.g. 3 • The coding function is then f (x) = x 3 mod 1189 i.e. take the remainder when x3is divided by 1189

  32. The inverse function is: f – 1 (x) = x 187 mod 1189 The number 187 has been calculated using 29 and 41. It is the number which, when it is multiplied by 3, gives an answer which is exactly one more than a multiple of the lowest common multiple of 28 (= 29 – 1) and 40 (= 41 – 1 ).

  33. A 30 tonne lorry travelling at 30 mph collides with a 1 tonne car travelling at 30 mph. Let v be the speed of the wreckage after the collision. 30 × 30 – 1 × 30 = 30v + 1v  870 = 31v  v = 28.1 mph

  34. The value of g is less on the equator (9.76 ms –2) than it is at the poles (9.86 ms –2 ), due to the greater distance to the centre of the earth (3963 miles v. 3949 miles) and also due to the earth’s rotation. A person is about ½ inch taller when they get up than when they go to bed. So to minimize your body mass index, you should measure your height and weight first thing in the morning on the equator. An anorexic should consider taking the measurements at the Pole just before retiring.

  35. Taking g = 10 may not produce accuracy to 1 significant place. e.g. v = u + at with u = 5.5 and t = 7 With g = 10, we obtain v = 75.5 or v = 80 (1 s.f.) With g = 9.8, we obtain v =74.1 or v = 70 (1 s.f.)

  36. “You will be given a surprise test in one of your lessons next week.” When the students enter Friday’s lesson, if the test has not been given, it will not be a surprise when they get it. So the surprise test can’t be on Friday. So when they enter Thursday’s lesson, if the test has not been given, it will not be a surprise when they get it.

  37. This sentence is false This sentence is true

  38. and its graph.

  39. The graph of y = sin 47x on Autograph,

  40. The graph of y = sin 47x on Autograph, and on the Texas TI-82.

  41. The word sine is from the Latin word sinus for breast. This is due to a mistranslation of the Hindu word for chord-half into Arabic.

  42. Suppose sin A = 3/5 and sin B = 5/13 - then cos A = 4/5 and cos B = 12/13 - and sin (A + B) = 3/5 × 12/13 + 4/5 × 5/13 = 56/65 cos (A + B) = 4/5 × 12/13 – 3/5 × 5/13 = 33/65 33, 56, 65 is a Pythagorean triplet. All Pythagorean triplets are of the form m 2 – n 2 , 2mn , m 2 + n 2 for integers m ,n.

  43. Quintics and higher powered polynomials cannot generally be solved. This was proved for quintics by Niels Abel in 1825. Evariste Galois proved it true for all polynomials with higher powers, though this wasn’t clear until rewritten by Camille Jordan in 1870.

  44. Pierre Wantzel resolved a couple of famous Greek problems in 1837: - an angle cannot be trisected using only compasses and a straight edge; - a cube cannot be doubled using only ruler and compasses. That a circle cannot be squared i.e. it is impossible to construct a square with the same area as a given circle using only compasses and a straight edge, followed the proof that  is transcendental in 1882.

  45. The question arises as to whether such numbers as e +  , e ×  , e e , e  , e etc are transcendental, and in most cases the answer is not known. An exception is e  which was shown to be transcendental by Alexandr Gelfond in 1934. It is also known that at least one of e e and e e² is transcendental.

  46. The number e is the number such that

  47. The number e is the number such that This can be obtained on a calculator thus:

  48. The coefficients in the binomial expansion of (1 + x) 5. The coefficient of x 6 in the expansion of (1 + x) 49 is 49 C 6 , the number of ways of winning the jackpot on the National Lottery.

  49. The number of ways of winning the jackpot on the National Lottery is 13 983 816. 13 983 816 two pence pieces laid end to end would stretch 220 miles – from London to Paris. 13 983 816 seconds is 161 days – from 13th April until 21st September.

More Related